如何根據文中的前三位數字找出10位數字？ Python

Question

我需要在文本中找到以某個數字系列開頭的所有 10 位數字。 有一個例子：

a_string = "一些文本 6401104219 和 6401104202 和 2201104202"

匹配 = ["240", "880", "898", "910", "920", "960", "209", "309", "409", "471", "640"]

結果是：6401104219, 6401104202

Answer 1

您可以使用正則表達式和str.startswith ：

import re

result = [s for s in re.findall(r"\d{10}", a_string) if any(map(s.startswith, matches))]
# ['6401104219', '6401104202']

如果你知道前綴都是 3 位數長，你可以做得更好：

matches = set(matches)

result = [s for s in re.findall(r"\d{10}", a_string) if s[:3] in matches]

如果您想排除可能的 10 位數字前綴的較長數字，則必須將正則表達式更改為r"\b(\d{10})\b" 。

Answer 2

你可以直接使用 re.

import re
a_string = "Some text 6401104219 and 6401104202 and 2201104202 and    640110420212"
matches = ["240", "880", "898", "910", "920", "960", "209", "309", "409", "471", "640"]
result = re.findall(r"\b(?:" + r"|".join(matches)+r")\d{7}\b", a_string)

print(result)
# ['6401104219', '6401104202']

Answer 3

你可以正則表達式。

• 找出所有的 10 位數字
  
• 過濾掉匹配列表中給定元素開始的數字

代碼：

import re

a_string = "Some text 6401104219 and 6401104202 and 2201104202"

matches = ["240", "880", "898", "910", "920",
           "960", "209", "309", "409", "471", "640"]

match = re.findall(r'\d{10}', a_string)  # finding all the 10 digit numbers

# filtering out the numbers which starts from the given elements in matches


ans = [i for i in match if any(map(i.startswith, matches))]
# OR 
# ans = [i for i in match if i[:3] in matches] # if lenght is 3 only then simply check its existence in list
print(ans)
# ['6401104219', '6401104202'] 

Answer 4

a_string = "Some text 6401104219 and 6401104202 and 2201104202 and    640110420212"
matches = ["240", "880", "898", "910", "920", "960", "209", "309", "409", "471", "640"]

a_string_list=a_string.split(' ')
for i in a_string_list:
    for j in matches:
        if i.startswith(j) and len(i)==10:
            print(i)
            break