以读模式打开文件。以字符串格式读取文件。替换文本。关闭文件。再次以写模式打开文件。最后，将替换后的文本写入相同的文件。

try:
    with open("file_name", "r+") as text_file:
        texts = text_file.read()
        texts = texts.replace("to_replace", "replace_string")
    with open(file_name, "w") as text_file:
        text_file.write(texts)
except FileNotFoundError as f:
    print("Could not find the file you are trying to read.")

你可以这样做替换

f1 = open('file1.txt', 'r')
f2 = open('file2.txt', 'w')
for line in f1:
    f2.write(line.replace('old_text', 'new_text'))
f1.close()
f2.close()

我稍微修改了Jayram Singh的帖子，以替换每一个'!'字符转换为一个数字，我想在每个实例中增加这个数字。我想这对那些想要修改每行出现不止一次的字符并且想要迭代的人可能会有帮助。希望这能帮助到别人。PS-我在编码方面很新，所以如果我的帖子在任何方面都不合适，我很抱歉，但这对我来说是有效的。

f1 = open('file1.txt', 'r')
f2 = open('file2.txt', 'w')
n = 1  

# if word=='!'replace w/ [n] & increment n; else append same word to     
# file2

for line in f1:
    for word in line:
        if word == '!':
            f2.write(word.replace('!', f'[{n}]'))
            n += 1
        else:
            f2.write(word)
f1.close()
f2.close()

我也有同样的问题。问题是，当你在变量中加载一个.txt时，你把它当作一个字符串数组使用，而它是一个字符数组。

swapString = []
with open(filepath) as f: 
    s = f.read()
for each in s:
    swapString.append(str(each).replace('this','that'))
s = swapString
print(s)

像这样:

def find_and_replace(file, word, replacement):
  with open(file, 'r+') as f:
    text = f.read()
    f.write(text.replace(word, replacement))

我试过用readline代替read

with open('dummy.txt','r') as file:
    list = file.readlines()
print(f'before removal {list}')
for i in list[:]:
        list.remove(i)

print(f'After removal {list}')
with open('dummy.txt','w+') as f:
    for i in list:
        f.write(i)