Question

我有一个包含数字的文本字符串。另外，我有一个号码清单。我想按字符串和列表的顺序用列表中的数字替换字符串中的数字。

通过使用正则表达式，我从字符串中提取了现有数字，并将它们也传递到列表中，现在我在原始数字和备用数字之间有了匹配。但是，仍不清楚如何找到调整项并按顺序进行替换。

这一行，我从给定的字符串中提取数字：

    list_of_numbers_in_string = [int(x) for x in re.findall('\d+', str)]

现在我想知道如何使用它，或者是通过这种输入获得所需结果的另一种方法：

    data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
    new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]

获取此输出：

    corrected_data = 'readingOrder {index:24;} person {offset:0; length:12;} textStyle {offset:0; length:12; underlined:true;} place {offset:58; length:14;} textStyle {offset:58; length:14; underlined:true;}'

Answer 1

接受的答案实际上是不正确的（现已删除）。 data.replace()将替换数字的第一个出现，但它并不总是正确的。例如，当您尝试将8替换为14时，实际上是将514替换为58。

这是我的解决方法：

import re

data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]

offset = 0
for index, match in enumerate(re.finditer('\d+', data)):
    data = data[:match.start() + offset] + str(new_numbers[index]) + data[match.end() + offset:]
    offset += len(str(new_numbers[index])) - match.end() + match.start()

Answer 2

如果您一次以一个data直接对string（作为new number）进行操作（即in a for loop，您执行data = operate on data），则它会时间复杂度很可能是O(len(new_numbers) * len(data))。

在O(len(data))时间内完成此操作的一种有效方法是对a list of characters进行操作：

def replace_numbers(data, new_numbers):
    new_numbers_idx = 0
    data_as_char_list = []

    skip = False

    for data_ch in data:
        if data_ch.isdigit():
            if not skip:
                # When we encounter any number's first digit in data, we will add the new number, and in the next iterations we will skip rest of the digits in data.
                # e.g. data = 'hi123hi', new_numbers = [444], then when we encounter `1` we will add ['4', '4', '4'] and skip the rest of the digits '2' and '3' from data by setting skip = True.
                new_number_as_char_list = list(str(new_numbers[new_numbers_idx]))
                data_as_char_list.extend(new_number_as_char_list)
                new_numbers_idx += 1
                skip = True
        else:
            data_as_char_list.append(data_ch)
            skip = False

    return ''.join(data_as_char_list)


data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]
data = replace_numbers(data, new_numbers)

corrected_data = 'readingOrder {index:24;} person {offset:0; length:12;} textStyle {offset:0; length:12; underlined:true;} place {offset:58; length:14;} textStyle {offset:58; length:14; underlined:true;}'
assert data == corrected_data

Answer 3

替代

import re

data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]
x = re.findall("\d+", data)
data = data.replace("{","{{").replace("}","}}")
for n in x:
    data = data.replace(n,"{}",1)
data = data.format(*new_numbers)
print(data)

[出]：

readingOrder {index：24;}人{offset：0;长度：12;} textStyle   {offset：0;长度：12;带下划线的：true;}地方{offset：58;长度：14;}   textStyle {offset：58;长度：14;下划线：true;}

如何用另一个列表中的数字替换字符串中的数字？

3 个答案: