我是一个初学者。我想读取文件夹中的所有文件。例如,文件名为1.csv,2.csv…….10.csv,11.csv…….20.csv。同样,它应该读取为1.csv,2.csv……但对我来说,它读取为1.csv,10.csv,11.csv…….19.csv,2.csv
我正在使用代码:
import glob
path = 'C://test//08October2014//DATA_INTV_NEW//October082014//*.sec.gz'
files=glob.glob(path)
for list in sorted(files):
print list
输出:
C://test//08October2014//DATA_INTV_NEW//October082014\1.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\10.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\11.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\12.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\13.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\14.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\15.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\16.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\17.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\18.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\19.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\2.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\20.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\21.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\22.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\23.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\24.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\25.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\26.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\27.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\28.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\29.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\3.sec.gz
C://test//08October2014//DATA_INTV_NEW//October082014\30.sec.gz
在字符串比较中,
"0" < "1" < "10" < "2" < "20"
(字典顺序)。您必须将文件命名为"01"
,"02"
,"09"
,"10"
。。。让他们无需额外努力就能正确排序如果你不能做到这一点,你要找的就是“自然排序”。这里有一个模块可以做到这一点:https://pypi.python.org/pypi/natsort
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