Pandas new column from indexing list by row value

Question

I am looking to create a new column in a Pandas data frame with the value of a list filtered by the df row value.

df = pd.DataFrame({'Index': [0,1,3,2], 'OtherColumn': ['a', 'b', 'c', 'd']})

   Index OtherColumn
      0           a
      1           b
      3           c
      2           d

l = [1000, 1001, 1002, 1003]

Desired output:

  Index OtherColumn  Value
      0           a   -
      1           b   -
      3           c   1003
      2           d   - 

My code:

df.loc[df.OtherColumn == 'c', 'Value'] = l[df.Index]

Which returns an error since 'df.Index' is not recognised as a int but as a list (not filter by OtherColumn == 'c').

For R users, I'm looking for:

df[OtherColumn == 'c', Value := l[Index]]

Thanks.

Answer 1

Convert list to numpy array for indexing and then filter by mask in both sides:

m = df.OtherColumn == 'c'
df.loc[m, 'Value'] = np.array(l)[df.Index][m]
print (df)
   Index OtherColumn   Value
0      0           a     NaN
1      1           b     NaN
2      3           c  1003.0
3      2           d     NaN

Or use numpy.where:

m = df.OtherColumn == 'c'
df['Value'] = np.where(m, np.array(l)[df.Index], '-')
print (df)
   Index OtherColumn Value
0      0           a     -
1      1           b     -
2      3           c  1003
3      2           d     -

Or:

df['value'] = np.where(m, df['Index'].map(dict(enumerate(l))), '-')

Answer 2

Use Series.where + Series.map:

df['value']=df['Index'].map(dict(enumerate(l))).where(df['OtherColumn']=='c','-')
print(df)

   Index OtherColumn value
0      0           a     -
1      1           b     -
2      3           c  1003
3      2           d     -