这个答案使用了DataFrame。过滤器方法来做这个没有列表理解:

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)

只输出'spike-2'。你也可以使用regex，正如一些人在上面的评论中建议的那样:

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

将输出两个列:['spike-2'， 'hey spke']

根据开始、包含和结束获取名称和子设置:

# 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)

你也可以使用下面的代码:

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

这将输出列名:'spike-2'， 'spike- in'

关于pandas.Series.str.contains的更多信息。

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

这个答案使用了DataFrame。过滤器方法来做这个没有列表理解:

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)

只输出'spike-2'。你也可以使用regex，正如一些人在上面的评论中建议的那样:

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

将输出两个列:['spike-2'， 'hey spke']