How to select columns from dataframe by regex

You can use DataFrame.filter this way:

import pandas as pddf = pd.DataFrame(np.array([[2,4,4],[4,3,3],[5,9,1]]),columns=['d','t','didi'])>>   d  t  didi0  2  4     41  4  3     32  5  9     1df.filter(regex=("d.*"))>>   d  didi0  2     41  4     32  5     1

The idea is to select columns by regex

Use select:

import pandas as pddf = pd.DataFrame([[10, 14, 12, 44, 45, 78]], columns=['a', 'b', 'c', 'd1', 'd2', 'd3'])df.select(lambda col: col.startswith('d'), axis=1)

Result:

   d1  d2  d30  44  45  78

This is a nice solution if you're not comfortable with regular expressions.

On a larger dataset especially, a vectorized approach is actually MUCH FASTER (by more than two orders of magnitude) and is MUCH more readable.I'm providing a screenshot as proof. (Note: Except for the last few lines I wrote at the bottom to make my point clear with a vectorized approach, the other code was derived from the answer by @Alexander.)

Here's that code for reference:

import pandas as pdimport numpy as npn = 10000cols = ['{0}_{1}'.format(letters, number)         for number in range(n) for letters in ('d', 't', 'didi')]df = pd.DataFrame(np.random.randn(30000, n * 3), columns=cols)%timeit df[[c for c in df if c[0] == 'd']]%timeit df[[c for c in df if c.startswith('d')]]%timeit df.select(lambda col: col.startswith('d'), axis=1)%timeit df.filter(regex=("d.*"))%timeit df.filter(like='d')%timeit df.filter(like='d', axis=1)%timeit df.filter(regex=("d.*"), axis=1)%timeit df.columns.map(lambda x: x.startswith("d"))columnVals = df.columns.map(lambda x: x.startswith("d"))%timeit df.filter(columnVals, axis=1)