Question :
Apologies if this has been asked before, but I looked extensively without results.
import pandas as pd
import numpy as np
df = pd.DataFrame(data = np.random.randint(1,10,10),columns=['a'])
a
0 7
1 8
2 8
3 3
4 1
5 1
6 2
7 8
8 6
9 6
I’d like to create a new column b that maps several values of a according to some rule, say a=[1,2,3] is 1, a = [4,5,6,7] is 2, a = [8,9,10] is 3. one-to-one mapping is clear to me, but what if I want to map by a list of values or a range?
I tought along these lines…
df['b'] = df['a'].map({[1,2,3]:1,range(4,7):2,[8,9,10]:3})
Answer #1:
There are a few alternatives.
Pandas via pd.cut / NumPy via np.digitize
You can construct a list of boundaries, then use specialist library functions. This is described in @EdChum’s solution, and also in this answer.
NumPy via np.select
df = pd.DataFrame(data=np.random.randint(1,10,10), columns=['a'])
criteria = [df['a'].between(1, 3), df['a'].between(4, 7), df['a'].between(8, 10)]
values = [1, 2, 3]
df['b'] = np.select(criteria, values, 0)
The elements of criteria are Boolean series, so for lists of values, you can use df['a'].isin([1, 3]), etc.
Dictionary mapping via range
d = {range(1, 4): 1, range(4, 8): 2, range(8, 11): 3}
df['c'] = df['a'].apply(lambda x: next((v for k, v in d.items() if x in k), 0))
print(df)
a b c
0 1 1 1
1 7 2 2
2 5 2 2
3 1 1 1
4 3 1 1
5 5 2 2
6 4 2 2
7 4 2 2
8 9 3 3
9 3 1 1
Answer #2:
IIUC you could use cut to achieve this:
In[33]:
pd.cut(df['a'], bins=[0,3,7,11], right=True, labels=False)+1
Out[33]:
0 2
1 3
2 3
3 1
4 1
5 1
6 1
7 3
8 2
9 2
Here you’d pass the cutoff values to cut, and this will categorise your values, by passing labels=False it will give them an ordinal value (zero-based) so you just +1 to them
Here you can see how the cuts were calculated:
In[34]:
pd.cut(df['a'], bins=[0,3,7,11], right=True)
Out[34]:
0 (3, 7]
1 (7, 11]
2 (7, 11]
3 (0, 3]
4 (0, 3]
5 (0, 3]
6 (0, 3]
7 (7, 11]
8 (3, 7]
9 (3, 7]
Name: a, dtype: category
Categories (3, interval[int64]): [(0, 3] < (3, 7] < (7, 11]]
