通过数字获取字符串[Python]

Question

我有一个以数字结尾的字符串列表，我需要根据给定的数字找到一个。

strings = ["..1", "..2" , "..6", "..16","..26"]
tosearch = 6
for mystr in strings:
    if tosearch in mystr:
        print("FIND!")

但是，如果我搜索“ 6”，我会发现3个字符串（.. 6，.. 16，.. 26）如何才能通过给定的数字精确地找到一个字符串（如果tosearch是26，我只会得到“ .. 26” ）

Answer 1

完成搜索的一种方法是更改​​搜索字符串：

strings = ["..1", "..2" , "..6", "..16","..26"]
tosearch = 6
tosearch = '..{}'.format(tosearch)  # ==> '..6'
for mystr in strings:
    ...

Answer 2

如果字符串中的所有字符都是数字，则isdigit（）方法返回True ，否则返回False 。 split（）方法在使用指定的分隔符将给定字符串断开后，返回字符串列表。

例如

strings = ["..1", "..2" , "..6", "..16","..26"]
tosearch = 6
for mystr in strings:
    elm = [int(x) for x in mystr.split(".") if x.isdigit()][0]
    if elm == tosearch:
        print("FIND!")

O / P：

FIND!

Answer 3

import re
strings = ["..1", "..2" , "..6", "..16","..26"]
pattern=re.compile(r"\d+")
stringnumbers = [int(pattern.findall(i)[0]) for i in strings]
tosearch = 6
if tosearch in stringnumbers:
    print("FIND!")

---

strings = ["..1", "..2" , "..6", "..6","..16","..26"]
nstr =[int(s.replace('.','')) for s in strings]
tosearch = 6
for mystr in nstr:
if tosearch == mystr: 
    print("FIND!")
    break 

输出：

FIND!