大约有 47,000 项符合查询结果(耗时:0.0459秒) [XML]
What is the most efficient way to create a dictionary of two pandas Dataframe columns?
...ies(df.Letter.values,index=df.Position).to_dict()
Out[9]: {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e'}
Speed comparion (using Wouter's method)
In [6]: df = pd.DataFrame(randint(0,10,10000).reshape(5000,2),columns=list('AB'))
In [7]: %timeit dict(zip(df.A,df.B))
1000 loops, best of 3: 1.27 ms per loo...
What's the difference between eval, exec, and compile?
...
3 Answers
3
Active
...
Is there a foreach in MATLAB? If so, how does it behave if the underlying data changes?
...in other languages. This means that the following code always prints 1, 2, 3, 4, 5 regardless of the value of B.
A = 1:5;
for i = A
A = B;
disp(i);
end
If you want to be able to respond to changes in the data structure during iterations, a WHILE loop may be more appropriate --- you'll be...
How is a CRC32 checksum calculated?
Maybe I'm just not seeing it, but CRC32 seems either needlessly complicated, or insufficiently explained anywhere I could find on the web.
...
How to find the sum of an array of numbers
Given an array [1, 2, 3, 4] , how can I find the sum of its elements? (In this case, the sum would be 10 .)
43 Answers
...
List comprehension: Returning two (or more) items for each item
...: 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):
yield f(x)
yield g(x)
print timeit(stmt='list(chain.from_iterable((f(x), g(x)) for x in range(3)))',...
What are the correct version numbers for C#?
...numbers for C#? What came out when? Why can't I find any answers about C# 3.5 ?
12 Answers
...
How do I detect what .NET Framework versions and service packs are installed?
A similar question was asked here , but it was specific to .NET 3.5. Specifically, I'm looking for the following:
13 Answe...
Add Variables to Tuple
...
388
Tuples are immutable; you can't change which variables they contain after construction. Howeve...