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Fast way of counting non-zero bits in positive integer
...'d use a lookup table. You can generate it at runtime:
counts = bytes(bin(x).count("1") for x in range(256)) # py2: use bytearray
Or just define it literally:
counts = (b'\x00\x01\x01\x02\x01\x02\x02\x03\x01\x02\x02\x03\x02\x03\x03\x04'
b'\x01\x02\x02\x03\x02\x03\x03\x04\x02\x03\x03\x...
What is a plain English explanation of “Big O” notation?
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Most popular screen sizes/resolutions on Android phones [closed]
...othing of how common a resolution is... just a table describing which one exists/categories?
– Ted
Mar 18 '14 at 15:03
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u'\ufeff' in Python string
...ecode the web page using the right codec, Python will remove it for you. Examples:
#!python2
#coding: utf8
u = u'ABC'
e8 = u.encode('utf-8') # encode without BOM
e8s = u.encode('utf-8-sig') # encode with BOM
e16 = u.encode('utf-16') # encode with BOM
e16le = u.encode('utf-16le') # e...
Python base64 data decode
...g piece of base64 encoded data, and I want to use python base64 module to extract information from it. It seems that module does not work. Can anyone tell me how?
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Fill remaining vertical space with CSS using display:flex
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Make it simple : DEMO
section {
display: flex;
flex-flow: column;
height: 300px;
}
header {
background: tomato;
/* no flex rules, it will grow */
}
div {
flex: 1; /* 1 and it will fill whole space left if no flex value are set to other children*...
How to access the last value in a vector?
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The nice thing with tail is that it works on dataframes too, unlike the x[length(x)] idiom.
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Looping in a spiral
...n need of an algorithm that would let him loop through the elements of an NxM matrix (N and M are odd). I came up with a solution, but I wanted to see if my fellow SO'ers could come up with a better solution.
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List comprehension: Returning two (or more) items for each item
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>>> from itertools import chain
>>> f = lambda x: x + 2
>>> g = lambda x: x ** 2
>>> list(chain.from_iterable((f(x), g(x)) for x in range(3)))
[2, 0, 3, 1, 4, 4]
Timings:
from timeit import timeit
f = lambda x: x + 2
g = lambda x: x ** 2
def fg(x):
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~x + ~y == ~(x + y) is always false?
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Assume for the sake of contradiction that there exists some x and some y (mod 2n) such that
~(x+y) == ~x + ~y
By two's complement*, we know that,
-x == ~x + 1
<==> -1 == ~x + x
Noting this result, we have,
~(x+y) == ~x + ~y
<==> ~(x+y) + (x+...