Find column whose name contains a specific string

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?

Asked By: erikfas

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Source

Answers:

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
Answered By: Alvaro Fuentes

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’]

Answered By: Ben

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.

Answered By: Michael James Kali Galarnyk

You also can use this code:

spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]
Answered By: Yury Wallet
# select columns containing 'spike'
df.filter(like='spike', axis=1)

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

Answered By: Manny
df.loc[:,df.columns.str.contains("spike")]
Answered By: DhanushNayak

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)
Answered By: vasili111

Another solution that returns a subset of the df with the desired columns:

df[df.columns[df.columns.str.contains("spike|spke")]]
Answered By: rambutan