根据行和列条件设置熊猫数据框值

Question

I have a fairly specific algorithm I want to follow.

Basically I have a dataframe as follows:

        month   taken   score
1       1       2       23
2       1       1       34
3       1       2       12
4       1       2       59
5       2       1       12
6       2       2       23
7       2       1       43
8       2       2       45
9       3       1       43
10      3       2       43
11      4       1       23
12      4       2       94

I want to make it so that the 'score' column is changed to 100 on days where taken == 2 continuously until the end of that month. So, not all occurrences of taken == 2 have their score set to 100, if any day following during that month has a taken == 1.

So the result I'd want is:

        month   taken   score
1       1       2       23
2       1       1       34
3       1       2       100
4       1       2       100
5       2       1       12
6       2       2       23
7       2       1       43
8       2       2       100
9       3       1       43
10      3       2       43
11      3       1       23
12      3       2       100
13      4       1       32
14      4       2       100

I've written this code which I feel should do it:

#iterate through months
for month in range(12):
    #iterate through scores
    for score in range(len(df_report.loc[df_report['month'] == month+1])):
        #starting from the bottom, of that month, if 'taken' == 2...
        if df_report.loc[df_report.month==month+1, 'taken'].iloc[-score-1] == 2:
            #then set the score to 100
            df_report.loc[df_report.month==month+1, 'score'].iloc[-score-2] = 100
        #if you run into a 'taken' == 1, move on to next month
        else: break

However, this doesn't appear to change any values, despite not throwing an error... it also doesn't give me an error about setting values to a copied dataframe.

Could anyone explain what I'm doing wrong?

Answer 1

The reason for your values not being updated is that assignment to iloc updates the copy returned by the preceding loc call, so the original is not touched.

---

Here's how I'd tackle this. First, define a function foo .

def foo(df):
    for i in reversed(df.index):
        if df.loc[i, 'taken'] != 2:
            break
        df.loc[i, 'score'] = 100
        i -= 1
    return df

Now, groupby month and call foo :

df = df.groupby('month').apply(foo)
print(df) 
    month  taken  score
1       1      2     23
2       1      1     34
3       1      2    100
4       1      2    100
5       2      1     12
6       2      2     23
7       2      1     43
8       2      2    100
9       3      1     43
10      3      2    100
11      4      1     23
12      4      2    100

Obviously, apply has its shortcomings, but I cannot think of a vectorised approach to this problem.

Answer 2

You can do

import numpy as np
def get_value(x):
    s = x['taken']
    # Get a mask of duplicate sequeence and change values using np.where
    mask = s.ne(s.shift()).cumsum().duplicated(keep=False)
    news = np.where(mask,100,x['score'])

    # if last number is 2 then change the news value to 100
    if s[s.idxmax()] == 2: news[-1] = 100 
    return pd.Series(news)

df['score'] = df.groupby('month').apply(get_value).values

Output :

month  taken  score
1       1      2     23
2       1      1     34
3       1      2    100
4       1      2    100
5       2      1     12
6       2      2     23
7       2      1     43
8       2      2    100
9       3      1     43
10      3      2    100
11      4      1     23
12      4      2    100

Almost identical speed but @coldspeed is winner

ndf = pd.concat([df]*10000).reset_index(drop=True)

%%timeit
ndf['score'] = ndf.groupby('month').apply(foo)
10 loops, best of 3: 40.8 ms per loop


%%timeit  
ndf['score'] = ndf.groupby('month').apply(get_value).values
10 loops, best of 3: 42.6 ms per loop