在 pandas 列中，将重复数字替换为除第一个之外的 NAN 值

Question

I have a data frame like this,

df
col1    col2
  1       A
  2       A
  3       B
  4       C
  5       C
  6       C
  7       B
  8       B
  9       A

Now we can see that there is continuous occurrence of A, B and C. I want only the rows where the occurrence is starting. And the other values of the same occurrence will be nan.

The final data frame I am looking for will look like,

df
col1    col2
  1       A
  2       NA
  3       B
  4       C
  5       NA
  6       NA
  7       B
  8       NA
  9       A

I can do it using for loop and comparing, But the execution time will be more. I am looking for pythonic way to do it. Some panda shortcuts may be.

Answer 1

Compare by Series.shift ed values and missing values by Series.where or numpy.where :

df['col2'] = df['col2'].where(df['col2'].ne(df['col2'].shift()))
#alternative
#df['col2'] = np.where(df['col2'].ne(df['col2'].shift()), df['col2'], np.nan)

Or by DataFrame.loc with inverted condition by ~ :

df.loc[~df['col2'].ne(df['col2'].shift()), 'col2'] = np.nan

Or thanks @Daniel Mesejo - use eq for == :

df.loc[df['col2'].eq(df['col2'].shift()), 'col2'] = np.nan

---

print (df)
   col1 col2
0     1    A
1     2  NaN
2     3    B
3     4    C
4     5  NaN
5     6  NaN
6     7    B
7     8  NaN
8     9    A

Detail :

print (df['col2'].ne(df['col2'].shift()))
0     True
1    False
2     True
3     True
4    False
5    False
6     True
7    False
8     True
Name: col2, dtype: bool