python - 如何通过传递参数来重新排列数据框中的列

Question

我有以下列表

('EmpId', 'EMpName', 'Age', 'SerialNo')

我想将 Age,SerialNo 放在起始位置，其余列如下

ColumnsToRearrange = ['年龄', '序列号']

通过对数据帧内的列进行硬编码，我可以实现如下所示。我是否有可能将其作为参数传递

df = df[['Age', 'SerialNo'] + [col for col in df.columns if col not in ['Age', 'SerialNo']]] 

但我想将其作为参数传递

df = df[Header + [col for col in df.columns if col not in Header]] 

我收到以下错误消息

Error:TypeError: can only concatenate tuple (not "list") to tuple

Answer 1

下面是对 DataFrame 的列进行重新排序的四种方法

import pandas as pd
df =pd.DataFrame({'EmpId':[1,2,3,4],'EMpName':[2,4,6,8],'Age':[3,7,6,8],'SerialNo':[2,4,8,2]})
"""
    EmpId  EMpName  Age  SerialNo
        0      1        2    3         2
        1      2        4    7         4
        2      3        6    6         8
        3      4        8    8         2 
"""

 Original- ('EmpId', 'EMpName', 'Age', 'SerialNo')
 Output -  ('Age', 'SerialNo','EmpId', 'EMpName')

两列交换

cols = list(df.columns)
a, b = cols.index('EmpId'), cols.index('Age')
cols[b], cols[a] = cols[a], cols[b]
df = df[cols]

重新排序列交换(2 次交换)

cols = list(df.columns)
a, b, c, d = cols.index('EmpId'), cols.index('Age'), cols.index('EMpName'), cols.index('SerialNo')
cols[a], cols[b], cols[c], cols[d] = cols[b], cols[a], cols[d], cols[c]
df = df[cols]

交换多个现在归结为如何使用列表切片 -

cols = list(df.columns)
colss = cols[2:] + cols[:2]
df = df[colss]
seq_list=['Age', 'SerialNo']

创建函数

def set_column_sequence(dataframe, seq, front=True):
    '''Takes a dataframe and a subsequence of its columns,
       returns dataframe with seq as first columns if "front" is True,
       and seq as last columns if "front" is False.
    '''
    cols = seq[:] # copy so we don't mutate seq
    for x in dataframe.columns:
        if x not in cols:
            if front: #we want "seq" to be in the front
                #so append current column to the end of the list
                cols.append(x)
            else:
                #we want "seq" to be last, so insert this
                #column in the front of the new column list
                #"cols" we are building:
                cols.insert(0, x)
    return dataframe[cols]
q=set_column_sequence(df,seq_list)