This question already has answers here:
and / or operators return value [duplicate]
(4 answers)
Closed 4 years ago.
I am trying to return a boolean in a function like this:
return mylist and any(condition(x) for x in mylist)
The behavior should be to return True if the list is empty or if any element in it meets the condition. I am using the first operand as a shortcircuit since any would return True if the list was empty, which is not what I am after.
I would expect [] and boolval to return False since the list is empty, but to my surprise it returns [] whether boolval is True or False. I would expect the first operand to be automatically evaluated as a boolean since it is involved in a comparison operation, and not whatever is happening.
I am not really asking how to solve my problem, which is easily done by an explicit type conversion: bool(mylist), but rather asking what is happening and why.
edit: when I ask "why" this is happening I am not looking for the "facts" only, as they are already explained in the linked duplicate question, but also the reasons behind the implementation of this behavior.
The and and or operators do not return True/False. They return the last thing evaluated (that's the case in other dynamic languages too, eg. javascript).
The official documentation describes that
for and, the first falsy value, or the last operand
for or, the first truthy value, or the last operand
That's by design, so you can create expressions like return username or 'guest'. So, if you want guarantee that a boolean value is returned, you have to
return bool(x or y)
instead of
return x or y
Because as khelwood said:
x and y gives x if x is falsey, otherwise it gives y.
That's the point, (and is not or :-)), so still best is:
return all([my_list,any(condition(x) for x in my_list)])
This has to do with how python evaluate the expression.
An empty list is considered as false by python, that means that the code after 'and' will not be executed, as this will not change the result.
Python does not need to convert the empty list into bool as it is not compared to anything, and just return it as empty list.
This shouldn't change anything for you, if you test the returned value of the function, it will be evaluate the same way as if the function did return False.
Related
I'm learning Python and I just started learning conditionals with booleans
I am very confused though as to the specific topic of "If Not". Could someone please explain to me the difference between :
x = False
if not x:
print("hello")
if x == False:
print("hello")
When testing this code on a Python compiler, I receive "hello" twice. I can assume this means that they both mean the same thing to the computer.
Could someone please explain to me why one would use one method over the other method?
It depends™. Python doesn't know what any of its operators should do. It calls magic methods on objects and lets them decide. We can see this with a simple test
class Foo:
"""Demonstrates the difference between a boolean and equality test
by overriding the operations that implement them."""
def __bool__(self):
print("bool")
return True
def __eq__(self, other):
print("eq", repr(other))
return True
x = Foo()
print("This is a boolean operation without an additional parameter")
if not x:
print("one")
print("This is an equality operation with a parameter")
if x == False:
print("two")
Produces
This is a boolean operation without an additional parameter
bool
This is an equality operation with a parameter
eq False
two
In the first case, python did a boolean test by calling __bool__, and in the second, an equality test by calling __eq__. What this means depends on the class. Its usually obvious but things like pandas may decide to get tricky.
Usually not x is faster than x == False because the __eq__ operator will typically do a second boolean comparison before it knows for sure. In your case, when x = False you are dealing with a builtin class written in C and its two operations will be similar. But still, the x == False comparison needs to do a type check against the other side, so it will be a bit slower.
There are already several good answers here, but none discuss the general concept of "truthy" and "falsy" expressions in Python.
In Python, truthy expressions are expression that return True when converted to bool, and falsy expressions are expressions that return False when converted to bool. (Ref: Trey Hunner's regular expression tutorial; I'm not affiliated with Hunner, I just love his tutorials.)
Truthy stuff:
What's important here is that 0, 0.0, [], None and False are all falsy.
When used in an if statement, they will fail the test, and they will pass the test in an if not statement.
Falsy stuff:
Non-zero numbers, non-empty lists, many objects (but read #tdelaney's answer for more details here), and True are all truthy, so they pass if and fail if not tests.
Equality tests
When you use equality tests, you're not asking about the truthiness of an expression, you're asking whether it is equal to the other thing you provide, which is much more restrictive than general truthiness or falsiness.
EDIT: Additional references
Here are more references on "Truthy" and "Falsy" values in Python:
Truth value testing in the Python manual
The exhaustive list of Falsy values
Truthy and Falsy tutorial from freeCodeCamp
In one case you are checking for equality with the value "False", on the other you are performing a boolean test on a variable. In Python, several variables pass the "if not x" test but only x = False passes "if x == False".
See example code:
x = [] # an empty list
if not x: print("Here!")
# x passes this test
if x == False: print("There!")
# x doesn't pass this test
Try it with x = None: not x would be True then and x == False would be False. Unlike with x = False when both of these are True. not statement also accounts for an empty value.
This question already has answers here:
How do Python's any and all functions work?
(10 answers)
Closed 4 years ago.
I accidentally found something in Numpy, which I can't really understand. If I check an empty Numpy array for any true value
np.array([]).any()
it will evaluate to false, whereas if I check all values to be true
np.array([]).all()
it evaluates to true. This appears weird to me since no value is true but at the same time all values are true...
This isn't a bug, it returns True because all values are not equal to zero which is the criteria for returning True see the following note:
Not a Number (NaN), positive infinity and negative infinity evaluate
to True because these are not equal to zero.
compare with the following:
In[102]:
np.array([True,]).all()
Out[102]: True
This would be equivalent to an array full of NaN which would return True
The logic you are seeing is not specific to NumPy. This is a Python convention which has been implemented in NumPy:
any returns True if any value is True. Otherwise False.
all returns True if no value is False. Otherwise True.
See the pseuedo-code in the docs to see the logic in pure Python.
