使用Python 3进行字符串练习_Python_Python 3.x

使用Python 3进行字符串练习

python python-3.x

使用Python 3进行字符串练习,python,python-3.x,Python,Python 3.x,编写一个名为remove_duplicates的函数，该函数将接受一个名为string的参数。此字符串输入将仅包含介于a-z之间的字符。函数应删除字符串中的所有重复字符，并返回一个包含两个值的元组：仅包含唯一排序字符的新字符串删除的重复项的总数例如：删除重复项（'aaabbbac'）应产生（'abc'）删除重复项（'a'）应产生（'a'，0）删除重复项（'thelexash'）应产生（'aehlstx'，2）我的代码： def remove_duplicates

编写一个名为

remove_duplicates

的函数，该函数将接受一个名为

string

的参数。此

字符串

输入将

仅包含介于a-z之间的字符

。函数应删除字符串中的所有重复字符，并返回一个包含两个值的元组：

仅包含唯一排序字符的新字符串
删除的重复项的总数

例如：

删除重复项（'aaabbbac'）

应产生

（'abc'）

```
删除重复项（'a'）
```
应产生
```
（'a'，0）
```

删除重复项（'thelexash'）

应产生

（'aehlstx'，2）

我的代码：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

预期产出：

我希望它打印

（{a，b，c}，6）

，而不是打印

（{a}，0）

我上面的代码有什么问题？为什么它不能产生我所期望的结果？

您从函数返回的第一个实例中找到了一个字符。所以它返回第一个“a”

请尝试以下方法：

def remove_duplicates(string):
    temp = set(string)
    return temp,len(string) - len(temp)


print(remove_duplicates("aaabbbccc"))

输出：

({'c', 'b', 'a'}, 6)

如果要删除除字母以外的所有内容（如您在评论中提到的），请尝试以下操作：

def remove_duplicates(string):
    a= set()
    for i in string:
        if i.isalpha() and i not in a:
            a.add(i)
    return a,len(string) - len(a)

如果不遍历字符串中的每个字符，将得到预期的结果

我已经对您的代码进行了注释，以便您可以看到您的脚本和我的脚本之间的差异

非工作注释代码：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

输出：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

工作代码：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

输出：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

p.s.：如果您希望结果井然有序，可以将

返回k，x

更改为

返回排序（k），x

，但随后输出将是一个列表

(['a', 'b', 'c'], 6)

编辑：如果希望代码仅在满足特定条件时运行（例如，仅在字符串没有任何数字时运行），则可以添加if/else子句：

示例代码：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

输出：

    def remove_duplicates(string):

        for string in "abcdefghijklmnopqrstuvwxyz":

            k = set(string)

            x = len(string) - len(set(string))

            return k, x

    print(remove_duplicates("aaabbbccc"))

def remove_duplicates(string):

    #loop through each char in "abcdefghijklmnopqrstuvwxyz" and call it "string"
    for string in "abcdefghijklmnopqrstuvwxyz":

        #create variable k that holds a set of 1 char because of the loop
        k = set(string)

        # create a variable x that holds the difference between 1 and 1 = 0
        x = len(string) - len(set(string))

        #return these values in each iteration
        return k, x

print(remove_duplicates("aaabbbccc"))

({'a'}, 0)

def remove_duplicates(string):

    #create variable k that holds a set of each unique char present in string
    k = set(string)

    # create a variable x that holds the difference between 1 and 1 = 0
    x = len(string) - len(set(string))

    #return these values
    return k, x

print(remove_duplicates("aaabbbccc"))

({'b', 'c', 'a'}, 6)

def remove_duplicates(s):

    if not s.isdigit():
        k = set(s)
        x = len(s) - len(set(s))
        return sorted(k), x
    else:
        msg = "This function only works with strings that doesn't contain any digits.."
        return msg


print(remove_duplicates("aaabbbccc"))
print(remove_duplicates("123123122"))

(['a', 'b', 'c'], 6)
This function only works with strings that doesn't contain any digits..

在代码中，函数将在迭代第一个字符后返回。 As

string

指的是输入字符串中的第一个字符。我认为您正在尝试逐字符迭代

字符串变量。
为此，您可以使用collections.Counter
，它可以更有效地执行相同的计算
但是，我们可以使用另一种解决方案，它不涉及计算给定字符串中每个字符的计数
def remove_duplicates(s):
    unique_characters = set(s) # extract the unique characters in the given string
    new_sorted_string = ''.join(sorted(unique_characters)) # create the sorted string with unique characters
    number_of_duplicates = len(s) - len(unique_characters) # compute the number of duplicates in the original string
    return new_sorted_string, number_of_duplicates

好的，如何将字符串限制为仅用于运行代码的字母表。像这样，您的代码也将为“22233377”运行。这就是我试图做的，通过“TTT中的xx”runI更新了我的答案。您确定要将集合作为输出而不是字符串吗？您先前的回答是可以的。我只需要增加一个子句，以确保它不接受“222333”作为字符串，而只接受“abc…z”检查我答案的第二部分。这对您不起作用吗？好的，我如何将字符串限制为仅用于运行代码的字母表。像这样，您的代码也将为“22233377”运行。这就是我想做的，通过“TTT中的xx”run@wapadunk我无法理解你试图用字符串和字母表指的是什么？字符串是Python类型
。前面的答案是可以的。我只需要额外的子句来确保它不接受“222333”作为字符串，而只接受“abc…z”好的，我如何将字符串限制为仅用于运行代码的字母表。像这样，您的代码也将为“22233377”运行。这就是我试图做的，通过“for xx in TTT”运行，您必须添加一个if/else
子句。我已经更新了我的答案，看一看。救命！我一直在尝试将此（['a'，'b'，'c']，6）的输出转换为（“abc”，6）。将return sorted（k），x
更改为return'。join（sorted（k）），x
工作正常。但请向我解释一下这种方法。您已将已排序集合并为字符串。所以结果变成一个字符串对吗？？？？？？