用熊猫中另一列的每个值替换一个列的值

Question

I have a csv file shown below:

ID,Number,Value
61745,three,11
61745,one,11
61745,one & two,12
61745,two,13
61743,one,41
61743,two,42
61741,one,21
61741,one & two,22
61715,one,31
61715,two,32
61715,three,33

What I am trying to achieve:

For each ID , if the Number column contains "one & two", I want all the Number column fields that contain "two" or "one" to be replaced with the "one & two" value. For example, for the "61745" ID I can see the "one & two" value at least once. But for the "61743" ID I cannot see this value. So, I want to return the following:

ID,Number,Value
61745,three,11
61745,one & two,11
61745,one & two,12
61745,one & two,13
61743,one,41
61743,two,42
61741,one & two,21
61741,one & two,22
61715,one,31
61715,two,32
61715,three,33

So far, I have tried this:

import pandas as pd
import os
import csv
import time
import dateutil.parser as dparser
import datetime

df = pd.read_csv("slack.csv")

for row in df.itertuples():
    if row[2] == "one & two":
        df.ix[df.Number.isin(['one & two','one','two']), 'Number'] = 'one & two'

and the result is that the script replaces all the "two" and "one" values in the Number column for every ID :

       ID     Number  Value
0   61745      three        11
1   61745  one & two        11
2   61745  one & two        12
3   61745  one & two        13
4   61743  one & two        41
5   61743  one & two        42
6   61741  one & two        21
7   61741  one & two        22
8   61715  one & two        31
9   61715  one & two        32
10  61715  pinterest        33

Answer 1

Use custom function with groupby with check if at least one value is one & two and then replace by dict :

def f(x):
    d = {'one':'one & two', 'two':'one & two'}
    if x.eq('one & two').any():
        return x.replace(d)
    else:
        return x

df['Number'] = df.groupby('ID')['Number'].apply(f)
print (df)
       ID     Number  Value
0   61745      three     11
1   61745  one & two     11
2   61745  one & two     12
3   61745  one & two     13
4   61743        one     41
5   61743        two     42
6   61741  one & two     21
7   61741  one & two     22
8   61715        one     31
9   61715        two     32
10  61715      three     33

Answer 2

Replace this line:

df.ix[df.Number.isin(['one & two','one','two']), 'Number'] = 'one & two'

with the following:

ids = df.ID[df.Number == 'one & two'].unique()
df.loc[df.ID.isin(ids) & df.Number.isin(['one', 'two']), 'Number'] = 'one & two'