python按顺序读取多个文件
我的文件顺序如下: H2_000.csv, H2_001.csv, H2_002.csv, H2_003.csv, H2_004.csv, H2_005.csvpython按顺序读取多个文件,python,file-io,Python,File Io,我的文件顺序如下: H2_000.csv, H2_001.csv, H2_002.csv, H2_003.csv, H2_004.csv, H2_005.csv import glob path = 'path/H2_*.csv' files=glob.glob(path) for file in files: f=open(file, 'r') print f output open file 'path/H2_003.csv', mode 'r' at 0x7f3ce9e
import glob
path = 'path/H2_*.csv'
files=glob.glob(path)
for file in files:
f=open(file, 'r')
print f
output
open file 'path/H2_003.csv', mode 'r' at 0x7f3ce9eca150,
open file 'path/H2_000.csv', mode 'r' at 0x7f3ce9eca1e0,
open file 'path/H2_004.csv', mode 'r' at 0x7f3ce9eca150,
open file 'path/H2_001.csv', mode 'r' at 0x7f3ce9eca1e0,
但是这个文件是随机读取的,
我希望文件按顺序打开。
有人能帮我吗。谢谢 您所需要做的就是对文件列表进行排序(并且始终将
与
一起使用)
请出示您的代码。请将您的代码粘贴到您的问题中。无论如何,对于排序中的元素(您的\u列表),解决方案肯定是
。现在您粘贴了,我确认:对于已排序的文件(文件)
应该做这项工作。您如何获得此列表..H2_000.csv,H2_001.csv,H2_002.csv,H2_003.csv,H2_004.csv,H2_005.csv?当您获得文件列表时,只需对其进行排序并使用循环打开文件,不要忘记使用
关闭文件,即使出现异常,它也会为您关闭文件
import glob
path = 'path/H2_*.csv'
files=glob.glob(path)
for file in sorted(files):
with open(file, 'r') as f:
print f