Python 如何通过正则表达式从数据框中选择列
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/30808430/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to select columns from dataframe by regex
提问by Yan Song
I have a dataframe in python pandas. The structure of the dataframe is as the following:
我在 python pandas 中有一个数据框。数据帧的结构如下：
a b c d1 d2 d3
10 14 12 44 45 78
I would like to select the columns which begin with d. Is there a simple way to achieve this in python .
我想选择以 d 开头的列。有没有一种简单的方法可以在 python 中实现这一点。
采纳答案by farhawa
You can use DataFrame.filterthis way:
你可以这样使用DataFrame.filter：
import pandas as pd
df = pd.DataFrame(np.array([[2,4,4],[4,3,3],[5,9,1]]),columns=['d','t','didi'])
>>
d t didi
0 2 4 4
1 4 3 3
2 5 9 1
df.filter(regex=("d.*"))
>>
d didi
0 2 4
1 4 3
2 5 1
The idea is to select columns by regex
这个想法是通过选择列 regex
回答by Alexander
You can use a list comprehension to iterate over all of the column names in your DataFrame dfand then only select those that begin with 'd'.
您可以使用列表理解来遍历 DataFrame 中的所有列名称df，然后仅选择以 'd' 开头的列名称。
df = pd.DataFrame({'a': {0: 10}, 'b': {0: 14}, 'c': {0: 12},
'd1': {0: 44}, 'd2': {0: 45}, 'd3': {0: 78}})
Use list comprehension to iterate over the columns in the dataframe and return their names (cbelow is a local variable representing the column name).
使用列表理解来迭代数据框中的列并返回它们的名称（c下面是表示列名称的局部变量）。
>>> [c for c in df]
['a', 'b', 'c', 'd1', 'd2', 'd3']
Then only select those beginning with 'd'.
然后只选择那些以“d”开头的。
>>> [c for c in df if c[0] == 'd'] # As an alternative to c[0], use c.startswith(...)
['d1', 'd2', 'd3']
Finally, pass this list of columns to the DataFrame.
最后，将此列列表传递给 DataFrame。
df[[c for c in df if c.startswith('d')]]
>>> df
d1 d2 d3
0 44 45 78
===========================================================================
================================================== ==========================
TIMINGS(added Feb 2018 per comments from devinbost claiming that this method is slow...)
时间（根据 devinbost 的评论于 2018 年 2 月添加，声称此方法很慢......）
First, lets create a dataframe with 30k columns:
首先，让我们创建一个包含 30k 列的数据框：
n = 10000
cols = ['{0}_{1}'.format(letters, number)
for number in range(n) for letters in ('d', 't', 'didi')]
df = pd.DataFrame(np.random.randn(3, n * 3), columns=cols)
>>> df.shape
(3, 30000)
>>> %timeit df[[c for c in df if c[0] == 'd']] # Simple list comprehension.
# 10 loops, best of 3: 16.4 ms per loop
>>> %timeit df[[c for c in df if c.startswith('d')]] # More 'pythonic'?
# 10 loops, best of 3: 29.2 ms per loop
>>> %timeit df.select(lambda col: col.startswith('d'), axis=1) # Solution of gbrener.
# 10 loops, best of 3: 21.4 ms per loop
>>> %timeit df.filter(regex=("d.*")) # Accepted solution.
# 10 loops, best of 3: 40 ms per loop
回答by gbrener
Use select:
使用select：
import pandas as pd
df = pd.DataFrame([[10, 14, 12, 44, 45, 78]], columns=['a', 'b', 'c', 'd1', 'd2', 'd3'])
df.select(lambda col: col.startswith('d'), axis=1)
Result:
结果：
d1 d2 d3
0 44 45 78
This is a nice solution if you're not comfortable with regular expressions.
如果您对正则表达式不满意，这是一个不错的解决方案。
回答by prafi
You can also use
你也可以使用
df.filter(regex='^d')
回答by devinbost
On a larger dataset especially, a vectorized approach is actually MUCH FASTER (by more than two orders of magnitude) and is MUCH more readable. I'm providing a screenshot as proof. (Note: Except for the last few lines I wrote at the bottom to make my point clear with a vectorized approach, the other code was derived from the answer by @Alexander.)
特别是在更大的数据集上，矢量化方法实际上更快（超过两个数量级）并且更具可读性。我提供截图作为证据。（注意：除了我在底部写的最后几行用矢量化方法明确我的观点外，其他代码来自@Alexander 的答案。）
Here's that code for reference:
这是供参考的代码：
import pandas as pd
import numpy as np
n = 10000
cols = ['{0}_{1}'.format(letters, number)
for number in range(n) for letters in ('d', 't', 'didi')]
df = pd.DataFrame(np.random.randn(30000, n * 3), columns=cols)
%timeit df[[c for c in df if c[0] == 'd']]
%timeit df[[c for c in df if c.startswith('d')]]
%timeit df.select(lambda col: col.startswith('d'), axis=1)
%timeit df.filter(regex=("d.*"))
%timeit df.filter(like='d')
%timeit df.filter(like='d', axis=1)
%timeit df.filter(regex=("d.*"), axis=1)
%timeit df.columns.map(lambda x: x.startswith("d"))
columnVals = df.columns.map(lambda x: x.startswith("d"))
%timeit df.filter(columnVals, axis=1)
