Question

我需要从以下 rows_string 变量获取字母字符串：

'Equity & 1,638 & \\$3,227,305 & \\$2,649,208 & \\$3,270,402 & \\$3,114,298 & \\$3,173,369 & \\$2,978,769 & \\$3,016,161 & \\$2,807,840\\\\\nFixed Income & 420 & \\$765,856 & \\$661,395 & \\$824,603 & \\$792,579 & \\$794,224 & \\$783,793 & \\$719,307 & \\$630,298\\\\\nCommodities & 119 & \\$72,911 & \\$66,302 & \\$81,649 & \\$81,633 & \\$79,296 & \\$76,450 & \\$64,136 & \\$63,667\\\\\nAsset Allocation & 63 & \\$10,190 & \\$9,275 & \\$10,684 & \\$10,089 & \\$10,371 & \\$9,829 & \\$9,619 & \\$8,880\\\\\nAlternatives & 55 & \\$5,601 & \\$6,023 & \\$6,715 & \\$6,279 & \\$6,365 & \\$6,645 & \\$6,757 & \\$6,243\\\\\nCurrency & 34 & \\$311 & \\$2,014 & \\$1,665 & \\$1,743 & \\$1,683 & \\$1,666 & \\$1,722 & \\$2,058\\\\\nTOTALS & 2,329 & \\$4,082,173 & \\$3,394,217 & \\$4,195,718 & \\$4,006,620 & \\$4,065,308 & \\$3,857,151 & \\$3,817,700 & \\$3,518,986\\\\'

例如，我需要以下列表：

[Equity, Fixed Income, Commodities, Asset Allocation, Alternatives, Currency, Total]

我尝试过：

re.findall(r'\\\\\n(\w+.*?) &', rows_string)

很好，但这省略了"equity"变量而且还给了我这个字符串变量的空列表

'Starting Portfolio & sell & 21.39\\% & -0.91\\% & 1.52\\% & 9.29\\% & 9.72\\% & 14.89\\% & 38.21\\% & 55.4\\% &  & 90.86\\%\\\\'

因此，对于第二个字符串，我需要['Starting Portfolio', 'sell'] 我想要的是抓住字符串变量中\\\\\n之后的第一项和'&'之前的第一项。谢谢

Answer 1

您只是缺少一个\。您不是要搜索字母\和n，而是要搜索换行符。因此，只需在正则表达式的开头添加广告\。此外，您还缺少自声明以来的第一个条目，即单词以\\\\\n开头。要获得第一，您可以使用^(\w+.*?)|[\\\\\n](\w+.*?) &例如

Answer 2

我认为没有理由专注于逃脱的换行符。这应该可以解决问题：

export const getUserSession = () => dispatch => {
    return AsyncStorage.getItem('userSession').then((data) => {
        console.log('Props at asynsstorage: ', data);
        // {"current_user":{"uid":"1","roles":["authenticated","administrator"], ...}
        dispatch(loading(false));
        dispatch(getSession(JSON.parse(data))); //convert to json here
    })
    .catch((err) => {
    })
}

Answer 3

尝试使用re.finditer()的这种模式：

pattern = r"(((?!\\\\\\\\\\n)([a-zA-Z\\s]+))|([a-zA-Z\\s]{2,}\\s?(?!\\&)))"
output_list = [i.group().strip() for i in re.finditer(pattern, rows_string) if i.group().strip()]

输入：

s1 = 'Equity & 1,638 & \\$3,227,305 & \\$2,649,208 & \\$3,270,402 & \\$3,114,298 & \\$3,173,369 & \\$2,978,769 & \\$3,016,161 & \\$2,807,840\\\\\nFixed Income & 420 & \\$765,856 & \\$661,395 & \\$824,603 & \\$792,579 & \\$794,224 & \\$783,793 & \\$719,307 & \\$630,298\\\\\nCommodities & 119 & \\$72,911 & \\$66,302 & \\$81,649 & \\$81,633 & \\$79,296 & \\$76,450 & \\$64,136 & \\$63,667\\\\\nAsset Allocation & 63 & \\$10,190 & \\$9,275 & \\$10,684 & \\$10,089 & \\$10,371 & \\$9,829 & \\$9,619 & \\$8,880\\\\\nAlternatives & 55 & \\$5,601 & \\$6,023 & \\$6,715 & \\$6,279 & \\$6,365 & \\$6,645 & \\$6,757 & \\$6,243\\\\\nCurrency & 34 & \\$311 & \\$2,014 & \\$1,665 & \\$1,743 & \\$1,683 & \\$1,666 & \\$1,722 & \\$2,058\\\\\nTOTALS & 2,329 & \\$4,082,173 & \\$3,394,217 & \\$4,195,718 & \\$4,006,620 & \\$4,065,308 & \\$3,857,151 & \\$3,817,700 & \\$3,518,986\\\\'
s2 = 'Starting Portfolio & sell & 21.39\\% & -0.91\\% & 1.52\\% & 9.29\\% & 9.72\\% & 14.89\\% & 38.21\\% & 55.4\\% &  & 90.86\\%\\\\'*

