将 file1 的每一行复制到 file2 的每一行（Python）

Question

对不起这个荒谬的标题；这可能是我在 Google 上找不到答案的原因。

我有 5 个文本文件要合并为 1 个。我想要这样的格式：

line1 of file1
line1 of file2
line1 of file3
line1 of file4
line1 of file5
line2 of file1
line2 of file2
line2 of file3
line2 of file4
line2 of file5

等等。

我尝试使用下面的 bash 命令，但对于 sed 或其他东西来说似乎太多了：它只是将文本插入第一行，而不是我正在调用的变量的行。

for ((num=1; num<=66; num++)) ; do
    queryline=$(sed -n "${num}p" "file2.txt")
    sed -i "${num}i ${queryline}" "file1.txt"
done

（我也试过了）

for ((num=1; num<=66; num++)) ; do
    numa=$((num + 1))
    queryline=$(sed -n "${num}p" "file2.txt")
    sed -i "${numa}i ${queryline}" "file1.txt"
done

我认为使用 python (3.4) 可能会更容易，但我不知道该怎么做。请任何人提供提示？

Answer 1

如果您确定您正好有 5 个文件，这将起作用。如果您需要对不同数量的文件进行这项工作，它会变得更加复杂。

with open("file1.txt") as f:
    file1 = f.readlines()
with open("file2.txt") as f:
    file2 = f.readlines()
with open("file3.txt") as f:
    file3 = f.readlines()
with open("file4.txt") as f:
    file4 = f.readlines()
with open("file5.txt") as f:
    file5 = f.readlines()
outfile = open("outfile.txt", "w")
for aline in [line for foo in zip(file1, file2, file3, file4, file5) for line in foo]:
    outfile.write(aline)
outfile.close()

Answer 2

使用 contextlib.ExitStack() 将输入文件作为一个组处理，并使用 zip 从所有文件中读取行：

import contextlib
import os

filenames = ['a','b','c','d','e']
output_file = 'fred'

# setup files for test
for filename in filenames:
    with open(filename, 'w') as fp:
        for i in range(10):
            fp.write('%s %d\n' % (filename, i))
if os.path.exists('fred'):
    os.remove('fred')

# open all the files and use zip to interleave the lines    
with open(output_file, 'w') as out_file, contextlib.ExitStack() as in_files:
    files = [in_files.enter_context(open(fname)) for fname in filenames]
    for lines in zip(*files):
        # if you're not sure last line has a \n
        for line in lines:
            out_file.write(line)
            if not line.endswith('\n'):
                out_file.write('\n')
        # if you are sure last line has a \n
        # out_file.write(''.join(lines))

print(open('fred').read())

Answer 3

您的 bash 不起作用，因为您试图插入在插入之前不存在的行中。

echo "\n" > file_to_insert.txt
for i in {1..5};do
  for((num=1;num<66;num++);do
    line_num=$((num*i)
    queryline=$(sed -n '${num}p' 'file${i}.txt'
    sed -i "${num}i '$queryline'" 'file_to_insert.txt'
done

Answer 4

这是一个gnu awk（gnu 对ARGIND（文件选择器）进行操作）

awk -v t=5 '{c=c<FNR?FNR:c; for (i=1;i<=t;i++) if (ARGIND==i) a[i FS FNR]=$0} END {for (i=1;i<=c;i++) for (j=1;j<=t;j++) print a[j FS i]}' file1 file2 file3 file4 file5

您将t 设置为文件数。

例子：

cat f1
file1 one
file1 two
file1 three
file1 four

cat f2
file2 one
file2 two
file2 three
file2 four

cat f3
file3 one
file3 two
file3 three
file3 four

awk -v t=3 '{c=c<FNR?FNR:c; for (i=1;i<=t;i++) if (ARGIND==i) a[i FS FNR]=$0} END {for (i=1;i<=c;i++) for (j=1;j<=t;j++) print a[j FS i]}' f1 f2 f3
file1 one
file2 one
file3 one
file1 two
file2 two
file3 two
file1 three
file2 three
file3 three
file1 four
file2 four
file3 four

---

它是如何工作的？

awk -v t=3 '                    # Set t to number of files
    {c=c<FNR?FNR:c              # Find the file with most records and store number in c
    for (i=1;i<=t;i++)      # Loop trough one and one file
        if (ARGIND==i)          # Test what file we are on
            a[i FS FNR]=$0}     # Stor data in array a
END {
    for (i=1;i<=c;i++)          # Loop trough line number
        for (j=1;j<=t;j++)      # Loop trough file number
            print a[j FS i]}    # Print data from array
' f1 f2 f3                      # Read the files

Answer 5

实现您想要的一个很好的可能性是坚持使用标准实用程序：这里推荐paste（由 POSIX 指定）：

paste -d '\n' file1 file2 file3 file4 file5

或者，如果你喜欢 Bashisms：

paste -d '\n' file{1..5}

这可以简单地推广到任意数量的文件。