Multiple 'In' operator in python
-4
arr = [1, True, 'a', 2]
print('a' in arr in arr) # False
Can you explain me why this code will output 'False'?
The question is closed.
Answer from @KlausD.: Actually it is a comparison operator chaining and will be interpreted as ('a' in arr) and (arr in arr).
python python-3.x operators in-operator
asked Nov 11 at 7:04
Matvey
13
|
show 1 more comment
-4
arr = [1, True, 'a', 2]
print('a' in arr in arr) # False
Can you explain me why this code will output 'False'?
The question is closed.
Answer from @KlausD.: Actually it is a comparison operator chaining and will be interpreted as ('a' in arr) and (arr in arr).
python python-3.x operators in-operator
asked Nov 11 at 7:04
Matvey
13
Just do'a' in arr
– W-B
Nov 11 at 7:05
I believe you're expectingTruebecause, you see'a' in arr in arrasTrue in arr?
– Austin
Nov 11 at 7:08
@Austin, just run this code. It output False.
– Matvey
Nov 11 at 7:23
Please explain waht your expected result was, and why. The question as it is, is unclear
– CIsForCookies
Nov 11 at 7:27
@CIsForCookies, What did you not understand? I wanna understand why the expression " 'a' in arr in arr " is False
– Matvey
Nov 11 at 7:32
|
show 1 more comment
-4
-4
-4
0
arr = [1, True, 'a', 2]
print('a' in arr in arr) # False
Can you explain me why this code will output 'False'?
The question is closed.
Answer from @KlausD.: Actually it is a comparison operator chaining and will be interpreted as ('a' in arr) and (arr in arr).
python python-3.x operators in-operator
asked Nov 11 at 7:04
Matvey
13
arr = [1, True, 'a', 2]
print('a' in arr in arr) # False
Can you explain me why this code will output 'False'?
The question is closed.
Answer from @KlausD.: Actually it is a comparison operator chaining and will be interpreted as ('a' in arr) and (arr in arr).
python python-3.x operators in-operator
python python-3.x operators in-operator
asked Nov 11 at 7:04
Matvey
13
asked Nov 11 at 7:04
Matvey
13
edited Nov 11 at 8:26
asked Nov 11 at 7:04
Matvey
13
asked Nov 11 at 7:04
Matvey
13
asked Nov 11 at 7:04
Matvey
13
13
Just do'a' in arr
– W-B
Nov 11 at 7:05
I believe you're expectingTruebecause, you see'a' in arr in arrasTrue in arr?
– Austin
Nov 11 at 7:08
@Austin, just run this code. It output False.
– Matvey
Nov 11 at 7:23
Please explain waht your expected result was, and why. The question as it is, is unclear
– CIsForCookies
Nov 11 at 7:27
@CIsForCookies, What did you not understand? I wanna understand why the expression " 'a' in arr in arr " is False
– Matvey
Nov 11 at 7:32
|
show 1 more comment
Just do'a' in arr
– W-B
Nov 11 at 7:05
I believe you're expectingTruebecause, you see'a' in arr in arrasTrue in arr?
– Austin
Nov 11 at 7:08
@Austin, just run this code. It output False.
– Matvey
Nov 11 at 7:23
Please explain waht your expected result was, and why. The question as it is, is unclear
– CIsForCookies
Nov 11 at 7:27
@CIsForCookies, What did you not understand? I wanna understand why the expression " 'a' in arr in arr " is False
– Matvey
Nov 11 at 7:32
Just do
'a' in arr– W-B
Nov 11 at 7:05
Just do
'a' in arr– W-B
Nov 11 at 7:05
I believe you're expecting
True because, you see 'a' in arr in arr as True in arr ?– Austin
Nov 11 at 7:08
I believe you're expecting
True because, you see 'a' in arr in arr as True in arr ?– Austin
Nov 11 at 7:08
@Austin, just run this code. It output False.
– Matvey
Nov 11 at 7:23
@Austin, just run this code. It output False.
