如何根据Python中的列表“另一列”值将列值的某些值设为零
S.No Num_1 Num_2 Num_3 Num_4 Num_5 id Final 0 2 4 2 1 1 1 1.1 1 2.5 4.6 2.7 1.2 1.1 1 1.2 2 2 4 2 1 1 1 1.3 3 2.5 4.6 2.7 1.2 1.1 2 1.4 4 2 4 2 1 1 2 1.5 5 2.5 4.6 2.7 1.2 1.1 2 1.6 6 2 4 2 1 1 3 1.7 7 2.5 4.6 2.7 1.2 1.1 4 1.8 8 2 4 2 1 1 4 1.9 9 2.5 4.6 2.7 1.2 1.1 5 2 10 2 4 2 1 1 5 2.1 11 2.5 4.6 2.7 1.2 1.1 6 2.2 12 2 4 2 1 1 6 2.3 13 2.5 4.6 2.7 1.2 1.1 6 2.4 14 2 4 2 1 1 6 2.5 15 2.5 4.6 2.7 1.2 1.1 7 2.6 16 2 4 2 1 1 7 2.7 17 2.5 4.6 2.7 1.2 1.1 7 2.8 18 2 4 2 1 1 7 2.9 19 2.5 4.6 2.7 1.2 1.1 8 3 20 2 4 2 1 1 8 3.1 21 2.5 4.6 2.7 1.2 1.1 8 3.2 22 2 4 2 1 1 10 3.3 23 2.5 4.6 2.7 1.2 1.1 10 3.4
和 id = [2,4,5,6,8]
id = [2,4,5,6,8]
无论我的id在哪里,最后的值都应该是0(零)
我怎么用蟒蛇或熊猫做这个?
试试这个:
id_ = [2,4,5,6,8] df['Final'].loc[df['id'].isin(id_)] = 0
你可以 mask 您的系列:
mask
ids = [2, 4, 5, 6, 8] df['Final'] = df['Final'].mask(df['id'].isin(ids), 0)
使用 np.where
np.where
import numpy as np id_ = [2,4,5,6,8] # as id is a keyword in Python df['Final'] = np.where(df['id'].isin(id_), 0, df['Final'])
