Python 3.x 从一组字符串中查找整数

嗨，我想解决一个问题，如下所示：- 如果I/P为“OzonePower”，O/P为012，即将0的字符串（零）与输入字符串进行比较，当找到时，它将出现在输出中，依此类推1和2。 为参考提供一组输入和输出：-

I/P:                               O/P:
WEIGHFOXTOURIST                    2468
OURNEONFOE                         114
ETHER                               3

我已经试过了，但这并不是所有情况下都有结果

def puzzle(dic_number,string,key):
    dic_values=0    
    length=len(dic_number)
    for i in dic_number:
        if i in string:
            dic_values+=1
    if dic_values ==length:
        print(key)


dic1={0:"ZERO",1:"ONE",2:"TWO",3:"THREE",4:"FOUR",5:"FIVE",6:"SIX",7:"SEVEN",8:"EIGHT",9:"NINE"}
string=input("Enter number")

for i,j in enumerate(dic1.values()):
    puzzle(j,string,i)

---

这里有一种实现方法：

numbers = {
    0: "ZERO",
    1: "ONE",
    2: "TWO",
    3: "THREE",
    4: "FOUR",
    5: "FIVE",
    6: "SIX",
    7: "SEVEN",
    8: "EIGHT",
    9: "NINE",
}


def puzzle(s):
    supper = s.upper()
    ret = []
    for n, chrs in numbers.items():
        if all(c in supper for c in chrs):
            ret.append(n)
    return ret


s = input("enter string ")

numbers_found = puzzle(s)
print('numbers found', numbers_found)

输出：

enter string WEIGHFOXTOURIST
numbers found [2, 3, 4, 6, 8]
enter string OURNEONFOE
numbers found [1, 4]
enter string ETHER
numbers found [3]
enter string OZONETOWER
numbers found [0, 1, 2]

enter string WEIGHFOXTOURIST
numbers found [2, 3, 6]
enter string OURNEONFOE
numbers found [1, 1, 4]
enter string ETHER
numbers found [3]
enter string OZONETOWER
numbers found [0, 1, 2]

注意

通过这种实现

• 对于输入

  3

  ，您会得到一个额外的

  3

  结果
• 对于输入

  OURNEONFOE

---

编辑

要获得重复匹配，拼图函数必须“使用”已使用的字符：

def puzzle(s):
    supper = s.upper()
    ret = []
    for n, chrs in numbers.items():
        while True:
            if all(c in supper for c in chrs):
                for c in chrs:
                    supper = supper.replace(c, '', 1)
                ret.append(n)
            else:
                break
    return ret

输出：

enter string WEIGHFOXTOURIST
numbers found [2, 3, 4, 6, 8]
enter string OURNEONFOE
numbers found [1, 4]
enter string ETHER
numbers found [3]
enter string OZONETOWER
numbers found [0, 1, 2]

enter string WEIGHFOXTOURIST
numbers found [2, 3, 6]
enter string OURNEONFOE
numbers found [1, 1, 4]
enter string ETHER
numbers found [3]
enter string OZONETOWER
numbers found [0, 1, 2]

注意

通过这种实现

• 对于输入

  8

  ，您不会得到

  8

  的结果

---

说明：

enter string WEIGHFOXTOURIST
numbers found [2, 3, 4, 6, 8]
enter string OURNEONFOE
numbers found [1, 4]
enter string ETHER
numbers found [3]
enter string OZONETOWER
numbers found [0, 1, 2]

enter string WEIGHFOXTOURIST
numbers found [2, 3, 6]
enter string OURNEONFOE
numbers found [1, 1, 4]
enter string ETHER
numbers found [3]
enter string OZONETOWER
numbers found [0, 1, 2]

在这个拼图中，当在输入字符串中找到一个数字的所有字母时，它被认为是一个“匹配”

为此，我们通过在字符串的字符上创建一个标记来逐个检查每个字符：

>>> chrs = 'FOUR'
>>> [c for c in chrs]
['F', 'O', 'U', 'R']

对于我们用来检查是否在大写输入字符串中找到的每个字符：

>>> chrs = 'FOUR'
>>> supper = 'OURNEONFOE'
>>> [c in supper for c in chrs]
[True, True, True, True]

如果未找到字符，它将在相应位置生成一个

False

：

>>> chrs = 'FOUR'
>>> supper = 'OXRNEONFOE'
>>> [c in supper for c in chrs]
[True, True, False, True]

然后，使用检查列表中的所有项目是否为

True

>>> chrs = 'FOUR'
>>> supper = 'OURNEONFOE'
>>> all([c in supper for c in chrs])
True
>>> supper = 'OXRNEONFOE'
>>> all([c in supper for c in chrs])
False

由于我们可以在

all

>>> chrs = 'FOUR'
>>> supper = 'OURNEONFOE'
>>> all(c in supper for c in chrs)
True

---

嗨，谢谢你的回答，效果很好。。你能解释一下这句话的确切含义吗：-```如果全部（c在晚餐中代表c在chrs中）：@shubham我在我的回答中添加了一个解释试图避免只使用代码的回答，并在你的回答中提供一些解释。