如何迭代多个函数参数？

Question

我试图使用局部变量()或*args来迭代多个函数参数。但是，我将函数参数定义为来自dataframe的列。如何编辑以下内容，使float_format函数对可变数量的参数进行迭代？

#! /usr/bin/env python3

import pandas as pd

def float_format(a, b, c, d, e, f): #Change to single *args function argument?
        for x in range(len(data[a])):
                data[a][x] = data[a][x].replace(' Mbps', '')
        for x in range(len(data[b])):
                data[b][x] = data[b][x].replace(' Mbps', '')
        for x in range(len(data[c])):
                data[c][x] = data[c][x].replace(' Mbps', '')                
        for x in range(len(data[d])):
                data[d][x] = data[d][x].replace(' Mbps', '')
        for x in range(len(data[e])):
                data[e][x] = data[e][x].replace(' Mbps', '')
        for x in range(len(data[f])):
                data[f][x] = data[f][x].replace(' Mbps', '')

file = r'Original_File.xls'
data = pd.read_excel(file, header=[2])

float_format('Average Receive bps',
               'Peak Receive bps',
               'Received Bandwidth',
               'Average Transmit bps',
               'Peak Transmit bps',
               'Transmit Bandwidth')

data.to_excel('results.xlsx', 'w+')

所以如果我尝试

def float_format(*iterate):
        for arg in iterate:
            for x in range(len(data[iterate])):
                    data[iterate][x] = data[iterate][x].replace(' Mbps', '')

在函数的运行方式上，我得到了跟踪错误。

示例df

>>> data
            Display Name Interface Name  ... Peak Transmit bps Transmit Bandwidth
0       1951 - LAB - FW1          port1  ...         0.56 Mbps         10.00 Mbps
1       1951 - LAB - FW1          port1  ...         0.37 Mbps         10.00 Mbps
2       1951 - LAB - FW1          port1  ...         0.34 Mbps         10.00 Mbps
3       1951 - LAB - FW1          port1  ...         0.36 Mbps         10.00 Mbps
4       1951 - LAB - FW1          port1  ...         0.83 Mbps         10.00 Mbps
5       1951 - LAB - FW1          port1  ...         0.55 Mbps         10.00 Mbps
6       1951 - LAB - FW1          port1  ...         0.27 Mbps         10.00 Mbps
7       1951 - LAB - FW1          port1  ...         0.41 Mbps         10.00 Mbps
8       1951 - LAB - FW1          port2  ...         0.00 Mbps       1000.00 Mbps
9       1951 - LAB - FW1          port2  ...         0.00 Mbps       1000.00 Mbps
10      1951 - LAB - FW1          port2  ...         0.00 Mbps       1000.00 Mbps
11      1951 - LAB - FW1          port2  ...         0.00 Mbps       1000.00 Mbps
12      1951 - LAB - FW1          port2  ...         0.00 Mbps       1000.00 Mbps
13      1951 - LAB - FW1          port2  ...         0.00 Mbps       1000.00 Mbps
14      1951 - LAB - FW1          port2  ...         0.19 Mbps       1000.00 Mbps
15      1951 - LAB - FW1          port2  ...         0.31 Mbps       1000.00 Mbps

Answer 1

这里没有必要使用*args或类似的东西，我们可以利用Pandas提供的操作。

设置代码

import numpy as np
import pandas as pd

df_1 = pd.DataFrame(data={'col_1': np.random.randint(0, 10, 10),
                          'col_2': np.random.randint(0, 50, 10),
                          'col_3': np.random.randint(0, 5, 10)})

df_1[['col_1', 'col_3']] = df_1[['col_1', 'col_3']].astype(str) + ' Mbps'

print(df_1)
print(df_1.dtypes)

输出：

    col_1  col_2   col_3
0  1 Mbps     45  0 Mbps
1  2 Mbps     34  1 Mbps
2  6 Mbps     46  2 Mbps
3  7 Mbps      2  1 Mbps
4  6 Mbps     36  0 Mbps
5  9 Mbps     36  3 Mbps
6  4 Mbps     39  1 Mbps
7  4 Mbps     26  1 Mbps
8  1 Mbps     10  1 Mbps
9  6 Mbps      1  1 Mbps
col_1    object
col_2     int64
col_3    object
dtype: object

方法1：Series.str.extract()

使用循环

cols_to_change = ['col_1', 'col_3']

for col_name in cols_to_change:
    df_1[col_name] = df_1[col_name].str.extract(r"(\d+) Mbps", expand=False).astype(int)

使用DataFrame.apply()的

cols_to_change = ['col_1', 'col_3']

df_1[cols_to_change] = df_1[cols_to_change].apply(lambda col: col.str.extract(r"(\d+) Mbps", expand=False)).astype(int)

方法2：Series.str.slice()

使用循环

cols_to_change = ['col_1', 'col_3']

for col_name in cols_to_change:
    df_1[col_name] = df_1[col_name].str.slice(stop=-5).astype(int)

使用DataFrame.apply()的

cols_to_change = ['col_1', 'col_3']

df_1[cols_to_change] = df_1[cols_to_change].apply(lambda col: col.str.slice(stop=-5)).astype(int)

结果

DataFrame内容：

   col_1  col_2  col_3
0      9     40      3
1      4      8      3
2      6     49      4
3      4     38      4
4      6     25      4
5      3      8      3
6      3     27      3
7      0     45      1
8      7     24      4
9      3     29      2

dtypes

col_1    int64
col_2    int64
col_3    int64
dtype: object

如果你有任何问题，请告诉我:)

Answer 2

下面这个怎么样？

def test(*argv):
    for arg in argv:
            print(arg)

test('hello', 'world')
# hello
# world