Question

我需要一些帮助来完成我的任务：

使用for循环，编写一个名为getMax4的函数，该函数接受数字列表。它确定并返回可被4整除的最大数字。如果参数为空列表，则函数返回-999 如果参数列表中没有数字可被4整除，则该函数返回0。

下面显示了调用该函数时的示例输出：

我已经设法在代码中执行以下操作：

def getMax4(list):
    highest = 0
    if len(list) == 0:
        return -999
    else:
        for number in list:
            if number % 4 == 0:
                if number < highest:
                    highest = number

                else:
                    highest = 0               
                return number
print(getMax4([]))
print(getMax4([1, 3, 9]))
print(getMax4([-4, 3, -12, -8, 13]))
print(getMax4([1, 16, 18, 12]))

对我来说输出如下：

我不明白为什么将值确定为None时将其确定为0。

Answer 1

让我们逐步执行代码。

def getMax4(list):
    highest = 0
    if len(list) == 0:
        return -999
    else:
        for number in list:
            if number % 4 == 0:
                if number < highest:
                    highest = number

                else:
                    highest = 0               
                return number

明显的情况是长度为零的列表，它可以正确处理它们。呜呼。

另一种明显的情况是数字被四整除的列表。让我们来看一下：

# imagine getMax4([[1, 16, 20, 12])
# note that this is different from your example, because your example doesn't
# highlight a meaningful bug!

# unroll the for loop:

number = 1
if number % 4 == 0:  # it doesn't, so skip
number = 16
if number % 4 == 0:
    if number < highest:  # it's not, but this should pass. Your comparison is backwards here!
    else:
        highest = 0  # what?? we reset it?? Why??!
    return number

因此，我们已经确定了几个问题，包括以下事实：一旦找到一个可被四整除的数字，即使正确的答案是20，我们也会以该数字作为解决方案。这显然是错误的！

让我们修复这些问题。

def getMax4(list):
    highest = 0
    if len(list) == 0:
        return -999
    else:
        for number in list:
            if number % 4 == 0:
                if number > highest:  # flip comparison
                    highest = number
                # removed the else clause
        return highest  # pulled this all the way out past the for loop's end

现在让我们用新代码遍历相同的呼叫

# getMax4([[1, 16, 20, 12])

# unroll the for loop
number = 1
if 1 % 4 == 0:  # it's not
number = 16
if 16 % 4 == 0:  # it is!
    if 16 > 0:  # it is!
        highest = 16  # remember it
number = 20
if 20 % 4 == 0:  # bingo!
    if 20 > 16:  # yup-a-roonie
        highest = 20  # remember this one now. We can forget 16
number = 12
if 12 % 4 == 0:  # batting 1.000
    if 12 > 20:  # not this time, buck-o.

# now we're all done with the for loop, so we step outside and find
return number  # which is 20

现在我们可以稍微清理一下：

def getMax4(list):
    highest = 0
    if len(list) == 0:
        return -999
    # whenever you have an early-exit conditional, there's no reason to
    # write the rest in an `else`. Just omit the `else` and dedent.
    for number in list:
        if number % 4 == 0:
            if number > highest:
                highest = number
    return highest

Answer 2

还有一点要提醒，不要使用内置名称作为变量，例如list，str，int，float，tuple，{ {1}}，如果稍后将这些内置函数调用为函数，则会给您带来讨厌的结果。因此，我建议您将列表更改为数字。

因为只有type会出现在if语句中，所以：

return number

要返回0：

将if number % 4 == 0:