查找名称包含特定字符串的列

Question

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).

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?

Answer 1

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

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:

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

Explanation:

1. df.columns returns a list of column names
2. [col for col in df.columns if 'spike' in col] iterates over the list df.columns with the variable col and adds it to the resulting list if col contains 'spike' . This syntax is list comprehension .

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

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

Output:

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

Answer 2

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

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:

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

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

Answer 3

You can also use 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'

More about pandas.Series.str.contains .

Answer 4

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

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

Answer 5

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

Answer 6

您也可以使用此代码：

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

Answer 7

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

# 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)

Answer 8

返回具有所需列的 df 子集的另一种解决方案：

df[df.columns[df.columns.str.contains("spike|spke")]]