Why is this codes output different?
lst = ["id1", "id2", "id3", "id4", "id5", "id6", "id7", "id8"]
for i in lst:
print(i[2])
print([i[2] for i in lst])
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1First is an iteration, where you print every index 2 of every string. The second is a list comprehension, where you create a new list with the same data as above and prints the list.Cow– Cow2022-02-18 08:11:49 +00:00Commented Feb 18, 2022 at 8:11
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3 Answers
First is an iteration, where you print every index 2 of every string. The second is a list comprehension, where you create a new list with the same data as above and prints the list.
First:
lst = ["id1", "id2", "id3", "id4", "id5", "id6", "id7", "id8"]
for i in lst:
print(i[2])
Result:
1
2
3
4
5
6
7
8
Second:
print([i[2] for i in lst])
Result:
['1', '2', '3', '4', '5', '6', '7', '8']
If you want the first version to imitate the same result as the second out, simply create a list and append to it:
newlst = []
for i in lst:
newlst.append(i[2])
print(newlst)
Result:
['1', '2', '3', '4', '5', '6', '7', '8']
If you want to print the list comprehension in one line, like the for loop, then use:
[print(i[2]) for i in lst]
Result:
1
2
3
4
5
6
7
8
If you need a type string for the comprehension, you could do this:
s = "\n".join([i[2] for i in lst])
print(s)
print(type(s))
Result:
1
2
3
4
5
6
7
8
<class 'str'>
8 Comments
Cow
@mvp I'm sorry, I do not understand what you are asking. What would you like with the code
print([i[2] for i in lst])? :-)Cow
@mvp Added example. :-)
Cow
@mvp Added too. :-)
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In the first case, you are doing multiple prints. In the second case you are building a list with all the data, then this said list is printed
Comments
lst = ["id1", "id2", "id3", "id4", "id5", "id6", "id7", "id8"]
# You are iterating over the list
for i in lst:
# prints 3rd character in each element (i)
print(i[2]) # be default this statement has newline at end
# To print with spaces try print(i[2], end=' ')
# here you are doing it using list comprehension so the output will be in a list
print([i[2] for i in lst])
