Mixing
Find out which condition breaks a logical and expression
I want to find an elegant way to get the component of the logical and expression below that is responsible if the if block is not being executed.
if test1(value) and test2(value) and test3(value):
print 'Yeeaah'
else:
print 'Oh, no!', 'Who is the first function that return false?'
In case the else block is entered, how do I find out whether test1, test2 or test3 is responsible by returning the first falsy value?
Salvo.
python boolean-expression
edited Nov 22 '18 at 12:25
timgeb
51k116693
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
add a comment |
I want to find an elegant way to get the component of the logical and expression below that is responsible if the if block is not being executed.
if test1(value) and test2(value) and test3(value):
print 'Yeeaah'
else:
print 'Oh, no!', 'Who is the first function that return false?'
In case the else block is entered, how do I find out whether test1, test2 or test3 is responsible by returning the first falsy value?
Salvo.
python boolean-expression
edited Nov 22 '18 at 12:25
timgeb
51k116693
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
add a comment |
I want to find an elegant way to get the component of the logical and expression below that is responsible if the if block is not being executed.
if test1(value) and test2(value) and test3(value):
print 'Yeeaah'
else:
print 'Oh, no!', 'Who is the first function that return false?'
In case the else block is entered, how do I find out whether test1, test2 or test3 is responsible by returning the first falsy value?
Salvo.
python boolean-expression
edited Nov 22 '18 at 12:25
timgeb
51k116693
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
I want to find an elegant way to get the component of the logical and expression below that is responsible if the if block is not being executed.
if test1(value) and test2(value) and test3(value):
print 'Yeeaah'
else:
print 'Oh, no!', 'Who is the first function that return false?'
In case the else block is entered, how do I find out whether test1, test2 or test3 is responsible by returning the first falsy value?
Salvo.
python boolean-expression
python boolean-expression
edited Nov 22 '18 at 12:25
timgeb
51k116693
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
edited Nov 22 '18 at 12:25
timgeb
51k116693
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
edited Nov 22 '18 at 12:25
timgeb
51k116693
edited Nov 22 '18 at 12:25
timgeb
51k116693
edited Nov 22 '18 at 12:25
timgeb
51k116693
51k116693
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
asked Nov 22 '18 at 11:55
Blackat.netBlackat.net
8116
8116
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You can use next and a generator expression:
breaker = next((test.__name__ for test in (test1, test2, test3) if not test(value)), None)
Demo:
>>> def test1(value): return True
>>> def test2(value): return False
>>> def test3(value): return True
>>> value = '_' # irrelevant for this demo
>>>
>>> tests = (test1, test2, test3)
>>> breaker = next((test.__name__ for test in tests if not test(value)), None)
>>> breaker
'test2'
>>> if not breaker:
...: print('Yeeaah')
...:else:
...: print('Oh no!')
...:
Oh no!
Note that test3 is never called in this code.
(Super corner case: use if breaker is not None over if not breaker if for reasons I cannot fathom pranksters reassigned the function's __name__ attribute to ''.)
~edit~
In case you want to know whether the first, second, or n'th test returned something falsy, you can use a similar generator expression with enumerate.
>>> breaker = next((i for i, test in enumerate(tests, 1) if not test(value)), None)
>>> breaker
2
If you want to count from zero, use enumerate(tests) and check if breaker is not None for entering the if block (because 0 is falsy like None).
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
1
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
add a comment |
We could store that test values in list and then check if they all are True and if not print index of first value that is False.
def test1(x):
return True
def test2(x):
return False
def test3(x):
return True
value = 'a'
statements = [test1(value), test2(value), test3(value)]
if all(statements):
print('Yeeaah')
else:
print('Oh, no! Function that first return False is {}'.format(statements.index(False)))
Output:
Oh, no! Function that first return False is 1
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
2
Note that this does evaluate all the tests, while the chainedandoperations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.
– timgeb
Nov 22 '18 at 12:04
add a comment |
Split it three-ways:
Using python3 syntax:
b1 = test1(value)
b2 = test2(value)
b3 = test3(value)
if b1 and b2 and b3:
print("Yeah")
else:
print("Nope! 1: {}, 2: {}, 3: {}".format(b1, b2, b3))
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
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
You can use next and a generator expression:
breaker = next((test.__name__ for test in (test1, test2, test3) if not test(value)), None)
Demo:
>>> def test1(value): return True
>>> def test2(value): return False
>>> def test3(value): return True
>>> value = '_' # irrelevant for this demo
>>>
>>> tests = (test1, test2, test3)
>>> breaker = next((test.__name__ for test in tests if not test(value)), None)
>>> breaker
'test2'
>>> if not breaker:
...: print('Yeeaah')
...:else:
...: print('Oh no!')
