I have a list of lists:
a = [[1, 3, 4], [2, 5, 7]]
I want the output in the following format:
1 3 4
2 5 7
I have tried it the following way , but the outputs are not in the desired way:
for i in a:
for j in i:
print[j, sep=' ']
Outputs:
1
3
4
2
5
7
While changing the print call to use end
instead:
for i in a:
for j in i:
print[j, end = ' ']
Outputs:
1 3 4 2 5 7
Any ideas?
asked Aug 10, 2016 at 11:35
0
Iterate through every sub-list in your original list and unpack it in the print call with *
:
a = [[1, 3, 4], [2, 5, 7]]
for s in a:
print[*s]
The separation is by default set to ' '
so there's no need to explicitly provide it. This prints:
1 3 4
2 5 7
In your approach you were iterating for every element in every sub-list and printing that individually. By using print[*s]
you unpack the list inside the print call, this essentially translates to:
print[1, 3, 4] # for s = [1, 3, 4]
print[2, 5, 7] # for s = [2, 5, 7]
answered Aug 10, 2016 at 11:43
1
oneliner:
print['\n'.join[' '.join[map[str,sl]] for sl in l]]
explanation:
you can convert list
into str
by using join function:
l = ['1','2','3']
' '.join[l] # will give you a next string: '1 2 3'
'.'.join[l] # and it will give you '1.2.3'
so, if you want linebreaks you should use new line symbol.
But join accepts only list of strings. For converting list of things to list of strings, you can apply str
function for each item in list:
l = [1,2,3]
' '.join[map[str, l]] # will return string '1 2 3'
And we apply this construction for each sublist sl
in list l
answered Aug 10, 2016 at 11:45
ailinailin
4715 silver badges15 bronze badges
0
You can do this:
>>> lst = [[1, 3, 4], [2, 5, 7]]
>>> for sublst in lst:
... for item in sublst:
... print item, # note the ending ','
... print # print a newline
...
1 3 4
2 5 7
answered Aug 10, 2016 at 12:09
nalzoknalzok
13.9k19 gold badges64 silver badges122 bronze badges
a = [[1, 3, 4], [2, 5, 7]]
for i in a:
for j in i:
print[j, end = ' ']
print['',sep='\n']
output:
1 3 4
2 5 7
SherylHohman
15k16 gold badges84 silver badges89 bronze badges
answered Jul 10, 2020 at 18:11
2
lst = [[1, 3, 4], [2, 5, 7]]
def f[lst ]:
yield from lst
for x in f[lst]:
print[*x]
using "yield from"...
Produces Output
1 3 4
2 5 7
[Program finished]
answered Feb 19, 2021 at 4:06
SubhamSubham
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There is an alternative method to display list rather than arranging them in sub-list:
tick_tack_display = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
loop_start = 0
count = 0
while loop_start < len[tick_tack_display]:
print[tick_tack_display[loop_start], end = '']
count +=1
if count - 3 == 0:
print["\n"]
count = 0
loop_start += 1
OUTPUT :
123
456
789
DuDa
3,6484 gold badges15 silver badges36 bronze badges
answered Dec 24, 2020 at 21:01
I believe this is pretty simple:
a = [[1, 3, 4], [2, 5, 7]] # defines the list
for i in range[len[a]]: # runs for every "sub-list" in a
for j in a[i]: # gives j the value of each item in a[i]
print[j, end=" "] # prints j without going to a new line
print[] # creates a new line after each list-in-list prints
output
1 3 4
2 5 7
DuDa
3,6484 gold badges15 silver badges36 bronze badges
answered Jun 5, 2020 at 17:38
def print_list[s]:
for i in range[len[s]]:
if isinstance[s[i],list]:
k=s[i]
print_list[k]
else:
print[s[i]]
s=[[1,2,[3,4,[5,6]],7,8]]
print_list[s]
you could enter lists within lists within lists ..... and yet everything will be printed as u expect it to be.
Output:
1
2
3
4
5
6
7
8
DuDa
3,6484 gold badges15 silver badges36 bronze badges
answered Jan 18, 2018 at 17:27
1
There's an easier one-liner way:
a = [[1, 3, 4], [2, 5, 7]] # your data
[print[*x] for x in a][0] # one-line print
And the result will be as you want it:
1 3 4
2 5 7
Make sure you add the [0]
to the end of the list comprehension, otherwise, the last line would be a list of None
values equal to the length of your list.
answered May 12, 2021 at 2:32