嗨,我想解决一个问题如下: - 如果I / P为“OZONETOWER”,则O / P为012,即将0的字符串(ZERO)与输入字符串进行比较,当找到时,它出现在输出中,依此类推为1和2.提供参考的输入和输出集: -
I/P: O/P:
WEIGHFOXTOURIST 2468
OURNEONFOE 114
ETHER 3
我已经尝试过这个但是这似乎没有给出所有情况的结果。
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)
以下是实现此目的的一种方法:
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]
注意
有了这个实现
3
的额外结果输入WEIGHFOXTOURIST
1, 4
而不是1, 1, 4
输入OURNEONFOE
编辑
要获得重复匹配,拼图功能必须“消耗”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, 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
有了这个实现
8
输入WEIGHFOXTOURIST
的结果说明:
在这个难题中,当在输入字符串中找到数字的所有字母时,它被认为是“匹配”。
为此,我们通过在字符串的字符上创建list comprehension逐个检查每个字符:
>>> chrs = 'FOUR'
>>> [c for c in chrs]
['F', 'O', 'U', 'R']
对于每个字符,我们使用in
来检查它是否在大写的输入字符串中找到:
>>> chrs = 'FOUR'
>>> supper = 'OURNEONFOE'
>>> [c in supper for c in chrs]
[True, True, True, True]
如果找不到某个角色,它将在相应的位置产生一个False
:
>>> chrs = 'FOUR'
>>> supper = 'OXRNEONFOE'
>>> [c in supper for c in chrs]
[True, True, False, True]
然后用all
检查列表中的所有项目是否都是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
因为PEP 289
我们可以直接在all
中使用生成器表达式
>>> chrs = 'FOUR'
>>> supper = 'OURNEONFOE'
>>> all(c in supper for c in chrs)
True