...:
Oh no!
Note that test3 is never called in this code.
(Super corner case: use if breaker is not None over if not breaker if for reasons I cannot fathom pranksters reassigned the function's __name__ attribute to ''.)
~edit~
In case you want to know whether the first, second, or n'th test returned something falsy, you can use a similar generator expression with enumerate.
>>> breaker = next((i for i, test in enumerate(tests, 1) if not test(value)), None)
>>> breaker
2
If you want to count from zero, use enumerate(tests) and check if breaker is not None for entering the if block (because 0 is falsy like None).
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
1
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
add a comment |
You can use next and a generator expression:
breaker = next((test.__name__ for test in (test1, test2, test3) if not test(value)), None)
Demo:
>>> def test1(value): return True
>>> def test2(value): return False
>>> def test3(value): return True
>>> value = '_' # irrelevant for this demo
>>>
>>> tests = (test1, test2, test3)
>>> breaker = next((test.__name__ for test in tests if not test(value)), None)
>>> breaker
'test2'
>>> if not breaker:
...: print('Yeeaah')
...:else:
...: print('Oh no!')
...:
Oh no!
Note that test3 is never called in this code.
(Super corner case: use if breaker is not None over if not breaker if for reasons I cannot fathom pranksters reassigned the function's __name__ attribute to ''.)
~edit~
In case you want to know whether the first, second, or n'th test returned something falsy, you can use a similar generator expression with enumerate.
>>> breaker = next((i for i, test in enumerate(tests, 1) if not test(value)), None)
>>> breaker
2
If you want to count from zero, use enumerate(tests) and check if breaker is not None for entering the if block (because 0 is falsy like None).
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
1
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
add a comment |
You can use next and a generator expression:
breaker = next((test.__name__ for test in (test1, test2, test3) if not test(value)), None)
Demo:
>>> def test1(value): return True
>>> def test2(value): return False
>>> def test3(value): return True
>>> value = '_' # irrelevant for this demo
>>>
>>> tests = (test1, test2, test3)
>>> breaker = next((test.__name__ for test in tests if not test(value)), None)
>>> breaker
'test2'
>>> if not breaker:
...: print('Yeeaah')
...:else:
...: print('Oh no!')
...:
Oh no!
Note that test3 is never called in this code.
(Super corner case: use if breaker is not None over if not breaker if for reasons I cannot fathom pranksters reassigned the function's __name__ attribute to ''.)
~edit~
In case you want to know whether the first, second, or n'th test returned something falsy, you can use a similar generator expression with enumerate.
>>> breaker = next((i for i, test in enumerate(tests, 1) if not test(value)), None)
>>> breaker
2
If you want to count from zero, use enumerate(tests) and check if breaker is not None for entering the if block (because 0 is falsy like None).
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
You can use next and a generator expression:
breaker = next((test.__name__ for test in (test1, test2, test3) if not test(value)), None)
Demo:
>>> def test1(value): return True
>>> def test2(value): return False
>>> def test3(value): return True
>>> value = '_' # irrelevant for this demo
>>>
>>> tests = (test1, test2, test3)
>>> breaker = next((test.__name__ for test in tests if not test(value)), None)
>>> breaker
'test2'
>>> if not breaker:
...: print('Yeeaah')
...:else:
...: print('Oh no!')
...:
Oh no!
Note that test3 is never called in this code.
(Super corner case: use if breaker is not None over if not breaker if for reasons I cannot fathom pranksters reassigned the function's __name__ attribute to ''.)
~edit~
In case you want to know whether the first, second, or n'th test returned something falsy, you can use a similar generator expression with enumerate.
>>> breaker = next((i for i, test in enumerate(tests, 1) if not test(value)), None)
>>> breaker
2
If you want to count from zero, use enumerate(tests) and check if breaker is not None for entering the if block (because 0 is falsy like None).