– Matvey
Nov 11 at 7:23
Please explain waht your expected result was, and why. The question as it is, is unclear
– CIsForCookies
Nov 11 at 7:27
Please explain waht your expected result was, and why. The question as it is, is unclear
– CIsForCookies
Nov 11 at 7:27
@CIsForCookies, What did you not understand? I wanna understand why the expression " 'a' in arr in arr " is False
– Matvey
Nov 11 at 7:32
@CIsForCookies, What did you not understand? I wanna understand why the expression " 'a' in arr in arr " is False
– Matvey
Nov 11 at 7:32
|
show 1 more comment
3 Answers
3
active
oldest
votes
0
I believe this is what you are trying to do:
arr = [1, True, 'a', 2]
print( 'a' in arr)
Output:
True
Or this:
arr = [1, True, 'a', 2]
print(bool(['a' in arr]) in arr)
Output:
True
answered Nov 11 at 7:06
Sanchit Kumar
31117
you can just doprint(('a' in arr) in arr)instead ofprint(bool(['a' in arr]) in arr)but that does not check if'a'is inarr. It check ifTrueis! Change arr toarr = ['a', 2]and verify
– CIsForCookies
Nov 11 at 7:21
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Legit. I thought OP only wants to check if'a' in arr. Either way, what actually being checked is ifarr in arr, and then if 'a' is in the result :-P
– CIsForCookies
Nov 11 at 7:26
bool(['a' in arr])is almost certainly not any part of what the questioner wanted, considering it's callingboolon a one-element list, and it'll returnTrueno matter whether or not'a'is inarr.
– user2357112
Nov 11 at 8:02
add a comment |
0
print('a' in arr in arr) // Falseis interpeted asprint('a' in arr in arr) // 0which throws ZeroDivisionError: integer division or modulo by zero error. If you meant to comment out theFalse, do it using"#", not"//"(e.g.print('a' in arr in arr) # False)
'a' in arr in arris read from right to left[1]: check ifarr in arr(False), and then check if'a' in False(False)
Using @Klaus D's helpful comment -print('a' in arr in arr)is evaluated asprint(('a' in arr) and (arr in arr))due to operator chaining. This, in turn is processed intoprint(True and False)->print(False)
To check if 'a' is in arr, just check print('a' in arr) # prints True
[1] Well, not exactly. As can seen from [ In which order is an if statement evaluated in Python ], the evaluation is right to left ,so this is what actually happens: (1) check if 'a' is in "something". (2) evaluate this "something" by checking if arr in arr. (3) use the reault of said something (which is False as sadly, arr isn't a member of itslef) and check if 'a' is inside that (meaning, check if 'a' in True, which again, is False[1]
answered Nov 11 at 7:11
CIsForCookies
6,68311546
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
2.'a' in Falseis incorrect expression. It will throws error.
– Matvey
Nov 11 at 7:36
2
1. Actually it is a comparison operator chaining and will be interpreted as('a' in arr) and (arr in arr).
– Klaus D.
Nov 11 at 7:48
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
add a comment |
0
It is False because 'a' is in 'arr' but 'arr' is not in 'arr'.
Meaning 'arr' can't be in itself.
answered Nov 11 at 9:40
Jack Herer
313112
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
I believe this is what you are trying to do:
arr = [1, True, 'a', 2]
print( 'a' in arr)
Output:
True
Or this:
arr = [1, True, 'a', 2]
print(bool(['a' in arr]) in arr)
Output:
True
answered Nov 11 at 7:06
Sanchit Kumar
31117
you can just doprint(('a' in arr) in arr)instead ofprint(bool(['a' in arr]) in arr)but that does not check if'a'is inarr. It check ifTrueis! Change arr toarr = ['a', 2]and verify
– CIsForCookies
Nov 11 at 7:21
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Legit. I thought OP only wants to check if'a' in arr. Either way, what actually being checked is ifarr in arr, and then if 'a' is in the result :-P
– CIsForCookies
Nov 11 at 7:26
bool(['a' in arr])is almost certainly not any part of what the questioner wanted, considering it's callingboolon a one-element list, and it'll returnTrueno matter whether or not'a'is inarr.
