What is the best way to compare two instances of some object for equality in Python? I'd like to be able to do something like
doc1 = ErrorDocument(path='/folder',title='Page')
doc2 = ErrorDocument(path='/folder',title='Page')
if doc1 == ...
The following code is causing me enormous headaches:
def extract_by_letters(letters, dictionary):
for word in dictionary:
for letter in letters:
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 ...
The following snippet is annotated with the output (as seen on ideone.com):
print "100" < "2" # True
print "5" > "9" ...
i will be comparing two values like this:\
I'm using a function in a card game, to check the value of each card, and see if it is higher than the last card played.
In python 3, int(50)<'2' causes a TypeError, and well it should. In python 2.x, however, int(50)<'2' returns True (this is also the case for other number formats, but int exists ...
It's well known that comparing floats for equality is a little fiddly due to rounding and precision issues.
What is the recommended way of deal with this in Python?
I have this code:
I am trying to figure out if there is a way to get a logical list through comparison in Python 3.
Basically what I want to input is this
x = [1,2,3,4,5,6,7,8,9]
xlist = ...
How does Python compare string and int?
widths = [image.width ...
What's the most efficient way to compare two python values both of which are probably strings, but might be integers. So far I'm using str(x)==str(y) but that feels inefficient:
>>> a ...
number1 = 54378
number2 = 54379
if number1 (is similar to) number2:
print (number1 + " " + number2)
input("what to do")
I found a strange thing in Python when I tried to compare lists composed of integers.
In : id(range(1,5)),id(range(1,15)),id(range(16,0,-1))
Out: (155687404, 155687404, 155687404)