Pandas replace dataframe value with other column values ​​in one row

Question

I have this pandas dataframe

BU       |   DATA1      DATA2
01-TT        zone 01   noData
02-FF        noData    zone 02
....

and I need to replace the string "noData" with the corresponding row in the BU column, but only using the first two characters and adding the word "zone"

BU       |   DATA1      DATA2
01-TT        zone 01    zone 01
02-FF        zone 02    zone 02
....

Thank you so much

Answer 1

You can use maskto replace the values Truewith the ones numpy arraycreated numpy.repeat:

df = df.set_index('BU')

arr = np.repeat('zone ' + df.index.str[:2], len(df.columns)).values.reshape(df.shape)
print (arr)
[['zone 01' 'zone 01']
 ['zone 02' 'zone 02']]

df = df.mask(df == 'noData', arr)
print (df.reset_index())
      BU    DATA1    DATA2
0  01-TT  zone 01  zone 01
1  02-FF  zone 02  zone 02

Delay

#[20000 rows x 3 columns]
df = pd.concat([df]*10000).reset_index(drop=True)
print (df)

df1 = df.copy()

def jez(df):
    df = df.set_index('BU')
    df = df.mask(df == 'noData', np.repeat('zone ' + df.index.str[:2], len(df.columns)).values.reshape(df.shape))
    return (df.reset_index())

def ed(df):
    cols = df.columns[df.columns.str.contains('DATA')]
    df[cols] = df[cols].mask(df[cols].apply(lambda x: x.str.contains('noData')), 'zone ' + df['BU'].str[:2], axis=0)
    return df


print (jez(df))
print (ed(df1))

In [219]: %timeit (jez(df))
100 loops, best of 3: 14.2 ms per loop

In [220]: %timeit (ed(df1))
10 loops, best of 3: 46.3 ms per loop

Answer 2

Common decision:

In [135]:
cols = df.columns[df.columns.str.contains('DATA')]
df[cols] = df[cols].mask(df[cols].apply(lambda x: x.str.contains('noData')), 'zone ' + df['BU'].str[:2], axis=0)
df

Out[135]:
      BU    DATA1    DATA2
0  01-TT  zone 01  zone 01
1  02-FF  zone 02  zone 02

cols, DATA, mask cols , ,