Python 查找名称包含特定字符串的列

Find column whose name contains a specific string

提问by erikfas

I have a dataframe with column names, and I want to find the one that contains a certain string, but does not exactly match it. I'm searching for 'spike'in column names like 'spike-2', 'hey spike', 'spiked-in'(the 'spike'part is always continuous).

我有一个带有列名的数据框，我想找到包含某个字符串但不完全匹配的那个。我'spike'在像'spike-2', 'hey spike', 之类的列名中搜索'spiked-in'（该'spike'部分始终是连续的）。

I want the column name to be returned as a string or a variable, so I access the column later with df['name']or df[name]as normal. I've tried to find ways to do this, to no avail. Any tips?

我希望将列名作为字符串或变量返回，因此我稍后可以使用df['name']或df[name]正常访问该列。我试图找到方法来做到这一点，但无济于事。有小费吗？

采纳答案by Alvaro Fuentes

Just iterate over DataFrame.columns, now this is an example in which you will end up with a list of column names that match:

只需迭代DataFrame.columns，现在这是一个示例，您将在其中得到匹配的列名列表：

import pandas as pd

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)

spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)

Output:

输出：

['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']

Explanation:

解释：

1. df.columnsreturns a list of column names
2. [col for col in df.columns if 'spike' in col]iterates over the list df.columnswith the variable coland adds it to the resulting list if colcontains 'spike'. This syntax is list comprehension.

1. df.columns返回列名列表
2. [col for col in df.columns if 'spike' in col]df.columns使用变量遍历列表，col如果colcontains则将其添加到结果列表中'spike'。这种语法是列表理解。

If you only want the resulting data set with the columns that match you can do this:

如果您只想要结果数据集与匹配的列，您可以这样做：

df2 = df.filter(regex='spike')
print(df2)

Output:

输出：

   spike-2  spiked-in
0        1          7
1        2          8
2        3          9

回答by Ben

This answeruses the DataFrame.filter method to do this without list comprehension:

此答案使用 DataFrame.filter 方法在没有列表理解的情况下执行此操作：

import pandas as pd

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)

print(df.filter(like='spike').columns)

Will output just 'spike-2'. You can also use regex, as some people suggested in comments above:

将只输出“spike-2”。您也可以使用正则表达式，正如一些人在上面的评论中建议的那样：

print(df.filter(regex='spike|spke').columns)

Will output both columns: ['spike-2', 'hey spke']

将输出两列：['spike-2', 'hey spke']

回答by Michael James Kali Galarnyk

You can also use df.columns[df.columns.str.contains(pat = 'spike')]

你也可以使用 df.columns[df.columns.str.contains(pat = 'spike')]

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)

colNames = df.columns[df.columns.str.contains(pat = 'spike')] 

print(colNames)

This will output the column names: 'spike-2', 'spiked-in'

这将输出列名： 'spike-2', 'spiked-in'

More about pandas.Series.str.contains.

更多关于pandas.Series.str.contains 的信息。

回答by Yury Wallet

You also can use this code:

您也可以使用此代码：

spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]

回答by Manny

# select columns containing 'spike'
df.filter(like='spike', axis=1)

You can also select by name, regular expression. Refer to: pandas.DataFrame.filter

您还可以按名称、正则表达式进行选择。参考：pandas.DataFrame.filter

回答by DhanushNayak

df.loc[:,df.columns.str.contains("spike")]

回答by vasili111

Getting name and subsetting based on Start, Contains, and Ends:

根据开始、包含和结束获取名称和子集：

# from: https://stackoverflow.com/questions/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html




import pandas as pd



data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)



print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist() 
print("Contains")
print(colNames_contains)



print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist() 
print("Starts")
print(colNames_starts)



print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist() 
print("Ends")
print(colNames_ends)



print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)



print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)



print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)