I want to sum a 2 dimensional array in python:
Here is what I have:
def sum1[input]:
sum = 0
for row in range [len[input]-1]:
for col in range[len[input[0]]-1]:
sum = sum + input[row][col]
return sum
print sum1[[[1, 2],[3, 4],[5, 6]]]
It displays 4 instead of 21 [1+2+3+4+5+6 = 21]. Where is my mistake?
asked May 23, 2012 at 3:43
1
I think this is better:
>>> x=[[1, 2],[3, 4],[5, 6]]
>>> sum[sum[x,[]]]
21
answered Nov 27, 2012 at 6:07
3
You could rewrite that function as,
def sum1[input]:
return sum[map[sum, input]]
Basically, map[sum, input] will return a list with the sums across all your rows, then, the outer most sum will add up that list.
Example:
>>> a=[[1,2],[3,4]]
>>> sum[map[sum, a]]
10
answered May 23, 2012 at 3:58
machowmachow
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1
This is yet another alternate Solution
In [1]: a=[[1, 2],[3, 4],[5, 6]]
In [2]: sum[[sum[i] for i in a]]
Out[2]: 21
answered May 14, 2015 at 16:44
AjayAjay
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0
And numpy solution is just:
import numpy as np
x = np.array[[[1, 2],[3, 4],[5, 6]]]
Result:
>>> b=np.sum[x]
print[b]
21
answered May 23, 2012 at 3:50
AkavallAkavall
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3
Better still, forget the index counters and just iterate over the items themselves:
def sum1[input]:
my_sum = 0
for row in input:
my_sum += sum[row]
return my_sum
print sum1[[[1, 2],[3, 4],[5, 6]]]
One of the nice [and idiomatic] features of Python is letting it do the counting for you. sum[] is a built-in and you should not use names of built-ins for your own identifiers.
answered May 23, 2012 at 3:59
mswmsw
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This is the issue
for row in range [len[input]-1]:
for col in range[len[input[0]]-1]:
try
for row in range [len[input]]:
for col in range[len[input[0]]]:
Python's range[x] goes from 0..x-1 already
range[...] range[[start,] stop[, step]] -> list of integers
Return a list containing an arithmetic progression of integers. range[i, j] returns [i, i+1, i+2, ..., j-1]; start [!] defaults to 0. When step is given, it specifies the increment [or decrement]. For example, range[4] returns [0, 1, 2, 3]. The end point is omitted! These are exactly the valid indices for a list of 4 elements.
answered May 23, 2012 at 3:45
dfbdfb
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range[] in python excludes the last element. In other words, range[1, 5] is [1, 5] or [1, 4]. So you should just use len[input] to iterate over the rows/columns.
def sum1[input]:
sum = 0
for row in range [len[input]]:
for col in range[len[input[0]]]:
sum = sum + input[row][col]
return sum
answered May 23, 2012 at 3:45
spinlokspinlok
3,46316 silver badges25 bronze badges
Don't put -1 in range[len[input]-1] instead use:
range[len[input]]
range automatically returns a list one less than the argument value so no need of explicitly giving -1
answered May 23, 2012 at 3:46
Kartik AnandKartik Anand
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def sum1[input]:
return sum[[sum[x] for x in input]]
answered Sep 13, 2018 at 22:49
J F FitchJ F Fitch
1161 silver badge3 bronze badges
Quick answer, use...
total = sum[map[sum,[array]]]
where [array] is your array title.
answered Apr 1, 2018 at 20:54
1
In Python 3.7
import numpy as np
x = np.array[[ [1,2], [3,4] ]]
sum[sum[x]]
outputs
10
answered Jan 21, 2019 at 14:51
It seems like a general consensus is that numpy is a complicated solution. In comparison to simpler algorithms. But for the sake of the answer being present:
import numpy as np
def addarrays[arr]:
b = np.sum[arr]
return sum[b]
array_1 = [
[1, 2],
[3, 4],
[5, 6]
]
print[addarrays[array_1]]
This appears to be the preferred solution:
x=[[1, 2],[3, 4],[5, 6]]
sum[sum[x,[]]]
answered Sep 26, 2019 at 0:14
peyopeyo
3393 silver badges14 bronze badges
def sum1[input]:
sum = 0
for row in input:
for col in row:
sum += col
return sum
print[sum1[[[1, 2],[3, 4],[5, 6]]]]
Sefan
7061 gold badge7 silver badges22 bronze badges
answered Aug 17, 2021 at 7:57
Speed comparison
import random
import timeit
import numpy
x = [[random.random[] for i in range[100]] for j in range[100]]
xnp = np.array[x]
Methods
print["Sum python array:"]
%timeit sum[map[sum,x]]
%timeit sum[[sum[i] for i in x]]
%timeit sum[sum[x,[]]]
%timeit sum[[x[i][j] for i in range[100] for j in range[100]]]
print["Convert to numpy, then sum:"]
%timeit np.sum[np.array[x]]
%timeit sum[sum[np.array[x]]]
print["Sum numpy array:"]
%timeit np.sum[xnp]
%timeit sum[sum[xnp]]
Results
Sum python array:
130 µs ± 3.24 µs per loop [mean ± std. dev. of 7 runs, 10000 loops each]
149 µs ± 4.16 µs per loop [mean ± std. dev. of 7 runs, 10000 loops each]
3.05 ms ± 44.8 µs per loop [mean ± std. dev. of 7 runs, 100 loops each]
2.58 ms ± 107 µs per loop [mean ± std. dev. of 7 runs, 100 loops each]
Convert to numpy, then sum:
1.36 ms ± 90.1 µs per loop [mean ± std. dev. of 7 runs, 1000 loops each]
1.63 ms ± 26.1 µs per loop [mean ± std. dev. of 7 runs, 1000 loops each]
Sum numpy array:
24.6 µs ± 1.95 µs per loop [mean ± std. dev. of 7 runs, 10000 loops each]
301 µs ± 4.78 µs per loop [mean ± std. dev. of 7 runs, 1000 loops each]
answered Apr 5 at 10:13
FasmoFasmo
215 bronze badges
1
def sum1[input]:
sum = 0
for row in range [len[input]-1]:
for col in range[len[input[0]]-1]:
sum = sum + input[row][col]
return sum
print [sum1[[[1, 2],[3, 4],[5, 6]]]]
You had a problem with parenthesis at the print command.... This solution will be good now The correct solution in Visual Studio Code
McLovin
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answered Aug 8 at 17:12
1