How to select all columns that start with "durations" or "shape"?

Question

How to select all columns that have header names starting with "durations" or "shape"? (instead of defining a long list of column names). I need to select these columns and substitute blank fields by 0.

column_names = ['durations.blockMinutes_x',
                'durations.scheduledBlockMinutes_y']
data[column_names] = data[column_names].fillna(0)

Answer 1

You could use str methods of dataframe startwith:

df = data[data.columns[data.columns.str.startwith('durations') | data.columns.str.startwith('so')]]
df.fillna(0)

Or you could use contains method:

df = data.iloc[:, data.columns.str.contains('durations.*'|'shape.*') ]
df.fillna(0)

Answer 2

Use my_dataframe.columns.values.tolist() to get the column names (based on Get list from pandas DataFrame column headers):

column_names = [x for x in data.columns.values.tolist() if x.startswith("durations") or x.startswith("shape")]

Answer 3

I would use the select method:

df.select(lambda c: c.startwith('durations') or c.startswith('shape'), axis=1)

Answer 4

A simple and easy way

data[data.filter(regex='durations|shape').columns].fillna(0)

Sample Screenshot