CODEWITHSUNDEEP
pythonpandasfor-loopbinning
AdrianC
AdrianC

Reputation: 393

Binning Categorical Columns Programatically Using Python

I am trying bin categorical columns programtically - any idea on how I can achieve this without manually hard-coding each value in that column

Essentially, what I would like is a function whereby it counts all values up to 80% [leaves the city name as is] and replaces the remaining 20% of city names with the word 'Other'

IE: if the first 17 city names make up 80% of that column, keep the city name as is, else return 'other'.

EG: 

0        Brighton
1        Yokohama
2           Levin
3       Melbourne
4     Coffeyville
5       Whakatane
6       Melbourne
7       Melbourne
8           Levin
9         Ashburn
10     Te Awamutu
11        Bishkek
12      Melbourne
13      Whanganui
14    Coffeyville
15       New York
16       Brisbane
17      Greymouth
18       Brisbane
19      Chuo City
20          Accra
21          Levin
22        Waiouru
23       Brisbane
24       New York
25      Chuo City
26        Lucerne
27      Whanganui
28    Los Angeles
29      Melbourne

df['city'].head(30).value_counts(ascending=False, normalize=True)*100

Melbourne      16.666667
Levin          10.000000
Brisbane       10.000000
Whanganui       6.666667
Coffeyville     6.666667
New York        6.666667
Chuo City       6.666667
Waiouru         3.333333
Greymouth       3.333333
Te Awamutu      3.333333
Bishkek         3.333333
Lucerne         3.333333
Ashburn         3.333333
Yokohama        3.333333
Whakatane       3.333333
Accra           3.333333
Brighton        3.333333
Los Angeles     3.333333

From Ashburn down - it should be renamed to 'other'

I have tried the below which is a start, but not exactly what I want:

city_map = dict(df['city'].value_counts(ascending=False, normalize=True)*100)
df['city_count']= df['city'].map(city_map)

def count(df):
    if df["city_count"] > 10:
        return "High"
    elif df["city_count"] < 0:
        return "Medium"
    else:
        return "Low"

df.apply(count, axis=1)

I'm not expecting any code - just some guidance on where to start or ideas on how I can achieve this

Upvotes: 0

Views: 453

Answers (1)

Erfan
Erfan

Reputation: 42926

We can groupby on city and get the size of each city. We divide those values by the length of our dataframe with len and calculate the cumsum. Last step is to check from which point we exceed the threshold, so we can broadcast the boolean series back to your dataframe with map.

threshold = 0.7
m = df['city'].map(df.groupby('city')['city'].size().sort_values(ascending=False).div(len(df)).cumsum().le(threshold))

df['city'] = np.where(m, df['city'], 'Other')

          city
0         Other
1         Other
2         Levin
3     Melbourne
4   Coffeyville
5         Other
6     Melbourne
7     Melbourne
8         Levin
9       Ashburn
10        Other
11      Bishkek
12    Melbourne
13        Other
14  Coffeyville
15     New York
16     Brisbane
17        Other
18     Brisbane
19    Chuo City
20        Other
21        Levin
22        Other
23     Brisbane
24     New York
25    Chuo City
26        Other
27        Other
28        Other
29    Melbourne

old method

If I understand you correctly you want calculate a cumulative sum with .cumsum and check when it exceeds your set threshold.

Then we use np.where to conditionally fill in the City name or Other.

threshold = 80

m  = df['Normalized'].cumsum().le(threshold)

df['City'] = np.where(m, df['City'], 'Other')

            City  Normalized
0       Auckland   40.399513
1   Christchurch   13.130783
2     Wellington   12.267604
3       Hamilton    4.026242
4       Tauranga    3.867353
5      (not set)    3.540075
6        Dunedin    2.044508
7          Other    1.717975
8          Other    1.632849
9          Other    1.520342
10         Other    1.255651
11         Other    1.173878
12         Other    1.040508
13         Other    0.988166
14         Other    0.880502
15         Other    0.766877
16         Other    0.601468
17         Other    0.539067
18         Other    0.471824
19         Other    0.440903
20         Other    0.440344
21         Other    0.405884
22         Other    0.365836
23         Other    0.321131
24         Other    0.306602
25         Other    0.280524
26         Other    0.237123
27         Other    0.207878
28         Other    0.186084
29         Other    0.167085
30         Other    0.163732
31         Other    0.154977

Note: this method assumed that your Normalized column is sorted descending.

Upvotes: 3

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