如何在python中用逗号分隔数字组？

Question

我有以下文字：

Cluster 7: {4, 15, 21, 28, 33, 35, 43, 47, 53, 57, 59, 66,
       69, 70, 74, 86, 87, 88, 90, 114, 136, 148, 201,
       202, 212, 220, 227, 250, 252, 253, 259, 262, 267,
       270, 282, 296, 318, 319, 323, 326, 341}
Cluster 8: {9, 10, 11, 20, 39, 55, 79, 101, 108, 143, 149,
       221, 279, 284, 285, 286, 287, 327, 333, 334, 335,
       336}
Cluster 9: {3, 64, 83, 93, 150, 153, 264, 269, 320, 321, 322}
Cluster 10: {94, 123, 147}

我想通过群集提取每组中的数字。

我尝试使用正则表达式没有太多运气

我努力了：

regex="(Cluster \d+): \{((\d+)[,\}][\n ]+)+|(?:(\d+),[\n ])"

但这些团体并不匹配。

我希望输出为：

["Cluster 7", '4', '15', '21', '28', '33', '35', '43', '47', '53', '57', '59', '66', '69', '70', '74', '86', '87', '88', '90', '114', '136', '148', '201', '202', '212', '220', '227', '250', '252', '253', '259', '262', '267', '270', '282', '296', '318', '319', '323', '326', '341', "Cluster 8", '9', '10', '11', '20', '39', '55', '79', '101', '108', '143', '149', '221', '279', '284', '285', '286', '287', '327', '333', '334', '335', '336', "Cluster 9", '3', '64', '83', '93', '150', '153', '264', '269', '320', '321', '322', "Cluster 10", "94", "123", "147"]

或者这可能不是最好的方法。

谢谢

Answer 1

我不会使用正则表达式。 您的文本在yaml规范内，可以直接使用保留订单的yaml加载程序（如oyaml）加载 。

import oyaml as yaml   # pip install oyaml
data = yaml.load(text)

要将该dict解压缩到所需的“平面”结构，它只是一个列表理解：

[x for (k, v) in data.items() for x in (k, *v)]

_{注意：我是oyaml的作者。}

Answer 2

您可以创建更通用的正则表达式：

import re
s = '\nCluster 7: {4, 15, 21, 28, 33, 35, 43, 47, 53, 57, 59, 66,\n       69, 70, 74, 86, 87, 88, 90, 114, 136, 148, 201,\n       202, 212, 220, 227, 250, 252, 253, 259, 262, 267,\n       270, 282, 296, 318, 319, 323, 326, 341}\nCluster 8: {9, 10, 11, 20, 39, 55, 79, 101, 108, 143, 149,\n       221, 279, 284, 285, 286, 287, 327, 333, 334, 335,\n       336}\nCluster 9: {3, 64, 83, 93, 150, 153, 264, 269, 320, 321, 322}\nCluster 10: {94, 123, 147}\n'
data = re.findall('Cluster \d+|\d+', s)

输出：

['Cluster 7', '4', '15', '21', '28', '33', '35', '43', '47', '53', '57', '59', '66', '69', '70', '74', '86', '87', '88', '90', '114', '136', '148', '201', '202', '212', '220', '227', '250', '252', '253', '259', '262', '267', '270', '282', '296', '318', '319', '323', '326', '341', 'Cluster 8', '9', '10', '11', '20', '39', '55', '79', '101', '108', '143', '149', '221', '279', '284', '285', '286', '287', '327', '333', '334', '335', '336', 'Cluster 9', '3', '64', '83', '93', '150', '153', '264', '269', '320', '321', '322', 'Cluster 10', '94', '123', '147']

Answer 3

请参阅此处使用的正则表达式

\w+(?: +\w+)?

• \\w+匹配一个或多个单词字符
• (?: +\\w+)? 可选择匹配以下内容
  • +匹配一个或多个空格
  • \\w+匹配一个或多个单词字符

请参阅此处使用的代码

import re

s = "Cluster 7: {4, 15, 21, 28, 33, 35, 43, 47, 53, 57, 59, 66,\n       69, 70, 74, 86, 87, 88, 90, 114, 136, 148, 201,\n       202, 212, 220, 227, 250, 252, 253, 259, 262, 267,\n       270, 282, 296, 318, 319, 323, 326, 341}\nCluster 8: {9, 10, 11, 20, 39, 55, 79, 101, 108, 143, 149,\n       221, 279, 284, 285, 286, 287, 327, 333, 334, 335,\n       336}\nCluster 9: {3, 64, \n3, 93, 150, 153, 264, 269, 320, 321, 322}\nCluster 10: {94, 123, 147}"
print(re.findall(r"\w+(?: +\w+)?", s))

结果：

['Cluster 7', '4', '15', '21', '28', '33', '35', '43', '47', '53', '57', '59', '66', '69', '70', '74', '86', '87', '88', '90', '114', '136', '148', '201', '202', '212', '220', '227', '250', '252', '253', '259', '262', '267', '270', '282', '296', '318', '319', '323', '326', '341', 'Cluster 8', '9', '10', '11', '20', '39', '55', '79', '101', '108', '143', '149', '221', '279', '284', '285', '286', '287', '327', '333', '334', '335', '336', 'Cluster 9', '3', '64', '83', '93', '150', '153', '264', '269', '320', '321', '322', 'Cluster 10', '94', '123', '147']