Drop columns whose name contains a specific string from pandas DataFrame

This can be done neatly in one line with:

df = df.drop(df.filter(regex='Test').columns, axis=1)

Cheaper, Faster, and Idiomatic: str.contains

In recent versions of pandas, you can use string methods on the index and columns. Here, str.startswith seems like a good fit.

To remove all columns starting with a given substring:

df.columns.str.startswith('Test')
# array([ True, False, False, False])

df.loc[:,~df.columns.str.startswith('Test')]

  toto test2 riri
0    x     x    x
1    x     x    x

---

For case-insensitive matching, you can use regex-based matching with str.contains with an SOL anchor:

df.columns.str.contains('^test', case=False)
# array([ True, False,  True, False])

df.loc[:,~df.columns.str.contains('^test', case=False)] 

  toto riri
0    x    x
1    x    x

if mixed-types is a possibility, specify na=False as well.

Here is one way to do this:

df = df[df.columns.drop(list(df.filter(regex='Test')))]

import pandas as pd

import numpy as np

array=np.random.random((2,4))

df=pd.DataFrame(array, columns=('Test1', 'toto', 'test2', 'riri'))

print df

      Test1      toto     test2      riri
0  0.923249  0.572528  0.845464  0.144891
1  0.020438  0.332540  0.144455  0.741412

cols = [c for c in df.columns if c.lower()[:4] != 'test']

df=df[cols]

print df
       toto      riri
0  0.572528  0.144891
1  0.332540  0.741412