Question

我有一对包含以下数字对的文件：

我想用＆＃34; 2000＆＃34;替换单个＆＃34; 0＆＃34; s。预期的输出应为：

然而，使用我的代码，它将所有0更改为2000s，我最终得到这个输出：

2000,21
2000,52
2000,464
120002000,478
1,120001
1,729
1,732000

我的代码是：

textToSearch = "0"
textToReplace = "2000"
fileToSearch = "example.csv"
with fileinput.FileInput(fileToSearch, inplace=True, backup='.bak') as file:
    for line in file:
        print(line.replace(textToSearch, textToReplace), end='')

另外：我从来不知道逗号左侧会有多少0，因此我无法限制要更改的0的数量。该文件是随机生成的，有时我有12个0，有时只有1个。我试过这个：

textToSearch = "0,"
textToReplace = "2000,"
fileToSearch = "example.csv"
with fileinput.FileInput(fileToSearch, inplace=True, backup='.bak') as file:
    for line in file:
        print(line.replace(textToSearch, textToReplace), end='')

但是，这次它不适用于"100"或"200"等数字，因为它们分别为"102000"和"202000"。

我该如何解决？

Answer 1

稍作修改，将数据保存到文件中。基本算法没有改变。

import re

temp = []

with open("example.csv", "r") as inputf:
    for line in inputf:
        line = line.strip("\n")
        match = re.search("^0", line)
        if match:
            list1 = line.split(",")
            list1[0] = 2000
            line = str(list1[0]) + "," + str(list1[1])
        temp.append(line)
inputf.close()

# overwrite original file

with open("example.csv", "w") as outputf:
    for item in temp:
        outputf.write(item + "\n")
outputf.close()

Answer 2

你应该尝试使用＆＃34;正则表达式＆＃34;导入重新模块如下。

import re    

fileToSearch = "example.csv"

    with open(fileToSearch) as file:
        for line in file:
            line = line.strip("\n")
            match = re.search("^0", line)
            if match:
                list1 = line.split(",")
                list1[0] = 2000
                line = str(list1[0]) + "," + str(list1[1])
            print(line)

这似乎产生了你想要的结果。

Answer 3

KISS方法。在玩regexp时，我提出了一个更简单，更简单的解决方案，您可能会考虑使用它。

import re        

with open("example.csv", "r") as inputf:

    randstr = inputf.read()
    regex = re.compile(r"\b0\b")
    result = re.sub(regex, "2000", randstr)

with open("example.csv", "w") as outputf:
    outputf.write(result)

如何在python中只替换文本中的一位数？

3 个答案: