sunny
sunny

Reputation: 489

pandas replace multiple values

Below is sample dataframe

>>> df = pd.DataFrame({'a': [1, 1, 1, 2, 2],  'b':[11, 22, 33, 44, 55]})
>>> df
      a   b
   0  1  11
   1  1  22
   2  1  33
   3  2  44
   4  3  55

Now I wanted to update/replace b values that are matched on a column from other dict based on index

ex:

match = {1:[111, 222], 2:[444, 555]}

output:

      a   b
   0  1  111
   1  1  222
   2  1  33  <-- ignores this bcz not enough values to replace in match dict for 1 
   3  2  444
   4  3  555

Thanks in advance

Upvotes: 4

Views: 1502

Answers (2)

Dani Mesejo
Dani Mesejo

Reputation: 61910

You could use the pop function of list:

import pandas as pd


def pop(default, lst):
    try:
        return lst.pop()
    except IndexError:
        return default


df = pd.DataFrame({'a': [1, 1, 1, 2, 2], 'b': [11, 22, 33, 44, 55]})

match = {1: [111, 222], 2: [444, 555]}

df['b'] = df[['a', 'b']].apply(lambda e: pop(e[1], match[e[0]]), axis=1)

print(df)

Output

   a    b
0  1  222
1  1  111
2  1   33
3  2  555
4  2  444

if the order must be preserved, you can always pop the first item:

def pop(default, lst):
    try:
        return lst.pop(0)
    except IndexError:
        return default

Output

   a    b
0  1  111
1  1  222
2  1   33
3  2  444
4  2  555

UPDATE

A faster (non-destructive) way is to use deque:

def pop(default, lst):
    try:
        return lst.popleft()
    except IndexError:
        return default

match_deque = {k: deque(v[:]) for k, v in match.items()}

df['b'] = df[['a', 'b']].apply(lambda e: pop(e[1], match_deque[e[0]]), axis=1)

print(df)

Upvotes: 4

jpp
jpp

Reputation: 164623

Here's one way. The idea is to calculate a cumulative count by group and use this to filter rows. Use itertools.chain to create a single array of values. Finally, use pd.DataFrame.loc and Boolean indexing to set values.

from itertools import chain

count = df.groupby('a').cumcount() + 1

m1 = df['a'].isin(match)
m2 = count.le(df['a'].map(match).map(len))
values = list(chain.from_iterable(match.values()))

df.loc[m1 & m2, 'b'] = values

print(df)

   a    b
0  1  111
1  1  222
2  1   33
3  2  444
4  2  555

Upvotes: 4

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