Question

假设我有两个文本文件1.txt和2.txt

1.txt的内容为：

[['Hi', 'I'], ['I'], ['_am']]

和

2.txt的内容为：

[['_a', 'good'], ['boy']]

如何以相同的方式将相同的内容连接起来并将其写入新文件，例如3.txt，它应该像这样：

[['Hi', 'I'], ['I'], ['_am'], ['_a', 'good'], ['boy']]

注意：我希望特殊字符（_）保持原样。

我尝试过Concatenation of Text Files (.txt files) with list in python?

中提到的堆栈溢出的先前答案

我尝试的方法如下：

global inputList
inputList = []
path = "F:/Try/"
def load_data():
    for file in ['1.txt', '2.txt']:
        with open(path + file, 'r', encoding = 'utf-8) as infile:
           inputList.extend(infile.readlines())
    print(inputList)
load_data()

但是如上所述，我没有得到所需的输出。我现在得到的输出如下：

["[['Hi', 'I'], ['I'], ['_am']]", "[['_a', 'good'], ['boy']]"]

为什么我的当前输出中有多余的（“”）。

请提出一些有益的建议？

谢谢。

Answer 1

您可能要使用：

import json

with open("1.txt", "r") as t1, open("2.txt", "r") as t2, open("3.txt", "w") as t3:
    t3.write(json.dumps([eval(t1.read().strip()), eval(t2.read().strip())]))

3.txt

[[["Hi", "I"], ["I"], ["_am"]], [["_a", "good"], ["boy"]]]

注意：

Answer 2

您的文件是否总是这样？然后，您只需删除“外部” [和]。

path = "F:/Try/"

def load_data():
    result = []
    for file in ['1.txt', '2.txt']:
        with open(path + file, 'r', encoding='utf-8') as infile:
            result.append(infile.readline().strip()[1:-1])
    return "[" + ", ".join(result) + "]"

print(load_data())

哪些印刷品

[['Hi', 'I'], ['I'], ['_am'], ['_a', 'good'], ['boy']]

Answer 3

xlist = ["[['Hi', 'I'], ['I'], ['_am']]", "[['_a', 'good'], ['boy']]"]
ylist = []
for x in xlist:
    if x.startswith('[') and x.endswith(']'):
        ylist.append(x[1:-1])
zstring =''
for y in ylist:
    if zstring == '':
        zstring = y
    else:
        zstring += ', ' + y

print (zstring)
#['Hi', 'I'], ['I'], ['_am'], ['_a', 'good'], ['boy']

Answer 4

import ast
x="[['Hi', 'I'], ['I'], ['_am']]"
ast.literal_eval(x)

输出：

[['Hi', 'I'], ['I'], ['_am']]

在您的情况下：

import ast
global inputList
inputList = []
path = "F:/Try/"
def load_data():
    for file in ['1.txt', '2.txt']:
        with open(path + file, 'r', encoding = 'utf-8') as infile:
           inputList.extend(ast.literal_eval(infile.readlines()))
    print(inputList)
load_data()

Answer 5

尝试一下（不要忘了加星号运算符）：

import ast
global inputList
inputList = []
path = "F:/Try/"
def load_data():
    for file in ['1.txt', '2.txt']:
        with open(path + file, 'r', encoding = 'utf-8') as infile:
            inputList.extend(ast.literal_eval(*infile.readlines()))
    print(inputList)
load_data()

Answer 6

想要真正的效率，把1.txt不带最后一个字符写到3.txt，再不带第一个字符写2.txt

def concat1():

    with open('3.txt', 'w') as outf:
        with open('1.txt') as inf:
            outf.write(inf.read()[:-1])

        outf.write(',')

        with open('2.txt') as inf:
            outf.write(inf.read()[1:])
            
concat1()

给予

[['Hi', 'I'], ['I'], ['_am'],['_a', 'good'], ['boy']]

由列表列表组成的文本文件的串联？

6 个答案: