Fill in same amount of characters where other column is NaN

Question

I have the following dummy dataframe:

df = pd.DataFrame({'Col1':['a,b,c,d', 'e,f,g,h', 'i,j,k,l,m'],
                   'Col2':['aa~bb~cc~dd', np.NaN, 'ii~jj~kk~ll~mm']})

        Col1            Col2
0    a,b,c,d     aa~bb~cc~dd
1    e,f,g,h             NaN
2  i,j,k,l,m  ii~jj~kk~ll~mm

The real dataset has shape 500000, 90.

I need to unnest these values to rows and I’m using the new explode method for this, which works fine.

The problem is the NaN, these will cause unequal lengths after the explode, so I need to fill in the same amount of delimiters as the filled values. In this case ~~~ since row 1 has three comma’s.

expected output

        Col1            Col2
0    a,b,c,d     aa~bb~cc~dd
1    e,f,g,h             ~~~
2  i,j,k,l,m  ii~jj~kk~ll~mm

Attempt 1:

df['Col2'].fillna(df['Col1'].str.count(',')*'~')

Attempt 2:

np.where(df['Col2'].isna(), df['Col1'].str.count(',')*'~', df['Col2'])

This works, but I feel like there’s an easier method for this:

characters = df['Col1'].str.replace('w', '').str.replace(',', '~')
df['Col2'] = df['Col2'].fillna(characters)

print(df)

        Col1            Col2
0    a,b,c,d     aa~bb~cc~dd
1    e,f,g,h             ~~~
2  i,j,k,l,m  ii~jj~kk~ll~mm

d1 = df.assign(Col1=df['Col1'].str.split(',')).explode('Col1')[['Col1']]
d2 = df.assign(Col2=df['Col2'].str.split('~')).explode('Col2')[['Col2']]

final = pd.concat([d1,d2], axis=1)
print(final)

  Col1 Col2
0    a   aa
0    b   bb
0    c   cc
0    d   dd
1    e     
1    f     
1    g     
1    h     
2    i   ii
2    j   jj
2    k   kk
2    l   ll
2    m   mm

Question: is there an easier and more generalized method for this? Or is my method fine as is.

Asked By: Erfan

||

Source

Answer 1

One way is using str.repeat and fillna() not sure how efficient this is though:

df.Col2.fillna(pd.Series(['~']*len(df)).str.repeat(df.Col1.str.count(',')))

0       aa~bb~cc~dd
1               ~~~
2    ii~jj~kk~ll~mm
Name: Col2, dtype: object

Answered By: anky

Answer 2

zip_longest can be useful here, given you don’t need the original Index. It will work regardless of which column has more splits:

from itertools import zip_longest, chain

df = pd.DataFrame({'Col1':['a,b,c,d', 'e,f,g,h', 'i,j,k,l,m', 'x,y'],
                   'Col2':['aa~bb~cc~dd', np.NaN, 'ii~jj~kk~ll~mm', 'xx~yy~zz']})
#        Col1            Col2
#0    a,b,c,d     aa~bb~cc~dd
#1    e,f,g,h             NaN
#2  i,j,k,l,m  ii~jj~kk~ll~mm
#3        x,y        xx~yy~zz

l = [zip_longest(*x, fillvalue='') 
     for x in zip(df.Col1.str.split(',').fillna(''), 
                  df.Col2.str.split('~').fillna(''))]

pd.DataFrame(chain.from_iterable(l))

    0   1
0   a  aa
1   b  bb
2   c  cc
3   d  dd
4   e    
5   f    
6   g    
7   h    
8   i  ii
9   j  jj
10  k  kk
11  l  ll
12  m  mm
13  x  xx
14  y  yy
15     zz

Answered By: ALollz

Answer 3

Just split the dataframe into two

df1=df.dropna()
df2=df.drop(df1.index)

d1 = df1['Col1'].str.split(',').explode()
d2 = df1['Col2'].str.split('~').explode()
d3 = df2['Col1'].str.split(',').explode()

final = pd.concat([d1, d2], axis=1).append(d3.to_frame(),sort=False)
Out[77]: 
  Col1 Col2
0    a   aa
0    b   bb
0    c   cc
0    d   dd
2    i   ii
2    j   jj
2    k   kk
2    l   ll
2    m   mm
1    e  NaN
1    f  NaN
1    g  NaN
1    h  NaN

Answered By: BENY

Answer 4

`pd.concat`

delims = {'Col1': ',', 'Col2': '~'}
pd.concat({
    k: df[k].str.split(delims[k], expand=True)
    for k in df}, axis=1
).stack()

    Col1 Col2
0 0    a   aa
  1    b   bb
  2    c   cc
  3    d   dd
1 0    e  NaN
  1    f  NaN
  2    g  NaN
  3    h  NaN
2 0    i   ii
  1    j   jj
  2    k   kk
  3    l   ll
  4    m   mm

This loops on columns in df. It may be wiser to loop on keys in the delims dictionary.

delims = {'Col1': ',', 'Col2': '~'}
pd.concat({
    k: df[k].str.split(delims[k], expand=True)
    for k in delims}, axis=1
).stack()

Same thing, different look

delims = {'Col1': ',', 'Col2': '~'}
def f(c): return df[c].str.split(delims[c], expand=True)
pd.concat(map(f, delims), keys=delims, axis=1).stack()

Answered By: piRSquared

Fill in same amount of characters where other column is NaN

Question:

Answers:

`pd.concat`

Same thing, different look