大约有 42,000 项符合查询结果（耗时：0.1311秒） [XML]
How does zip(*[iter(s)]*n) work in Python?
... arguments for a function call. Therefore you're passing the same iterator 3 times to zip(), and it pulls an item from the iterator each time.
x = iter([1,2,3,4,5,6,7,8,9])
print zip(x, x, x)
share
|
...
How can I obtain the element-wise logical NOT of a pandas Series?
... True, False, True])
In [8]: ~s
Out[8]:
0 False
1 False
2 True
3 False
dtype: bool
Using Python2.7, NumPy 1.8.0, Pandas 0.13.1:
In [119]: s = pd.Series([True, True, False, True]*10000)
In [10]: %timeit np.invert(s)
10000 loops, best of 3: 91.8 µs per loop
In [11]: %timeit ~s
10...
Generate all permutations of a list without adjacent equal elements
...
30
This is along the lines of Thijser's currently incomplete pseudocode. The idea is to take the m...
Which is faster in Python: x**.5 or math.sqrt(x)?
...
93
math.sqrt(x) is significantly faster than x**0.5.
import math
N = 1000000
%%timeit
for i in r...
How to filter Pandas dataframe using 'in' and 'not in' like in SQL
...andas as pd
>>> df
country
0 US
1 UK
2 Germany
3 China
>>> countries_to_keep
['UK', 'China']
>>> df.country.isin(countries_to_keep)
0 False
1 True
2 False
3 True
Name: country, dtype: bool
>>> df[df.country.isin(countries_to_ke...
How to update Python?
...will stop receiving official updates from python.org in 2020. Also, Python-3.7 has been released. Check out Python-Future on how to make your Python-2 code compatible with Python-3. For updating conda, the documentation now recommends using conda update --all in each of your conda environments to up...
Why is pow(a, d, n) so much faster than a**d % n?
...
|
edited Jan 3 '13 at 6:08
answered Jan 3 '13 at 6:03
...
Markdown: continue numbered list
In the following markdown code I want item 3 to start with list number 3. But because of the code block in between markdown starts this list item as a new list. Is there any way to prevent that behaviour?
...
How do you get the magnitude of a vector in Numpy?
...ray -- say x.norm() -- but oh well).
import numpy as np
x = np.array([1,2,3,4,5])
np.linalg.norm(x)
You can also feed in an optional ord for the nth order norm you want. Say you wanted the 1-norm:
np.linalg.norm(x,ord=1)
And so on.
...
jQuery select by attribute using AND and OR operators
...
AND operation
a=$('[myc="blue"][myid="1"][myid="3"]');
OR operation, use commas
a=$('[myc="blue"],[myid="1"],[myid="3"]');
As @Vega commented:
a=$('[myc="blue"][myid="1"],[myc="blue"][myid="3"]');
...