Python 按csv文件在我的驱动器上的显示顺序读取csv文件

我有一个名为

Edgelist_subgraphXXX.csv

的文件夹，其中

XXX

代表一个数字，从0到最后一个文件，例如：

Edgelist_subgraph0.csv
Edgelist_subgraph1.csv
Edgelist_subgraph124.csv
Edgelist_subgraph1156.csv
Edgelist_subgraph843.csv

我需要以正确的顺序读取这些文件，并将csv中的矩阵附加到列表中。我正在做：

path = r'Edgelist_subgraphs' # use your path
all_files = glob.glob(path + "/*.csv")
all_files.sort()

list_of_edgeList_matrices = []

for filename in all_files:
    df = pd.read_csv(filename, index_col=None, header=0)
    list_of_edgeList_matrices += [df]

但是我注意到文件的读取顺序错误。如果我打印

所有_文件的前几个元素

，我就会明白为什么：

['Edgelist_subgraphs/Edgelist_subgraph0.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1.csv',
 'Edgelist_subgraphs/Edgelist_subgraph10.csv',
 'Edgelist_subgraphs/Edgelist_subgraph100.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1000.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1001.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1002.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1003.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1004.csv',
 'Edgelist_subgraphs/Edgelist_subgraph1005.csv']

---

这类东西乱七八糟。有没有一种快速而肮脏的方法可以正确地对这些文件进行排序，无论是在python中，还是在bash中快速重命名它们，比如在结尾处使用

0001

而不是

1

？

您应该将

键

函数传递给

排序（）

，以便按数值排序，而不是按字母排序

将

所有文件.sort（）

更改为

所有文件.sort（key=lambda x:int（x[17:-4]）

17是

Edgelist_子图的len，-4是为了排除文件扩展名。
范例

输出

['Edgelist_subgraphs/Edgelist_subgraph2144.csv', 'Edgelist_subgraphs/Edgelist_subgraph2349.csv', 'Edgelist_subgraphs/Edgelist_subgraph3157.csv', 'Edgelist_subgraphs/Edgelist_subgraph3345.csv', 'Edgelist_subgraphs/Edgelist_subgraph3396.csv', 'Edgelist_subgraphs/Edgelist_subgraph3938.csv', 'Edgelist_subgraphs/Edgelist_subgraph3957.csv', 'Edgelist_subgraphs/Edgelist_subgraph5739.csv', 'Edgelist_subgraphs/Edgelist_subgraph6307.csv', 'Edgelist_subgraphs/Edgelist_subgraph6475.csv']

或者您可以使用
os.path

from os.path import basename, splitext
print(basename('Edgelist_subgraphs/Edgelist_subgraph6307.csv'))
spam = ['Edgelist_subgraphs/Edgelist_subgraph6307.csv', 'Edgelist_subgraphs/Edgelist_subgraph2144.csv',
        'Edgelist_subgraphs/Edgelist_subgraph3396.csv', 'Edgelist_subgraphs/Edgelist_subgraph6475.csv',
        'Edgelist_subgraphs/Edgelist_subgraph3157.csv', 'Edgelist_subgraphs/Edgelist_subgraph3345.csv', 
        'Edgelist_subgraphs/Edgelist_subgraph5739.csv', 'Edgelist_subgraphs/Edgelist_subgraph3957.csv', 
        'Edgelist_subgraphs/Edgelist_subgraph3938.csv', 'Edgelist_subgraphs/Edgelist_subgraph2349.csv'] 

spam.sort(key=lambda x:int(basename(x)[17:-4]))
print(spam)

出于某种原因，您的代码给了我ValueError:int（）的无效文本以10为基数：“s/Edgelist_subgraph6307”这是因为
s/
-您还可以获得路径，而不仅仅是文件名。您需要排除路径。我的示例仅使用文件名。如果路径是常量，如果我正确计算路径len，您可以将其更改为17到36。哦，它可以使用：all_files.sort（key=lambda x:int（x[36:-4]），以及路径。谢谢。建议：首先将所有文件名提取到一个列表中，然后将其放入数据框中，使用regex添加额外的列，并对其排序。然后您可以按该顺序提取数据吗？
from os.path import basename, splitext
print(basename('Edgelist_subgraphs/Edgelist_subgraph6307.csv'))
spam = ['Edgelist_subgraphs/Edgelist_subgraph6307.csv', 'Edgelist_subgraphs/Edgelist_subgraph2144.csv',
        'Edgelist_subgraphs/Edgelist_subgraph3396.csv', 'Edgelist_subgraphs/Edgelist_subgraph6475.csv',
        'Edgelist_subgraphs/Edgelist_subgraph3157.csv', 'Edgelist_subgraphs/Edgelist_subgraph3345.csv', 
        'Edgelist_subgraphs/Edgelist_subgraph5739.csv', 'Edgelist_subgraphs/Edgelist_subgraph3957.csv', 
        'Edgelist_subgraphs/Edgelist_subgraph3938.csv', 'Edgelist_subgraphs/Edgelist_subgraph2349.csv'] 

spam.sort(key=lambda x:int(basename(x)[17:-4]))
print(spam)