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i have a dataframe like thisenter image description here

i want to add new column and my desire format is like thisenter image description here

i use indexing eg dll.loc[:6,'category'] = "sectowise" dll.loc[7:11,'category'] = "productwise" dll.loc[12:16,'category'] = "collateral wise" but which is risky because data index can change anytime is there any method to do this?

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  • $\begingroup$ How about use a dictionary that maps items to categories and populate the new column based on the dictionary key values. $\endgroup$
    – dustin
    Oct 24, 2018 at 17:15
  • $\begingroup$ rows of dataset is large so how can i use dictionary key values ? can u give me example to solve this problem? $\endgroup$
    – subash poudel
    Oct 25, 2018 at 3:45

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Here is an example (I just through it together hastily). You can probably do this a little bit smoother.

import pandas as pd
import numpy as np
from random import shuffle

a = np.repeat(['a', 'b', 'c', 'd', 'e'], 6)
x = np.random.randn(30)
y = np.random.randn(30)
z = np.random.randn(30)

shuffle(a)

a = a.reshape(30, 1)
x = x.reshape(30, 1)
y = y.reshape(30, 1)
z = z.reshape(30, 1)

data = np.concatenate((a, x, y, z), axis=1)
df = pd.DataFrame(data, columns=['item', 'x', 'y', 'z'])

mapping = {'1': ['a', 'b'], '2': ['d'], '3': ['c', 'e']}

for k in mapping:
    df.loc[df[df['item'].mask(
        ~df.item.isin(mapping[k])).notnull()].index.tolist(), 'category'] = k
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  • $\begingroup$ As commented & illustrated by @dustin, using dictionaries can easily get this done. $\endgroup$
    – Random Nerd
    Nov 27, 2018 at 9:58

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