– user2357112
Nov 11 at 8:02
add a comment |
0
I believe this is what you are trying to do:
arr = [1, True, 'a', 2]
print( 'a' in arr)
Output:
True
Or this:
arr = [1, True, 'a', 2]
print(bool(['a' in arr]) in arr)
Output:
True
answered Nov 11 at 7:06
Sanchit Kumar
31117
you can just doprint(('a' in arr) in arr)instead ofprint(bool(['a' in arr]) in arr)but that does not check if'a'is inarr. It check ifTrueis! Change arr toarr = ['a', 2]and verify
– CIsForCookies
Nov 11 at 7:21
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Legit. I thought OP only wants to check if'a' in arr. Either way, what actually being checked is ifarr in arr, and then if 'a' is in the result :-P
– CIsForCookies
Nov 11 at 7:26
bool(['a' in arr])is almost certainly not any part of what the questioner wanted, considering it's callingboolon a one-element list, and it'll returnTrueno matter whether or not'a'is inarr.
– user2357112
Nov 11 at 8:02
add a comment |
0
0
0
I believe this is what you are trying to do:
arr = [1, True, 'a', 2]
print( 'a' in arr)
Output:
True
Or this:
arr = [1, True, 'a', 2]
print(bool(['a' in arr]) in arr)
Output:
True
answered Nov 11 at 7:06
Sanchit Kumar
31117
I believe this is what you are trying to do:
arr = [1, True, 'a', 2]
print( 'a' in arr)
Output:
True
Or this:
arr = [1, True, 'a', 2]
print(bool(['a' in arr]) in arr)
Output:
True
answered Nov 11 at 7:06
Sanchit Kumar
31117
answered Nov 11 at 7:06
Sanchit Kumar
31117
answered Nov 11 at 7:06
Sanchit Kumar
31117
answered Nov 11 at 7:06
Sanchit Kumar
31117
31117
you can just doprint(('a' in arr) in arr)instead ofprint(bool(['a' in arr]) in arr)but that does not check if'a'is inarr. It check ifTrueis! Change arr toarr = ['a', 2]and verify
– CIsForCookies
Nov 11 at 7:21
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Legit. I thought OP only wants to check if'a' in arr. Either way, what actually being checked is ifarr in arr, and then if 'a' is in the result :-P
– CIsForCookies
Nov 11 at 7:26
bool(['a' in arr])is almost certainly not any part of what the questioner wanted, considering it's callingboolon a one-element list, and it'll returnTrueno matter whether or not'a'is inarr.
– user2357112
Nov 11 at 8:02
add a comment |
you can just doprint(('a' in arr) in arr)instead ofprint(bool(['a' in arr]) in arr)but that does not check if'a'is inarr. It check ifTrueis! Change arr toarr = ['a', 2]and verify
– CIsForCookies
Nov 11 at 7:21
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Legit. I thought OP only wants to check if'a' in arr. Either way, what actually being checked is ifarr in arr, and then if 'a' is in the result :-P
– CIsForCookies
Nov 11 at 7:26
bool(['a' in arr])is almost certainly not any part of what the questioner wanted, considering it's callingboolon a one-element list, and it'll returnTrueno matter whether or not'a'is inarr.
– user2357112
Nov 11 at 8:02
you can just do
print(('a' in arr) in arr) instead of print(bool(['a' in arr]) in arr) but that does not check if 'a' is in arr. It check if True is! Change arr to arr = ['a', 2] and verify– CIsForCookies
Nov 11 at 7:21
you can just do
print(('a' in arr) in arr) instead of print(bool(['a' in arr]) in arr) but that does not check if 'a' is in arr. It check if True is! Change arr to arr = ['a', 2] and verify– CIsForCookies
Nov 11 at 7:21
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Question is not clear. I am assuming he wants to get the output from 'a' in arr as True and then check if True in arr.
