Hướng dẫn how do you subtract two elements from a list in python? - làm cách nào để trừ hai phần tử khỏi danh sách trong python?

Question

Nhiều giải pháp đã được đề xuất.

Nội dung chính Show

Làm thế nào để bạn trừ một giá trị từ một danh sách trong Python?
Tôi có thể trừ một danh sách từ một danh sách trong Python không?
Chúng ta có thể trừ 2 danh sách trong Python không?

Nếu tốc độ được quan tâm, thì đây là đánh giá các giải pháp khác nhau liên quan đến tốc độ (từ nhanh nhất đến chậm nhất)

import timeit
import operator

a = [2,2,2]
b = [1,1,1]  # we want to obtain c = [2,2,2] - [1,1,1] = [1,1,1

%timeit map(operator.sub, a, b)
176 ns ± 7.18 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit map(int.__sub__, a, b)
179 ns ± 4.95 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit map(lambda x,y: x-y, a,b)
189 ns ± 8.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit [a_i - b_i for a_i, b_i in zip(a, b)]
421 ns ± 18.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit [x - b[i] for i, x in enumerate(a)]
452 ns ± 17.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each

%timeit [a[i] - b[i] for i in range(len(a))]
530 ns ± 16.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit list(map(lambda x, y: x - y, a, b))
546 ns ± 16.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit np.subtract(a,b)
2.68 µs ± 80.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit list(np.array(a) - np.array(b))
2.82 µs ± 113 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit np.matrix(a) - np.matrix(b)
12.3 µs ± 437 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Sử dụng map rõ ràng là nhanh nhất. Đáng ngạc nhiên, numpy là chậm nhất. Nó chỉ ra rằng chi phí đầu tiên chuyển đổi danh sách a và

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

0 thành mảng numpy là một nút cổ chai lớn hơn bất kỳ hiệu quả nào đạt được từ vector hóa.

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Trừ hai danh sách bằng hàm zip () trong phương thức này, chúng tôi sẽ chuyển hai danh sách đầu vào cho hàm zip. Sau đó, lặp lại trên đối tượng zip bằng cách sử dụng cho vòng lặp. Trên mỗi lần lặp, chương trình sẽ lấy một yếu tố từ List1 và List2, trừ chúng và nối kết quả vào danh sách khác.
Examples:

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

Chúng ta có thể trừ 2 danh sách trong Python không? The naive approach is to traverse both list simultaneously and if the element in first list in greater than element in second list, then subtract it, else if the element in first list in smaller than element in second list, then return element of first list only.

Python3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

4

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

9

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

1__16

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

4

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

3__

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

5

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

6

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

7

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

8

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

0

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

1

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

2

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

3

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

5

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

6

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

7

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

8

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

2

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

0

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

1

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

5

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

3

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

6

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map1

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9map4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map7

Output:

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

Phương pháp 2: Sử dụng zip () Chúng tôi trừ nếu phần tử trong danh sách đầu tiên lớn hơn phần tử trong danh sách thứ hai, nếu không chúng tôi xuất phần tử của danh sách đầu tiên. & NBSP;Using zip() we subtract if element in first list is greater than element in second list, else we output element of first list.

Python3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

4

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

9

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

1__16

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

4

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

3__

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3__

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

6

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map1

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9map4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map7

Output:

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

Phương pháp 3: Sử dụng danh sách hiểu. & NBSP; Using list comprehension.

Python3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

4

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

9

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

1__16

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

4

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

3__

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3__

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9map4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

6

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map1

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9map4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map7

Output:

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

Phương pháp 3: Sử dụng danh sách hiểu. & NBSP; Using numpy() to complete the above task.

Python3

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

68

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

70

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

3

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

72

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

73

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

0

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

75

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

5

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

6

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

7

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

8__

Phương pháp 4: Sử dụng numpy () để hoàn thành nhiệm vụ trên. & Nbsp;

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

97

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

98

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

6

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

9

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4map1

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

4

Original list are :
[10, 20, 30, 40, 50, 60]
[60, 50, 40, 30, 20, 10]

Output list is
[10, 20, 30, 10, 30, 50]

9map4

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

2

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

3

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

01

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

5

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

7

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

9

%timeit a = np.array([2,2,2]); b=np.array([1,1,1])
1.55 µs ± 54.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

a = np.array([2,2,2])
b = np.array([1,1,1])
%timeit a - b
417 ns ± 12.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

6

Input:
l1 = [10, 20, 30, 40, 50, 60]
l2 = [60, 50, 40, 30, 20, 10]
Output:
[10, 20, 30, 10, 30, 50]

Input:
l1 = [15, 9, 10, 56, 23, 78, 5, 4, 9]
l2 = [9, 4, 5, 36, 47, 26, 10, 45, 87]
Output:
[6, 5, 5, 20, 23, 52, 5, 4, 9]

1__16

Output:

Original list are :
[10 20 30 40 50 60]
[60 50 40 30 20 10]

Output list is
[10 20 30 10 30 50]

Làm thế nào để bạn trừ một giá trị từ một danh sách trong Python?

Sử dụng một vòng lặp để trừ một giá trị từ mỗi số trong danh sách..

Số = [1, 2, 3].

Đối với I trong phạm vi (LEN (số)):.

số [i] = số [i] - 1 ..

Tôi có thể trừ một danh sách từ một danh sách trong Python không?

Trừ hai danh sách bằng hàm zip () trong phương thức này, chúng tôi sẽ chuyển hai danh sách đầu vào cho hàm zip.Sau đó, lặp lại trên đối tượng zip bằng cách sử dụng cho vòng lặp.Trên mỗi lần lặp, chương trình sẽ lấy một yếu tố từ List1 và List2, trừ chúng và nối kết quả vào danh sách khác. In this method, we'll pass the two input lists to the Zip Function. Then, iterate over the zip object using for loop. On every iteration, the program will take an element from list1 and list2, subtract them and append the result into another list.

Chúng ta có thể trừ 2 danh sách trong Python không?

Phương pháp 3: Sử dụng danh sách hiểu và đặt để tìm sự khác biệt giữa hai danh sách trong Python.Trong phương pháp này, chúng tôi chuyển đổi danh sách thành các bộ một cách rõ ràng và sau đó chỉ cần giảm cái này từ mẫu kia bằng toán tử trừ.Use a list comprehension and set to Find the Difference Between Two Lists in Python. In this method, we convert the lists into sets explicitly and then simply reduce one from the other using the subtract operator.

programming python List difference Python Subtract 2 list Np subtract

Hướng dẫn how do you subtract two elements from a list in python? - làm cách nào để trừ hai phần tử khỏi danh sách trong python?

Python3

Python3

Python3

Python3

Làm thế nào để bạn trừ một giá trị từ một danh sách trong Python?

Tôi có thể trừ một danh sách từ một danh sách trong Python không?

Chúng ta có thể trừ 2 danh sách trong Python không?

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