Reputation: 6159
I have a df,
A B
one six
two seven
three level
five one
and a dictionary
my_dict={1:"one,two",2:"three,four"}
I want to replace df.A with my_dict keys().
My desired output is,
A B
1 six
1 seven
2 level
five one
I tried df.A.replace(my_dict,regex=True)
but it doesn't work.
Upvotes: 4
Views: 9729
Reputation: 164663
Here is one solution via map
/ fillna
:
d = {v_i: k for k, v in my_dict.items() for v_i in v.split(',')}
df['A'] = df['A'].map(d).fillna(df['A'])
# A B
# 0 1 six
# 1 1 seven
# 2 2 level
# 3 five one
Upvotes: 2
Reputation: 862611
You need dict comprehension for separate each values to keys first:
my_dict={1:"one,two",2:"three,four"}
d = {k: oldk for oldk, oldv in my_dict.items() for k in oldv.split(',')}
print (d)
{'one': 1, 'three': 2, 'four': 2, 'two': 1}
df.A = df.A.replace(my_dict)
Upvotes: 5