– Sanchit Kumar
Nov 11 at 7:22
Legit. I thought OP only wants to check if
'a' in arr. Either way, what actually being checked is if arr in arr, and then if 'a' is in the result :-P– CIsForCookies
Nov 11 at 7:26
Legit. I thought OP only wants to check if
'a' in arr. Either way, what actually being checked is if arr in arr, and then if 'a' is in the result :-P– CIsForCookies
Nov 11 at 7:26
bool(['a' in arr]) is almost certainly not any part of what the questioner wanted, considering it's calling bool on a one-element list, and it'll return True no matter whether or not 'a' is in arr.– user2357112
Nov 11 at 8:02
bool(['a' in arr]) is almost certainly not any part of what the questioner wanted, considering it's calling bool on a one-element list, and it'll return True no matter whether or not 'a' is in arr.– user2357112
Nov 11 at 8:02
add a comment |
0
print('a' in arr in arr) // Falseis interpeted asprint('a' in arr in arr) // 0which throws ZeroDivisionError: integer division or modulo by zero error. If you meant to comment out theFalse, do it using"#", not"//"(e.g.print('a' in arr in arr) # False)
'a' in arr in arris read from right to left[1]: check ifarr in arr(False), and then check if'a' in False(False)
Using @Klaus D's helpful comment -print('a' in arr in arr)is evaluated asprint(('a' in arr) and (arr in arr))due to operator chaining. This, in turn is processed intoprint(True and False)->print(False)
To check if 'a' is in arr, just check print('a' in arr) # prints True
[1] Well, not exactly. As can seen from [ In which order is an if statement evaluated in Python ], the evaluation is right to left ,so this is what actually happens: (1) check if 'a' is in "something". (2) evaluate this "something" by checking if arr in arr. (3) use the reault of said something (which is False as sadly, arr isn't a member of itslef) and check if 'a' is inside that (meaning, check if 'a' in True, which again, is False[1]
answered Nov 11 at 7:11
CIsForCookies
6,68311546
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
2.'a' in Falseis incorrect expression. It will throws error.
– Matvey
Nov 11 at 7:36
2
1. Actually it is a comparison operator chaining and will be interpreted as('a' in arr) and (arr in arr).
– Klaus D.
Nov 11 at 7:48
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
add a comment |
0
print('a' in arr in arr) // Falseis interpeted asprint('a' in arr in arr) // 0which throws ZeroDivisionError: integer division or modulo by zero error. If you meant to comment out theFalse, do it using"#", not"//"(e.g.print('a' in arr in arr) # False)
'a' in arr in arris read from right to left[1]: check ifarr in arr(False), and then check if'a' in False(False)
Using @Klaus D's helpful comment -print('a' in arr in arr)is evaluated asprint(('a' in arr) and (arr in arr))due to operator chaining. This, in turn is processed intoprint(True and False)->print(False)
To check if 'a' is in arr, just check print('a' in arr) # prints True
[1] Well, not exactly. As can seen from [ In which order is an if statement evaluated in Python ], the evaluation is right to left ,so this is what actually happens: (1) check if 'a' is in "something". (2) evaluate this "something" by checking if arr in arr. (3) use the reault of said something (which is False as sadly, arr isn't a member of itslef) and check if 'a' is inside that (meaning, check if 'a' in True, which again, is False[1]
answered Nov 11 at 7:11
CIsForCookies
6,68311546
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
2.'a' in Falseis incorrect expression. It will throws error.
– Matvey
Nov 11 at 7:36
2
1. Actually it is a comparison operator chaining and will be interpreted as('a' in arr) and (arr in arr).
– Klaus D.
