python: 元组搜索列表
python: list of tuples search
这里需要一些帮助
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
如果最后 6 位数字位于 return 第一位数字。
示例 if 找到 10,11,29,30,36,47 return 2
您可以将 next
与条件生成器表达式一起使用:
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
search = 11
next(first for first, *rest in num if search in rest)
# 2
您可以使用 next
类似于用户的方法:
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
to_find = [10,11,29,30,36,47]
print(next(n for n, *nums in num if nums == to_find))
2
这里需要一些帮助
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
如果最后 6 位数字位于 return 第一位数字。
示例 if 找到 10,11,29,30,36,47 return 2
您可以将 next
与条件生成器表达式一起使用:
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
search = 11
next(first for first, *rest in num if search in rest)
# 2
您可以使用 next
类似于用户的方法:
num = [(1,4,5,30,33,41,52),(2,10,11,29,30,36,47),(3,15,25,37,38,58,59)]
to_find = [10,11,29,30,36,47]
print(next(n for n, *nums in num if nums == to_find))
2