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在python中按数字对字符串列表进行排序

[英]Sort list of Strings by numbers in python

I do want to read some .csv files and print the minimum and maximum value of the 5th column. 我确实想读取一些.csv文件并打印第5列的最小值和最大值。 I also want to print in which file I can find the max or min. 我还想打印可以在其中找到最大值或最小值的文件。

My code is right now: 我的代码现在是:

import pandas, glob
import numpy as np

path = "/home/path/to/log/"

fn = glob.glob(path + "*.csv") 
list_of_dfs = [pandas.read_csv(filename, header=None) for filename in fn]
k = len(list_of_dfs)
b = np.zeros((k, 1))
cnt = 0 
for i in list_of_dfs:
    b[cnt,0] = np.min(i[4])
    cnt = cnt + 1
print(np.min(b[:,0]))
print(np.argmin(b[:,0]))

My .csv Files are named like: 我的.csv文件的命名如下:

0.csv, 1.csv ... 10.csv ...2463.csv 0.csv,1.csv ... 10.csv ... 2463.csv

After seeing that my argmin doesn't show the correct file I realised that fn isn't sorted. 看到我的argmin没有显示正确的文件后,我意识到fn没有排序。 I found this Sort filenames in directory in ascending order , where the solution was: 在目录中按升序找到了排序文件名的方法,解决方法是:

fn.sort(key=lambda f: int(filter(str.isdigit, f)))

But with this line I get the following error: 但是在这行代码中,出现以下错误:

TypeError: int() argument must be a string, a bytes-like object or a number, not 'filter'

Any Suggestions? 有什么建议么?

I think need sorted with convert file names to integer s: 我认为需要使用将文件名转换为integer s sorted

fn = ['0.csv', '1.csv', '2463.csv', '10.csv']

fn = sorted(fn, key=lambda f: int(f.split('.')[0]))
print (fn)
['0.csv', '1.csv', '10.csv', '2463.csv']

If there are full paths: 如果有完整路径:

print (fn)
['files\\1.csv', 'files\\10.csv', 'files\\2.csv']

fn = sorted(fn, key=lambda f: int(os.path.basename(f).split('.')[0]))
print (fn)
['files\\1.csv', 'files\\2.csv', 'files\\10.csv']

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