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PHP DirectoryIterator isWritable () Function: StackOverflow Questions

os.path.dirname(__file__) returns empty

I want to get the path of the current directory under which a .py file is executed.

For example a simple file D: est.py with code:

import os

print os.getcwd()
print os.path.basename(__file__)
print os.path.abspath(__file__)
print os.path.dirname(__file__)

It is weird that the output is:

D:
test.py
D:	est.py
EMPTY

I am expecting the same results from the getcwd() and path.dirname().

Given os.path.abspath = os.path.dirname + os.path.basename, why

os.path.dirname(__file__)

returns empty?

Answer #1:

Because os.path.abspath = os.path.dirname + os.path.basename does not hold. we rather have

os.path.dirname(filename) + os.path.basename(filename) == filename

Both dirname() and basename() only split the passed filename into components without taking into account the current directory. If you want to also consider the current directory, you have to do so explicitly.

To get the dirname of the absolute path, use

os.path.dirname(os.path.abspath(__file__))

What is the difference between os.path.basename() and os.path.dirname()?

What is the difference between os.path.basename() and os.path.dirname()?

I already searched for answers and read some links, but didn"t understand. Can anyone give a simple explanation?

Answer #1:

Both functions use the os.path.split(path) function to split the pathname path into a pair; (head, tail).

The os.path.dirname(path) function returns the head of the path.

E.g.: The dirname of "/foo/bar/item" is "/foo/bar".

The os.path.basename(path) function returns the tail of the path.

E.g.: The basename of "/foo/bar/item" returns "item"

From: http://docs.python.org/3/library/os.path.html#os.path.basename

PHP DirectoryIterator isWritable () Function: StackOverflow Questions

Python"s equivalent of && (logical-and) in an if-statement

Question by delete

Here"s my code:

def front_back(a, b):
  # +++your code here+++
  if len(a) % 2 == 0 && len(b) % 2 == 0:
    return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] 
  else:
    #todo! Not yet done. :P
  return

I"m getting an error in the IF conditional.
What am I doing wrong?

Answer #1:

You would want and instead of &&.

Answer #2:

Python uses and and or conditionals.

i.e.

if foo == "abc" and bar == "bac" or zoo == "123":
  # do something

Answer #3:

I"m getting an error in the IF conditional. What am I doing wrong?

There reason that you get a SyntaxError is that there is no && operator in Python. Likewise || and ! are not valid Python operators.

Some of the operators you may know from other languages have a different name in Python. The logical operators && and || are actually called and and or. Likewise the logical negation operator ! is called not.

So you could just write:

if len(a) % 2 == 0 and len(b) % 2 == 0:

or even:

if not (len(a) % 2 or len(b) % 2):

Some additional information (that might come in handy):

I summarized the operator "equivalents" in this table:

+------------------------------+---------------------+
|  Operator (other languages)  |  Operator (Python)  |
+==============================+=====================+
|              &&              |         and         |
+------------------------------+---------------------+
|              ||              |         or          |
+------------------------------+---------------------+
|              !               |         not         |
+------------------------------+---------------------+

See also Python documentation: 6.11. Boolean operations.

Besides the logical operators Python also has bitwise/binary operators:

+--------------------+--------------------+
|  Logical operator  |  Bitwise operator  |
+====================+====================+
|        and         |         &          |
+--------------------+--------------------+
|         or         |         |          |
+--------------------+--------------------+

There is no bitwise negation in Python (just the bitwise inverse operator ~ - but that is not equivalent to not).

See also 6.6. Unary arithmetic and bitwise/binary operations and 6.7. Binary arithmetic operations.

The logical operators (like in many other languages) have the advantage that these are short-circuited. That means if the first operand already defines the result, then the second operator isn"t evaluated at all.

To show this I use a function that simply takes a value, prints it and returns it again. This is handy to see what is actually evaluated because of the print statements:

>>> def print_and_return(value):
...     print(value)
...     return value

>>> res = print_and_return(False) and print_and_return(True)
False

As you can see only one print statement is executed, so Python really didn"t even look at the right operand.

This is not the case for the binary operators. Those always evaluate both operands:

>>> res = print_and_return(False) & print_and_return(True);
False
True

But if the first operand isn"t enough then, of course, the second operator is evaluated:

>>> res = print_and_return(True) and print_and_return(False);
True
False

To summarize this here is another Table:

+-----------------+-------------------------+
|   Expression    |  Right side evaluated?  |
+=================+=========================+
| `True` and ...  |           Yes           |
+-----------------+-------------------------+
| `False` and ... |           No            |
+-----------------+-------------------------+
|  `True` or ...  |           No            |
+-----------------+-------------------------+
| `False` or ...  |           Yes           |
+-----------------+-------------------------+

The True and False represent what bool(left-hand-side) returns, they don"t have to be True or False, they just need to return True or False when bool is called on them (1).

