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Accepted
vladionair
Asked:2020-08-31 19:27:20 +0000 UTC2020-08-31 19:27:20 +0000 UTC

从字符串/列表中提取大于 9 的数字

  • 772

需要从字符串/列表中提取大于一阶的数字，而不使用标准的拆分连接方法等等。最多十个，与 numbers = '0123456789' 的比较有效，比方说，但接下来呢？帮帮我！）我的代码版本如下，但它不区分10（

a = input()
b = []
numbers = '0123456789'
for i in a:
    if i in numbers:
        b.append(int(i))
    else:
        continue
print(b)

вход: 3,5,7,9,10
выход: [3, 5, 7, 9, 1, 0]
python

3 个回答

  1. Best Answer
    numbers = '0123456789'

def qwerty(stri):
    (a, la, stri) = ('', 0, (' ' + stri))
    for (e, s) in enumerate(stri):
        if s in numbers:
            la += 1
            if not a:
                a = e
        elif la > 0:
            if la > 1:
                yield stri[a:e]
            (a, la) = ([], 0)
        continue

    if la > 1:
        yield stri[a:(e+1)]
    return

print(*qwerty('11,3,5,7,9,10,11,2,12,1'))  # 11 10 11 12
    • 1 
  2. input_data = '1A23D56R4ASDF12345S112'
result = []

add = False
base = 1
for ch in reversed(input_data):
    try:
        num = int(ch) * base
        if add:
            result[0] += num
        else:
            result.insert(0, num)
            add = True
        base *= 10
    except ValueError:
        add = False
        base = 1

print(result)  # [1, 23, 56, 4, 12345, 112]
    • 0

  3. 一切都变得容易多了。它也适用于负面的。

    a = input()
b = []
s = str()
numbers = '-0123456789'
for i in a:
    if i in numbers:
        s += i
        continue
    else:
        if len(s) > 0:
            b.append(int(s))
            s = str()
if len(s) > 0:
    b.append(int(s))
print(b)

вход: 0.[0000[/-1//1000,,[12345ю6789,,
выход: [0, 0, -1, 1000, 12345, 6789]
    • -1

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