输出：

['Equity', 'Fixed Income', 'Commodities', 'Asset Allocation', 'Alternatives', 'Currency', 'TOTALS']
['Starting Portfolio', 'sell']

Answer 4

要获取值，您可以使用替代方式来匹配字符串开头的单词或获取 &之前的单词

(?:^[A-Za-z]+(?: [A-Za-z]+)*|[A-Za-z]+(?: [A-Za-z]+)*(?= &))

(?:非捕获组
- ^行的开头
- [A-Za-z]+(?: [A-Za-z]+)*仅将字符A-Za-z匹配1个以上的单词
- |或
- [A-Za-z]+(?: [A-Za-z]+)*(?= &)匹配单词，后跟 &
)关闭群组

Regex demo | Python demo

例如

import re

pattern = r'(?:^[A-Za-z]+(?: [A-Za-z]+)*|[A-Za-z]+(?: [A-Za-z]+)*(?= &))'
rows_string = 'Equity & 1,638 & \\$3,227,305 & \\$2,649,208 & \\$3,270,402 & \\$3,114,298 & \\$3,173,369 & \\$2,978,769 & \\$3,016,161 & \\$2,807,840\\\\\nFixed Income & 420 & \\$765,856 & \\$661,395 & \\$824,603 & \\$792,579 & \\$794,224 & \\$783,793 & \\$719,307 & \\$630,298\\\\\nCommodities & 119 & \\$72,911 & \\$66,302 & \\$81,649 & \\$81,633 & \\$79,296 & \\$76,450 & \\$64,136 & \\$63,667\\\\\nAsset Allocation & 63 & \\$10,190 & \\$9,275 & \\$10,684 & \\$10,089 & \\$10,371 & \\$9,829 & \\$9,619 & \\$8,880\\\\\nAlternatives & 55 & \\$5,601 & \\$6,023 & \\$6,715 & \\$6,279 & \\$6,365 & \\$6,645 & \\$6,757 & \\$6,243\\\\\nCurrency & 34 & \\$311 & \\$2,014 & \\$1,665 & \\$1,743 & \\$1,683 & \\$1,666 & \\$1,722 & \\$2,058\\\\\nTOTALS & 2,329 & \\$4,082,173 & \\$3,394,217 & \\$4,195,718 & \\$4,006,620 & \\$4,065,308 & \\$3,857,151 & \\$3,817,700 & \\$3,518,986\\\\'
print(re.findall(pattern, rows_string, re.M))


rows_string2 = 'Starting Portfolio & sell & 21.39\\% & -0.91\\% & 1.52\\% & 9.29\\% & 9.72\\% & 14.89\\% & 38.21\\% & 55.4\\% &  & 90.86\\%\\\\'
print(re.findall(pattern, rows_string2, re.M))

输出

['Equity', 'Fixed Income', 'Commodities', 'Asset Allocation', 'Alternatives', 'Currency', 'TOTALS']
['Starting Portfolio', 'sell']

如果所有匹配项后面都应加上 &，则可以将模式简化为

[A-Za-z]+(?: [A-Za-z]+)*(?= &)

Regex demo

Answer 5

假设目标字符串（财务关键字）在换行符（或字符串的开头）之后并且在&之前，您可以这样做：

>>> re.findall(r'(?:\n|^)([A-Za-z ]+)\s&', s)
['Equity', 'Fixed Income', 'Commodities', 'Asset Allocation', 'Alternatives', 'Currency', 'TOTALS']

这使用了一些快捷方式，但是取决于您是否具有更复杂的字符串，例如“ P＆E”，“ Misc。Expenses”等，以上可能就足够了。

正则表达式与多个条件匹配

5 个答案: