3

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.

Patrick Yoder
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Pyd
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2 Answers2

4

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)
jezrael
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2

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
jpp
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