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Python 3.x 如何连续生成文本文件数组?_Python 3.x - Fatal编程技术网

Python 3.x 如何连续生成文本文件数组?

Python 3.x 如何连续生成文本文件数组?,python-3.x,Python 3.x,aTxt文件夹有3000个文本文件。我的代码以随机顺序读取这些.txt文件,如(1.txt、10.txt、1000.txt,…)。我如何连续读取这些文本文件(比如先读取1.txt,然后读取2.txt,等等,最高可达3000.txt)并将这些文本文件附加到数组中 from pickle import dump import glob textlist = [] textfiles = glob.glob('D:/qrt/aTxt/*.txt') for x in textfiles: x1

aTxt文件夹有3000个文本文件。我的代码以随机顺序读取这些.txt文件,如(1.txt、10.txt、1000.txt,…)。我如何连续读取这些文本文件(比如先读取1.txt,然后读取2.txt,等等,最高可达3000.txt)并将这些文本文件附加到数组中

from pickle import dump
import glob
textlist = []
textfiles = glob.glob('D:/qrt/aTxt/*.txt')
for x in textfiles:
    x1 = open(x, 'r')
    x2 = x1.read()
    textlist.append(x2)
    x1.close()
dump(textlist, open('textlists.pkl', 'wb'))

如果您指的是词法顺序,您可以使用内置函数对从
glob
获得的文件路径列表进行排序
sorted()

请定义“与文件夹中的顺序相同”?你是说词法顺序,创建日期,索引节点号,还是其他?
from pickle import dump
import glob
textlist = []
textfiles = glob.glob('D:/qrt/aTxt/*.txt')
for filepath in sorted(textfiles):
    with open(filepath , 'rt') as finput:
        content = finput.read()
        textlist.append(content)

dump(textlist, open('textlists.pkl', 'wb'))