两个如何在 .txt 文件中按列写两个嵌套列表？

Question

我有两个列表列表如下所示：

sentences = [['its', 'a', 'great', 'show'], ['nice', 'movie'], ['good', 'series']]
labels = [['O', 'O', 'O', 'B_A'], ['O', 'B_A'], ['O', 'B_A']]

我想在 txt 文件中保存这对列表按列，因此每个元素对应该用白色分隔 space，每个列表对应该用一个空行分隔。

所需的输出应如下所示：

its O
a O
great O
show B_A

nice O
movie B_A

good O
series B_A

我试过这个：

filename = 'data.txt'

with open(filename, 'w') as f:
    for sen in sentences:
        for lab in labels:
            line = sen + ' ' + lab
            f.write(line)

我有以下错误：

TypeError: can only concatenate list (not "str") to list

更新：使用第一个答案，我试图定义一个函数，它接受两个嵌套列表和新文件名，如下所示：


def create_txt(ls1, ls2,file_name):
    with open(file_name, 'w') as f:
        for sen, lab in zip(ls1,ls2):
            for i, j in zip(sen, lab):
                f.write(f'{i} {j}')
            f.write('\n')
    return file_name

但它 returns 提供的文件名作为字符串:

create_txt(sentences, labels,'data_n.txt')

Output: 'data_n.txt'

我在这里做的逻辑问题是什么？

提前致谢！

Answer 1

您可以为此使用 csv 模块。

import csv

with open("file.txt", "w", newline="\n") as fp:
    writer = csv.writer(fp, delimiter=" ")
    for sentence, label in zip(sentences, labels):
        for i, j in zip(sentence, label):
            writer.writerow([i, j])
        fp.write('\n')

不使用任何附加模块

with open("file.txt", "w") as fp:
    for sentence, label in zip(sentences, labels):
        for i, j in zip(sentence, label):
            fp.write(f'{i} {j}\n')
        fp.write('\n')

Answer 2

另一个有效的答案，略有不同，带有一些解释性注释：

sentences = [['its', 'a', 'great', 'show'], ['nice', 'movie'], ['good', 'series']]
labels = [['O', 'O', 'O', 'B_A'], ['O', 'B_A'], ['O', 'B_A']]

filename = "data.txt"
outputstring = ""

# Construct the output string with zip.

# First we're zipping the elements of the source lists,
# which gives a sequence of pairs like this:
# (sentences[0], labels[0]), (sentences[1], labels[1]), etc.

# Then we iterate over that sequence and zip up the contents of
# each pair of lists in the same way, and concatenate those strings
# with the outputstring, followed by a single newline character.
# After that, an extra newline is added to break up the groups.

for sentence, label in zip(sentences, labels):
    for i, j in zip(sentence, label):
        outputstring += i + " " + j + "\n"
    outputstring += "\n"

# This removes the extra whitespace at the end.

outputstring = outputstring.rstrip()

# Finally, you can just write the string to your output file.

with open(filename, "w") as f:
    f.write(outputstring)

这是不使用 zip 的第二个示例：

sentences = [['its', 'a', 'great', 'show'], ['nice', 'movie'], ['good', 'series']]
labels = [['O', 'O', 'O', 'B_A'], ['O', 'B_A'], ['O', 'B_A']]

filename = "data.txt"
outputstring = ""

# Check the length of each list of lists and make sure they're the same:

sentenceslen = len(sentences)
labelslen = len(labels)

if sentenceslen != labelslen:
    print("Malformed data!")
    raise SystemExit

# Iterate over both lists using their lengths to define the range of indices.

for i in range(sentenceslen):
    
    # Check the lengths of each pair of sublists and make sure they're the same:
    
    subsentenceslen = len(sentences[i])
    sublabelslen = len(labels[i])
    
    if subsentenceslen != sublabelslen:
        print("Malformed data!")
        raise SystemExit
    
    # Iterate over each pair of sublists using their lengths to define the range of indices:
    
    for j in range(subsentenceslen):
        
        # Construct the outputstring by using both indices to drill down to the right strings,
        # ending with newline:
        
        outputstring += sentences[i][j] + " " + labels[i][j] + "\n"
    
    # Break up groups with newline again:
    
    outputstring += "\n"

# Remove whitespace at end:

outputstring = outputstring.rstrip()

# Write outputstring to file:

with open(filename, "w") as f:
    f.write(outputstring)

我不建议实际使用第二个示例中的代码。它不必要地复杂，但我将其包括在内是为了完整性，并说明使用上面的 zip 函数如何节省工作量。 zip 函数也不关心你是否给它提供不同长度的列表，所以如果你尝试这样做但不检查它，你的脚本不会崩溃；它会吐出成对的值，直到较小列表的长度，并忽略较大列表之后的值。

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