如何根据另一列中的多个条件创建新列

Question

In pandas, How can I create a new column B based on a column A in df , such that:

• B(i)=1 if A_(i-1)-A_(i) >= 0 when A_(i) <= 10
• B(i)=1 if A_(i-1)-A_(i) >= 2 when 10 < A_(i) <= 20
• B(i)=1 if A_(i-1)-A_(i) >= 5 when 20 < A_(i)
• B(i)=0 for any other case

However, the first B_i value is always two

Example:

A	B
5	2 (the first B_i)
12	0
14	0
22	0
20	1
33	0
11	1
8	1
15	0
11	1

Answer 1

You can use Pandas.shift for creating A_(i-1) and use Numpy.select for checking multiple conditions like below:

import pandas as pd
import numpy as np

df = pd.DataFrame({'A':[5,12,14,22,20,33,11,8,15,11]})
df['A_prv'] = df['A'].shift(1)

conditions = [
    (df.index==0),
    ((df['A_prv'] - df['A'] >= 0) & (df['A'].le(10))),
    ((df['A_prv'] - df['A'] >= 2) & (df['A'].between(10, 20, inclusive='right'))),
                                     # ^^^  10 < df['A'] <= 20 ^^^
    ((df['A_prv'] - df['A'] >= 5) & (df['A'].ge(20)))
]
choices = [2, 1, 1, 1]
df['B'] = np.select(conditions, choices, default=0)
print(df)

Output:

    A  A_prv  B
0   5    NaN  2
1  12    5.0  0
2  14   12.0  0
3  22   14.0  0
4  20   22.0  1
5  33   20.0  0
6  11   33.0  1
7   8   11.0  1
8  15    8.0  0
9  11   15.0  1

Answer 2

The most intuitive way is to iterate trough the lines testing all the three conditions in a single-line if-else (as B(i) is 1 for all the true conditions).

import pandas as pd

df = pd.DataFrame({'A':[5,12,14,22,20,33,11,8,15,11]})
B = [2]
for i in range(1,len(df['A'])):
    newvalue = 1 if (df['A'][i-1]-df['A'][i]>=0 and df['A'][i]<=10) or (df['A'][i-1]-df['A'][i]>=2 and df['A'][i]>10 and df['A'][i]<=20) or (df['A'][i-1]-df['A'][i]>=5 and df['A'][i]>20) else 0
    B.append(newvalue)
df['B'] = B
print(df)

Output:

    A   B
0   5   2
1   12  0
2   14  0
3   22  0
4   20  1
5   33  0
6   11  1
7   8   1
8   15  0
9   11  1