在 Python 中拆分 .text 文件行

Question

我有 .txt 文件：

x1 x2 x3 y1 y2 y3 z1 z2 z3
x1 x2 x3 y1 y2 y3 z1 z2 z3

例如：

   1.9849207e-01   1.9993099e-01   2.0150793e-01   9.6169322e-02   9.6354487e-02   1.0630896e-01   1.5000000e-02   1.6090730e-02   1.5000000e-02
   1.9993099e-01   2.0261176e-01   2.0150793e-01   9.6354487e-02   1.0536750e-01   1.0630896e-01   1.6090730e-02   1.6090730e-02   1.5000000e-02

我使用了以下片段：

from itertools import *

with open('path/snip.txt', 'r') as f_input, open('path/snip_output.txt', 'w') as f_output:
    values = []
    for line in f_input:
        values.extend(line.split())

    ivalues = iter(values)

    for triple in iter(lambda: list(islice(ivalues, 3)), []):
        f_output.write(' '.join(triple) + '\n')

得到了：

x1 x2 x3
y1 y2 y3
z1 z2 z3
x1 x2 x3
y1 y2 y3
z1 z2 z3

不过这是我想要的，但我不知道如何处理 islice 中的 2 个步骤：

x1 y1 z1
x2 y2 z2
x3 y3 z3
x1 y1 z1
x2 y2 z2
x3 y3 z3

Answer 1

with open('path/snip.txt', 'r') as f_input, open('path/snip_output.txt', 'w') as f_output:
    for line in f_input:
        xyz = line.split()
        x, y, z = xyz[0:3], xyz[3:6], xyz[6:9]
        for triple in zip(x, y, z):
            f_output.write(' '.join(triple) + '\n')

写

1.9849207e-01 9.6169322e-02 1.5000000e-02
1.9993099e-01 9.6354487e-02 1.6090730e-02
2.0150793e-01 1.0630896e-01 1.5000000e-02
1.9993099e-01 9.6354487e-02 1.6090730e-02
2.0261176e-01 1.0536750e-01 1.6090730e-02
2.0150793e-01 1.0630896e-01 1.5000000e-02

Answer 2

你可以这样做：

n = 3
values = []
with open("log.txt") as f, open("output.txt", 'w') as g:
    for line in f:
        lst = line.rstrip().split()
        for i in zip(*[lst[i:i + n] for i in range(0, len(lst), n)]):
            print(*i, file=g)

output.txt的内容：

1.9849207e-01 9.6169322e-02 1.5000000e-02
1.9993099e-01 9.6354487e-02 1.6090730e-02
2.0150793e-01 1.0630896e-01 1.5000000e-02
1.9993099e-01 9.6354487e-02 1.6090730e-02
2.0261176e-01 1.0536750e-01 1.6090730e-02
2.0150793e-01 1.0630896e-01 1.5000000e-02

您需要将返回的列表从 line.rstrip().split() 拆分为 n 块。然后使用 zip 您可以并行迭代它们。

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