In the case of np.array([]).any() or any([]), there are no True values, because you have a 0-dimensional array or a 0-length list. Therefore, the result is False.
In the case of np.array([]).all() or all([]), there are no False values, because you have a 0-dimensional array or a 0-length list. Therefore, the result is True.
This is a normal behavior.
It is not possible to find a value that is true, so np.array([]).any() is False
For every value in the array, this value is False (It is easy to check, because there are no values in the array, so you don't have to check anything).
Can anybody please tell me here what is the exact difference between if and if not here in the code.
def postordertraversse(self,top):
m=[]
if(top):
if not self.postordertraversse(top.left):
m.append(top.root)
top_most=m.pop(0)
conv=createlist();
conv.postordertraversse(conv.top)
What i can understand is if top means if top object instance exists then move inside the block and check if not i.e till top.left is not null keep appending.
if x: means "if x is truthy".
if not x: means "if x is falsey".
Whether something is truthy or falsey depends on what kind of object it is.
For numbers, 0 is falsey, and all other values are truthy.
For boolean values, True is truthy and False is falsey (obviously!)
For collections (lists, tuples, dictionaries, strings, etc), empty ones are falsey and non-empty ones are truthy.
So in your example code, the two if statements are saying:
if top is truthy:
if the result of self.postordertraversse(top.left) is falsey:
not in python is like negation in other programming languages.
'not' statement just converts further expression. For example:
not True - False
not False - True
>What's wrong with this..
Code-1
def first_last6(nums):
if nums[0]==6 or nums[len(nums)-1] == 6:
return True
else:
return False
Code-2
def first_last6(nums):
return (nums[0]==6 or nums[-1]== 6)
How come both True?
There seem to be two questions inside, so I’ll answer both.
First of all, why are nums[len(nums)-1] and nums[-1] the same? When specifying an index, Python allows you to use negative numbers that are interpreted like this: if i in nums[i] is negative, then the index len(nums)+i is returned. So, basically, [-1] will get the last element, [-2] the second to last etc.
The second question is why the two formats are identical:
if expression:
return True
else
return False
and
return expression
expression in this case is an expression that returns a boolean type, so either True or False. The if statements checks exactly that; if the expression equals to true, it will return true, otherwise (if the expression equals to false) it will return false.
So you can (and should, to make it cleaner) just return the expression itself, as it is already true or false.
In the case expression itself is not a boolean expression, the if statement will still check to what boolean type it would evaluate (for example a non-empty string would be true, or a number other than 0 would be true too). To keep the short syntax, you can then explicitely convert the expression to a boolean value, using bool(expression), as larsmans mentioned in the comments.
nums[-k] is a shorthand for nums[len(nums)-k]. To get the k-th last element you use the notation nums[-k]. Usually it is clear what the notation stands for and the compiler knows how to turn that python code into machine code, which is why certain language contructs are possible and others are not. Other short hands include nums[:k] to get the first k elements, nums[:-k] to get all elements up to the k-th last element etc. Via google, python docs, you will find much more on this. List operations are a great strength of the python.
http://www.diveintopython.net/native_data_types/lists.html
uh, because both are exactly the same and they both evaluate to True.
This question already has answers here:
Why does comparing strings using either '==' or 'is' sometimes produce a different result?
(15 answers)
Closed 9 years ago.
I noticed a Python script I was writing was acting squirrelly, and traced it to an infinite loop, where the loop condition was while line is not ''. Running through it in the debugger, it turned out that line was in fact ''. When I changed it to !='' rather than is not '', it worked fine.
Also, is it generally considered better to just use '==' by default, even when comparing int or Boolean values? I've always liked to use 'is' because I find it more aesthetically pleasing and pythonic (which is how I fell into this trap...), but I wonder if it's intended to just be reserved for when you care about finding two objects with the same id.
For all built-in Python objects (like
strings, lists, dicts, functions,
etc.), if x is y, then x==y is also
True.
Not always. NaN is a counterexample. But usually, identity (is) implies equality (==). The converse is not true: Two distinct objects can have the same value.
Also, is it generally considered better to just use '==' by default, even
when comparing int or Boolean values?
You use == when comparing values and is when comparing identities.
When comparing ints (or immutable types in general), you pretty much always want the former. There's an optimization that allows small integers to be compared with is, but don't rely on it.
For boolean values, you shouldn't be doing comparisons at all. Instead of:
if x == True:
# do something
write:
if x:
# do something
For comparing against None, is None is preferred over == None.
I've always liked to use 'is' because
I find it more aesthetically pleasing
and pythonic (which is how I fell into
this trap...), but I wonder if it's
intended to just be reserved for when
you care about finding two objects
with the same id.
Yes, that's exactly what it's for.
I would like to show a little example on how is and == are involved in immutable types. Try that:
a = 19998989890
b = 19998989889 +1
>>> a is b
False
>>> a == b
True
is compares two objects in memory, == compares their values. For example, you can see that small integers are cached by Python:
c = 1
b = 1
>>> b is c
True
You should use == when comparing values and is when comparing identities. (Also, from an English point of view, "equals" is different from "is".)
The logic is not flawed. The statement
if x is y then x==y is also True
should never be read to mean
if x==y then x is y
It is a logical error on the part of the reader to assume that the converse of a logic statement is true. See http://en.wikipedia.org/wiki/Converse_(logic)
See This question
Your logic in reading
For all built-in Python objects (like
strings, lists, dicts, functions,
etc.), if x is y, then x==y is also
True.
is slightly flawed.
If is applies then == will be True, but it does NOT apply in reverse. == may yield True while is yields False.