Question

我正在尝试创建一个与3个不同文本文件组合在一起的新文本文件。我会用代码提供更多细节。

text a file = z
              a
              y

text b file = s
              x
              d

text c file = 1
              3
              2

所以有a，b，c文本文件，我想做：

text newfile: z s 1
              y d 2
              a x 3

如上所示，我希望newfile按'C'文件的顺序排列。以下是我的成就。

def main():
a = open("text1","r")
b = open("text2","r")
c = open("text3","r")
text1list = []
text2list = []
text3list = []
for line1 in a:
    line1 = line1.strip()
    text1list.append(line1)
for line2 in b:
    line2 = line2.strip()
    text2list.append(line2)
for line3 in c:
    line3 = line3.strip()
    text3list.append(line3)
aa,bb,cc = zip(*sorted(zip(text3list, text1list, text2list)))
combine = list(zip(bb,cc,aa))
with open("finalfiles", 'w') as zzzz:
    for item in combine:
        zzzz.write("{}\n".format(item))

问题是，现在，我的输出是

('z','s','1')
('y','d','2')
('a','x','3')

我的分类工作正常，但它与我的预期不同。我不想要那些''和（）。我想是因为那些是名单......？我陷入了困境。另外，请告诉我，我的排序看起来很好！

Answer 1

您的combine变量是元组列表。加入每个元组（item）并写入文件

zzzz.write(" ".join(item))

您的代码有点缩短和DRY版本：

files = ["text1", "text2", "text3"]
groups = []
for each_file in files:
    with open(each_file, 'r') as fo:
        groups.append(sorted(fo.read().split('\n')))
with open("finalfiles", 'w') as out_file:
    for each_group in zip(*groups):
        out_file.write("{} {} {}\n".format(*each_group))

Python）将三个文本文件（列表）合并为一个排序

1 个答案: