pandas 如何根据现有列熊猫的多个条件创建新列
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How to create new column based on multiple conditions from existing column pandas
提问by e9e9s
So I have a df column of 9 digit IDs. There are no duplicates and each ID starts with a different number that ranges from 1-6 -- depending on the number each ID starts with I want to create a separate column with the "name" that the first number of the ID represents. (e.g. IDs that start with 1 represent Maine, IDs that start with 2 represent California... and so on)
所以我有一个 df 列的 9 位 ID。没有重复，并且每个 ID 都以不同的数字开头，范围从 1 到 6 - 根据每个 ID 开头的数字我想创建一个单独的列，其中包含 ID 的第一个数字代表的“名称”。（例如，以 1 开头的 ID 代表缅因州，以 2 开头的 ID 代表加利福尼亚州......等等）
This works if it was only 2 conditions:
如果只有 2 个条件，则此方法有效：
df['id_label'] = ['name_1' if name.startswith('1') else 'everything_else' for name in df['col_1']]
I couldn't figure out how to create a multi line line comprehension for what I need so I thought this would work, but it only creates the id_labelcolumn from the last iteration of the loop (i.e. the id_labelcolumn will only contain 'name_5):
我无法弄清楚如何为我需要的内容创建多行理解，所以我认为这会起作用，但它只id_label从循环的最后一次迭代创建列（即该id_label列将只包含'name_5）：
for col in df['col_1']:
if col.startswith('1'):
df['id_label'] = 'name_1'
if col.startswith('2'):
df['id_label'] = 'name_2'
if col.startswith('3'):
df['id_label'] = 'name_3'
if col.startswith('4'):
df['id_label'] = 'name_4'
if col.startswith('5'):
df['id_label'] = 'name_5'
if col.startswith('6'):
df['id_label'] = 'name_5'
My question is how can I create a new column from an old column based on multiple conditional statements?
我的问题是如何根据多个条件语句从旧列创建新列？
回答by jezrael
I think you can convert column to strby astype, select first value and last mapby dict:
我认为您可以将列转换为strby astype，选择第一个值和最后一个mapby dict：
df = pd.DataFrame({'col_1':[133,255,36,477,55,63]})
print (df)
d = {'1':'Maine', '2': 'California', '3':'a', '4':'f', '5':'r', '6':'r'}
df['id_label'] = df['col_1'].astype(str).str[0].map(d)
print (df)
col_1 id_label
0 133 Maine
1 255 California
2 36 a
3 477 f
4 55 r
5 63 r
回答by Bharath
You can use applyin case you have lot of if elses
apply如果你有很多 if elses，你可以使用
def ifef(col):
col = str(col)
if col.startswith('1'):
return 'name_1'
if col.startswith('2'):
return 'name_2'
if col.startswith('3'):
return 'name_3'
if col.startswith('4'):
return'name_4'
if col.startswith('5'):
return 'name_5'
if col.startswith('6'):
return 'name_5'
df = pd.DataFrame({'col_1':[133,255,36,477,55,63]})
df['id_label'] = df['col_1'].apply(ifef)
col_1 id_label 0 133 name_1 1 255 name_2 2 36 name_3 3 477 name_4 4 55 name_5 5 63 name_5
In case if you have a dictionaary you can use
如果你有字典，你可以使用
df = pd.DataFrame({'col_1':[133,255,36,477,55,63]})
d = {'1':'M', '2': 'C', '3':'a', '4':'f', '5':'r', '6':'s'}
def ifef(col):
col = str(col)
return d[col[0]]
df['id_label'] = df['col_1'].apply(ifef)
print(df)
col_1 id_label 0 133 M 1 255 C 2 36 a 3 477 f 4 55 r 5 63 s
回答by Preetham
Can you check this and let me know whether its suitable for your question.
你能检查一下吗，让我知道它是否适合你的问题。
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.DataFrame({'col_1':[133,255,36,477,55,63]})
df['col_2'] = df['col_1'].astype(str).str[0]
condlist = [df['col_2'] == "1",
df['col_2'] == "2",
df['col_2'] == "3",
df['col_2'] == "4",
((df['col_2'] == "5") | (df['col_2'] == "6")),
]
choicelist = ['Maine','California','India', 'Frnace','5/6']
df['id_label'] = np.select(condlist, choicelist)
print(df)
#### Output ####
col_1 col_2 id_label
0 133 1 Maine
1 255 2 California
2 36 3 India
3 477 4 Frnace
4 55 5 5/6
5 63 6 5/6
PS: Thanks for @ALollzwho introduced me to np.select
PS：感谢@ALollz向我介绍 np.select