Nov 11 at 7:48
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
add a comment |
0
0
0
print('a' in arr in arr) // Falseis interpeted asprint('a' in arr in arr) // 0which throws ZeroDivisionError: integer division or modulo by zero error. If you meant to comment out theFalse, do it using"#", not"//"(e.g.print('a' in arr in arr) # False)
'a' in arr in arris read from right to left[1]: check ifarr in arr(False), and then check if'a' in False(False)
Using @Klaus D's helpful comment -print('a' in arr in arr)is evaluated asprint(('a' in arr) and (arr in arr))due to operator chaining. This, in turn is processed intoprint(True and False)->print(False)
To check if 'a' is in arr, just check print('a' in arr) # prints True
[1] Well, not exactly. As can seen from [ In which order is an if statement evaluated in Python ], the evaluation is right to left ,so this is what actually happens: (1) check if 'a' is in "something". (2) evaluate this "something" by checking if arr in arr. (3) use the reault of said something (which is False as sadly, arr isn't a member of itslef) and check if 'a' is inside that (meaning, check if 'a' in True, which again, is False[1]
answered Nov 11 at 7:11
CIsForCookies
6,68311546
print('a' in arr in arr) // Falseis interpeted asprint('a' in arr in arr) // 0which throws ZeroDivisionError: integer division or modulo by zero error. If you meant to comment out theFalse, do it using"#", not"//"(e.g.print('a' in arr in arr) # False)
'a' in arr in arris read from right to left[1]: check ifarr in arr(False), and then check if'a' in False(False)
Using @Klaus D's helpful comment -print('a' in arr in arr)is evaluated asprint(('a' in arr) and (arr in arr))due to operator chaining. This, in turn is processed intoprint(True and False)->print(False)
To check if 'a' is in arr, just check print('a' in arr) # prints True
[1] Well, not exactly. As can seen from [ In which order is an if statement evaluated in Python ], the evaluation is right to left ,so this is what actually happens: (1) check if 'a' is in "something". (2) evaluate this "something" by checking if arr in arr. (3) use the reault of said something (which is False as sadly, arr isn't a member of itslef) and check if 'a' is inside that (meaning, check if 'a' in True, which again, is False[1]
answered Nov 11 at 7:11
CIsForCookies
6,68311546
edited Nov 11 at 8:13
answered Nov 11 at 7:11
CIsForCookies
6,68311546
answered Nov 11 at 7:11
CIsForCookies
6,68311546
answered Nov 11 at 7:11
CIsForCookies
6,68311546
6,68311546
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
2.'a' in Falseis incorrect expression. It will throws error.
– Matvey
Nov 11 at 7:36
2
1. Actually it is a comparison operator chaining and will be interpreted as('a' in arr) and (arr in arr).
– Klaus D.
Nov 11 at 7:48
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
add a comment |
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
2.'a' in Falseis incorrect expression. It will throws error.
– Matvey
Nov 11 at 7:36
2
1. Actually it is a comparison operator chaining and will be interpreted as('a' in arr) and (arr in arr).
– Klaus D.
Nov 11 at 7:48
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
1. Yeah, i'm sorry. There must be a "#"
– Matvey
Nov 11 at 7:35
2.
'a' in False is incorrect expression. It will throws error.– Matvey
Nov 11 at 7:36
2.
'a' in False is incorrect expression. It will throws error.– Matvey
Nov 11 at 7:36
2
2
1. Actually it is a comparison operator chaining and will be interpreted as
('a' in arr) and (arr in arr).– Klaus D.
Nov 11 at 7:48
1. Actually it is a comparison operator chaining and will be interpreted as
('a' in arr) and (arr in arr).– Klaus D.
Nov 11 at 7:48
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
@KlausD, You're right. Thanks, man. It's correct answer for my question.
– Matvey
Nov 11 at 7:52
add a comment |
0
It is False because 'a' is in 'arr' but 'arr' is not in 'arr'.
Meaning 'arr' can't be in itself.
answered Nov 11 at 9:40
Jack Herer
313112
add a comment |
0
It is False because 'a' is in 'arr' but 'arr' is not in 'arr'.
Meaning 'arr' can't be in itself.
answered Nov 11 at 9:40
Jack Herer
313112
add a comment |
0
0
0
It is False because 'a' is in 'arr' but 'arr' is not in 'arr'.
Meaning 'arr' can't be in itself.
answered Nov 11 at 9:40
Jack Herer
313112
It is False because 'a' is in 'arr' but 'arr' is not in 'arr'.
Meaning 'arr' can't be in itself.
answered Nov 11 at 9:40
Jack Herer
313112
answered Nov 11 at 9:40
Jack Herer
313112
answered Nov 11 at 9:40
Jack Herer
313112
answered Nov 11 at 9:40
Jack Herer
313112
313112
add a comment |
add a comment |
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Just do
'a' in arr– W-B
Nov 11 at 7:05
I believe you're expecting
Truebecause, you see'a' in arr in arrasTrue in arr?– Austin
Nov 11 at 7:08
@Austin, just run this code. It output False.
– Matvey
Nov 11 at 7:23
Please explain waht your expected result was, and why. The question as it is, is unclear
– CIsForCookies
Nov 11 at 7:27
@CIsForCookies, What did you not understand? I wanna understand why the expression " 'a' in arr in arr " is False
– Matvey
Nov 11 at 7:32