CODEWITHSUNDEEP
pythonpandasdataframe
Franck Dernoncourt
Franck Dernoncourt

Reputation: 83437

Counting changes of value in each column in a data frame in pandas

Is there any neat way to count the number of changes of value in each column in a data frame in pandas?

I don't want to have to loop myself over each column, e.g.:

import pandas as pd

frame = pd.DataFrame({
    'time':[1234567000,1234567005,1234567009],
    'X1':[96.32,96.01,96.05],
    'X2':[23.88,23.96,23.96]
},columns=['time','X1','X2']) 

print(frame)

changes = []
for column_name in frame.columns.values:
    print('column_name: {0}'.format(column_name))
    changes.append(sum(frame[column_name]!=frame[column_name].shift(1)))

print('changes: {0}'.format(changes))

returns:

         time     X1     X2
0  1234567000  96.32  23.88
1  1234567005  96.01  23.96
2  1234567009  96.05  23.96
column_name: time
column_name: X1
column_name: X2
changes: [3, 3, 2]

Upvotes: 3

Views: 7242

Answers (1)

unutbu
unutbu

Reputation: 880797

If the values are numeric you could take the differences between adjacent rows and test if the difference is non-zero. Then take a sum down each column to count the number of changes in value:

In [48]: (frame.diff(axis=0) != 0).sum(axis=0)
Out[48]: 
time    3
X1      3
X2      2
dtype: int64

If the values are not necessarily numeric, then a more general way would be to compare the frame against itself shift-ed down by one row -- this is similar to the code you posted, except the operation is done on the entire DataFrame instead of column-by-column:

In [50]: (frame != frame.shift(axis=0)).sum(axis=0)
Out[50]: 
time    3
X1      3
X2      2
dtype: int64

The numeric version is faster, the shifted version is more robust.

Upvotes: 12

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