Reputation: 1161
I have some data in dictionary like and a pandas dataframe like:
s_dict = {('A1','B1'):100, ('A3','B3'):300}
df = pd.DataFrame(data={'A': ['A1', 'A2'], 'B': ['B1', 'B2'],
'C': ['C1', 'C2'], 'count':[1,2]})
# A B C count
#0 A1 B1 C1 1
#1 A2 B2 C2 2
I want to replace count column of "df" if data exist in s_dict. So I want following output:
# A B C count
#0 A1 B1 C1 100
#1 A2 B2 C2 2
Upvotes: 2
Views: 69
Reputation: 18906
Here is one way using zip()
which is generally faster than .apply()
.
import pandas as pd
s_dict = {('A1','B1'):100, ('A3','B3'):300}
df = pd.DataFrame(data={'A': ['A1', 'A2'], 'B': ['B1', 'B2'],
'C': ['C1', 'C2'], 'count':[1,2]})
# Create a map
m = pd.Series(list(zip(df['A'],df['B']))).map(s_dict).dropna()
# Assign to the index that are not nan
df.loc[m.index, 'count'] = m
Inspired by filling na with the column values you could do: (seems to be the quickest)
df['count'] = pd.Series(list(zip(df['A'],df['B']))).map(s_dict).fillna(df['count'])
Timings
df['count'] = pd.Series(list(zip(df['A'],df['B']))).map(s_dict).fillna(df['count'])
# 1.52 ms ± 85.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df['count'] = df[['A', 'B']].apply(tuple, axis=1).map(s_dict).fillna(df['count'])
# 1.88 ms ± 100 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
dropna and loc (2 row-operation above)
# 1.93 ms ± 55.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Upvotes: 1
Reputation: 76297
You can use:
df['count'] = df[['A', 'B']].apply(tuple, axis=1).map(s_dict).fillna(df['count'])
apply(tuple, axis=1)
creates a tuple of the relevant columns' values.map(s_dict)
maps the tuples to the values in s_dict
.fillna(df['count'])
fills missing values with those of count
.Upvotes: 3