Pandas "diff()" with string

Question

How can I flag a row in a dataframe every time a column change its string value?

Ex:

Input

ColumnA   ColumnB
1            Blue
2            Blue
3            Red
4            Red
5            Yellow


#  diff won't work here with strings....  only works in numerical values
dataframe['changed'] = dataframe['ColumnB'].diff()        


ColumnA   ColumnB      changed
1            Blue         0
2            Blue         0
3            Red          1
4            Red          0
5            Yellow       1

Asked By: guilhermecgs

||

Source

Answer 1

Use .shift and compare:

dataframe['changed'] = dataframe['ColumnB'] == dataframe['ColumnB'].shift(1).fillna(dataframe['ColumnB'])

Answered By: Kartik

Answer 2

For me works compare with shift, then NaN was replaced 0 because before no value:

df['diff'] = (df.ColumnB != df.ColumnB.shift()).astype(int)
df.ix[0,'diff'] = 0
print (df)
   ColumnA ColumnB  diff
0        1    Blue     0
1        2    Blue     0
2        3     Red     1
3        4     Red     0
4        5  Yellow     1

Edit by timings of another answer – fastest is use ne:

df['diff'] = (df.ColumnB.ne(df.ColumnB.shift())).astype(int)
df.ix[0,'diff'] = 0

Answered By: jezrael

Answer 3

I get better performance with ne instead of using the actual != comparison:

df['changed'] = df['ColumnB'].ne(df['ColumnB'].shift().bfill()).astype(int)

Timings

Using the following setup to produce a larger dataframe:

df = pd.concat([df]*10**5, ignore_index=True)

I get the following timings:

%timeit df['ColumnB'].ne(df['ColumnB'].shift().bfill()).astype(int)
10 loops, best of 3: 38.1 ms per loop

%timeit (df.ColumnB != df.ColumnB.shift()).astype(int)
10 loops, best of 3: 77.7 ms per loop

%timeit df['ColumnB'] == df['ColumnB'].shift(1).fillna(df['ColumnB'])
10 loops, best of 3: 99.6 ms per loop

%timeit (df.ColumnB.ne(df.ColumnB.shift())).astype(int)
10 loops, best of 3: 19.3 ms per loop

Answered By: root

Pandas "diff()" with string

Question:

Answers: