按csv文件在我的驱动器上出现的顺序读取它们

Question

我有一个文件夹，有将近7000个名为Edgelist_subgraphXXX.csv的csv文件，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]

然而，我注意到文件是按错误的顺序读取的。如果我打印all_files的前几个元素，我就知道为什么：

['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']

这是一种完全的混乱。是否有一种快速而肮脏的方法来正确排序这些文件，无论是在python中，还是在bash中快速重命名它们，类似于0001而不是1？

Answer 1

您应该将key函数传递给sort()，以便按数值进行排序，而不是按字母顺序排序。

将all_files.sort()更改为all_files.sort(key=lambda x:int(x[17:-4]) 17是Edgelist_subgraph的len，-4是为了排除文件扩展名。示例

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(x[36:-4]))
print(spam)

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中的一些函数

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)