pandas 如何根据部分匹配选择DataFrame列？
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How to select DataFrame columns based on partial matching?
提问by Michele Ancis
I was struggling this afternoon to find a way of selecting few columns of my Pandas DataFrame, by checking the occurrence of a certain pattern in their name (label?).
今天下午我正在努力寻找一种方法来选择我的 Pandas DataFrame 的几列，方法是检查它们名称（标签？）中某个模式的出现。
I had been looking for something like containsor isinfor nd.arrays/ pd.series, but got no luck.
我一直在寻找类似contains或isin为nd.arrays/ 的东西pd.series，但没有运气。
This frustrated me quite a bit, as I was already checking the columns of my DataFramefor occurrences of specific string patterns, as in:
这让我很沮丧，因为我已经在检查我的列中DataFrame是否出现了特定的字符串模式，如下所示：
hp = ~(df.target_column.str.contains('some_text') | df.target_column.str.contains('other_text'))
df_cln= df[hp]
However, no matter how I banged my head, I could not apply .str.contains()to the object returned bydf.columns- which is an Index- nor the one returned by df.columns.values- which is an ndarray. This works fine for what is returned by the "slicing" operation df[column_name], i.e. a Series, though.
但是，无论我怎么敲我的头，我都无法应用.str.contains()到由df.columns-返回的对象- 这是一个Index- 也不能应用到-返回的对象df.columns.values- 是一个ndarray. 不过，这对于“切片”操作返回的内容（df[column_name]即 a Series）很有效。
My first solution involved a forloop and the creation of a help list:
我的第一个解决方案涉及一个for循环和帮助列表的创建：
ll = []
for a in df.columns:
if a.startswith('start_exp1') | a.startswith('start_exp2'):
ll.append(a)
df[ll]
(one could apply any of the strfunctions, of course)
（str当然，可以应用任何功能）
Then, I found the mapfunction and got it to work with the following code:
然后，我找到了该map函数并使其与以下代码一起工作：
import re
sel = df.columns.map(lambda x: bool(re.search('your_regex',x))
df[df.columns[sel]]
Of course in the first solution I could have performed the same kind of regex checking, because I can apply it to the strdata type returned by the iteration.
当然，在第一个解决方案中，我可以执行相同类型的正则表达式检查，因为我可以将它应用于str迭代返回的数据类型。
I am very new to Python and never really programmed anything so I am not too familiar with speed/timing/efficiency, but I tend to think that the second method - using a map - could potentially be faster, besides looking more elegant to my untrained eye.
我对 Python 很陌生，从来没有真正编程过任何东西，所以我对速度/时间/效率不太熟悉，但我倾向于认为第二种方法 - 使用地图 - 可能会更快，除了对我未经训练的人来说看起来更优雅眼睛。
I am curious to know what you think of it, and what possible alternatives would be. Given my level of noobness, I would really appreciate if you could correct any mistakes I could have made in the code and point me in the right direction.
我很想知道您对此有何看法，以及可能的替代方案是什么。鉴于我的菜鸟水平，如果您能纠正我在代码中可能犯的任何错误并为我指明正确的方向，我将不胜感激。
Thanks, Michele
谢谢，米歇尔
EDIT: I just found the Indexmethod Index.to_series(), which returns - ehm - a Seriesto which I could apply .str.contains('whatever').
However, this is not quite as powerful as a true regex, and I could not find a way of passing the result of Index.to_series().strto the re.search()function..
编辑：我刚刚发现的Index方法Index.to_series()，它的回报- EHM -一Series到，我可以申请.str.contains('whatever')。但是，这并不像真正的正则表达式那么强大，而且我找不到将 的结果传递Index.to_series().str给re.search()函数的方法。
回答by Robert Smith
Your solution using mapis very good. If you really want to use str.contains, it is possible to convert Index objects to Series (which have the str.containsmethod):
您使用的解决方案map非常好。如果您真的想使用 str.contains，则可以将 Index 对象转换为 Series（具有该str.contains方法）：
In [1]: df
Out[1]:
x y z
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
In [2]: df.columns.to_series().str.contains('x')
Out[2]:
x True
y False
z False
dtype: bool
In [3]: df[df.columns[df.columns.to_series().str.contains('x')]]
Out[3]:
x
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
UPDATEI just read your last paragraph. From the documentation, str.containsallows you to pass a regex by default (str.contains('^myregex'))
更新我刚读了你的最后一段。从文档中，str.contains允许您默认传递正则表达式 ( str.contains('^myregex'))
回答by Philipp Schwarz
Select column by partial string, can simply be done, via:
按部分字符串选择列，可以简单地完成，通过：
df.filter(like='hello') # select columns which contain the word hello
And to select rows by partial string match, you can pass axis=0 to filter:
并且要通过部分字符串匹配来选择行，您可以通过 axis=0 来过滤：
df.filter(like='hello', axis=0)
回答by Geeocode
I think df.keys().tolist()is the thing you're searching for.
我认为df.keys().tolist()是你正在寻找的东西。
A tiny example:
from pandas import DataFrame as df
d = df({'somename': [1,2,3], 'othername': [4,5,6]})
names = d.keys().tolist()
for n in names:
print n
print type(n)
Output:
输出：
othername
type 'str'
somename
type 'str'
Then with the strings you got, you can do any string operation you want.
然后用你得到的字符串，你可以做任何你想做的字符串操作。