So in Pseudo-Code(!) the and and or functions work like these:

def and(expr1, expr2):
    left = evaluate(expr1)
    if bool(left):
        return evaluate(expr2)
    else:
        return left

def or(expr1, expr2):
    left = evaluate(expr1)
    if bool(left):
        return left
    else:
        return evaluate(expr2)

Note that this is pseudo-code not Python code. In Python you cannot create functions called and or or because these are keywords. Also you should never use "evaluate" or if bool(...).

Customizing the behavior of your own classes

This implicit bool call can be used to customize how your classes behave with and, or and not.

To show how this can be customized I use this class which again prints something to track what is happening:

class Test(object):
    def __init__(self, value):
        self.value = value

    def __bool__(self):
        print("__bool__ called on {!r}".format(self))
        return bool(self.value)

    __nonzero__ = __bool__  # Python 2 compatibility

    def __repr__(self):
        return "{self.__class__.__name__}({self.value})".format(self=self)

So let"s see what happens with that class in combination with these operators:

>>> if Test(True) and Test(False):
...     pass
__bool__ called on Test(True)
__bool__ called on Test(False)

>>> if Test(False) or Test(False):
...     pass
__bool__ called on Test(False)
__bool__ called on Test(False)

>>> if not Test(True):
...     pass
__bool__ called on Test(True)

If you don"t have a __bool__ method then Python also checks if the object has a __len__ method and if it returns a value greater than zero. That might be useful to know in case you create a sequence container.

See also 4.1. Truth Value Testing.

NumPy arrays and subclasses

Probably a bit beyond the scope of the original question but in case you"re dealing with NumPy arrays or subclasses (like Pandas Series or DataFrames) then the implicit bool call will raise the dreaded ValueError:

>>> import numpy as np
>>> arr = np.array([1,2,3])
>>> bool(arr)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>> arr and arr
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

>>> import pandas as pd
>>> s = pd.Series([1,2,3])
>>> bool(s)
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
>>> s and s
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

In these cases you can use the logical and function from NumPy which performs an element-wise and (or or):

>>> np.logical_and(np.array([False,False,True,True]), np.array([True, False, True, False]))
array([False, False,  True, False])
>>> np.logical_or(np.array([False,False,True,True]), np.array([True, False, True, False]))
array([ True, False,  True,  True])

If you"re dealing just with boolean arrays you could also use the binary operators with NumPy, these do perform element-wise (but also binary) comparisons:

>>> np.array([False,False,True,True]) & np.array([True, False, True, False])
array([False, False,  True, False])
>>> np.array([False,False,True,True]) | np.array([True, False, True, False])
array([ True, False,  True,  True])

(1)

That the bool call on the operands has to return True or False isn"t completely correct. It"s just the first operand that needs to return a boolean in it"s __bool__ method:

class Test(object):
    def __init__(self, value):
        self.value = value

    def __bool__(self):
        return self.value

    __nonzero__ = __bool__  # Python 2 compatibility

    def __repr__(self):
        return "{self.__class__.__name__}({self.value})".format(self=self)

>>> x = Test(10) and Test(10)
TypeError: __bool__ should return bool, returned int
>>> x1 = Test(True) and Test(10)
>>> x2 = Test(False) and Test(10)

That"s because and actually returns the first operand if the first operand evaluates to False and if it evaluates to True then it returns the second operand:

>>> x1
Test(10)
>>> x2
Test(False)

Similarly for or but just the other way around:

>>> Test(True) or Test(10)
Test(True)
>>> Test(False) or Test(10)
Test(10)

However if you use them in an if statement the if will also implicitly call bool on the result. So these finer points may not be relevant for you.

How do you get the logical xor of two variables in Python?

Question by Zach Hirsch

How do you get the logical xor of two variables in Python?

For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):

str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
    print "ok"
else:
    print "bad"

The ^ operator seems to be bitwise, and not defined on all objects:

>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: "str" and "str"

Answer #1:

If you"re already normalizing the inputs to booleans, then != is xor.

bool(a) != bool(b)

Answer #2:

You can always use the definition of xor to compute it from other logical operations:

(a and not b) or (not a and b)

But this is a little too verbose for me, and isn"t particularly clear at first glance. Another way to do it is:

bool(a) ^ bool(b)

The xor operator on two booleans is logical xor (unlike on ints, where it"s bitwise). Which makes sense, since bool is just a subclass of int, but is implemented to only have the values 0 and 1. And logical xor is equivalent to bitwise xor when the domain is restricted to 0 and 1.

So the logical_xor function would be implemented like:

def logical_xor(str1, str2):
    return bool(str1) ^ bool(str2)

Credit to Nick Coghlan on the Python-3000 mailing list.