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
edited Nov 22 '18 at 12:51
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
answered Nov 22 '18 at 12:02
timgebtimgeb
51k116693
51k116693
1
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
add a comment |
1
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
1
1
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
This seems to me the most pythonic answer, and most easily generalized
– planetmaker
Nov 22 '18 at 12:04
add a comment |
We could store that test values in list and then check if they all are True and if not print index of first value that is False.
def test1(x):
return True
def test2(x):
return False
def test3(x):
return True
value = 'a'
statements = [test1(value), test2(value), test3(value)]
if all(statements):
print('Yeeaah')
else:
print('Oh, no! Function that first return False is {}'.format(statements.index(False)))
Output:
Oh, no! Function that first return False is 1
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
2
Note that this does evaluate all the tests, while the chainedandoperations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.
– timgeb
Nov 22 '18 at 12:04
add a comment |
We could store that test values in list and then check if they all are True and if not print index of first value that is False.
def test1(x):
return True
def test2(x):
return False
def test3(x):
return True
value = 'a'
statements = [test1(value), test2(value), test3(value)]
if all(statements):
print('Yeeaah')
else:
print('Oh, no! Function that first return False is {}'.format(statements.index(False)))
Output:
Oh, no! Function that first return False is 1
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
2
Note that this does evaluate all the tests, while the chainedandoperations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.
– timgeb
Nov 22 '18 at 12:04
add a comment |
We could store that test values in list and then check if they all are True and if not print index of first value that is False.
def test1(x):
return True
def test2(x):
return False
def test3(x):
return True
value = 'a'
statements = [test1(value), test2(value), test3(value)]
if all(statements):
print('Yeeaah')
else:
print('Oh, no! Function that first return False is {}'.format(statements.index(False)))
Output:
Oh, no! Function that first return False is 1
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
We could store that test values in list and then check if they all are True and if not print index of first value that is False.
def test1(x):
return True
def test2(x):
return False
def test3(x):
return True
value = 'a'
statements = [test1(value), test2(value), test3(value)]
if all(statements):
print('Yeeaah')
else:
print('Oh, no! Function that first return False is {}'.format(statements.index(False)))
Output:
Oh, no! Function that first return False is 1
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
edited Nov 22 '18 at 12:01
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 22 '18 at 11:57
Filip MłynarskiFilip Młynarski
1,7961413
1,7961413
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
2
Note that this does evaluate all the tests, while the chainedandoperations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.
– timgeb
Nov 22 '18 at 12:04
add a comment |
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
2
Note that this does evaluate all the tests, while the chainedandoperations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.
– timgeb
Nov 22 '18 at 12:04
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
Maybe at least some text would benefit your answer
– user8408080
Nov 22 '18 at 12:00
2
2
Note that this does evaluate all the tests, while the chained
and operations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.– timgeb
Nov 22 '18 at 12:04
Note that this does evaluate all the tests, while the chained
and operations are lazy and short circuit. So in theory if the tests have side effects, this code is not strictly equivalent.– timgeb
Nov 22 '18 at 12:04
add a comment |
Split it three-ways:
Using python3 syntax:
b1 = test1(value)
b2 = test2(value)
b3 = test3(value)
if b1 and b2 and b3:
print("Yeah")
else:
print("Nope! 1: {}, 2: {}, 3: {}".format(b1, b2, b3))
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
add a comment |
Split it three-ways:
Using python3 syntax:
b1 = test1(value)
b2 = test2(value)
b3 = test3(value)
if b1 and b2 and b3:
print("Yeah")
else:
print("Nope! 1: {}, 2: {}, 3: {}".format(b1, b2, b3))
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
add a comment |
Split it three-ways:
Using python3 syntax:
b1 = test1(value)
b2 = test2(value)
b3 = test3(value)
if b1 and b2 and b3:
print("Yeah")
else:
print("Nope! 1: {}, 2: {}, 3: {}".format(b1, b2, b3))
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
Split it three-ways:
Using python3 syntax:
b1 = test1(value)
b2 = test2(value)
b3 = test3(value)
if b1 and b2 and b3:
print("Yeah")
else:
print("Nope! 1: {}, 2: {}, 3: {}".format(b1, b2, b3))
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
answered Nov 22 '18 at 11:59
planetmakerplanetmaker
4,66421729
4,66421729
add a comment |
add a comment |
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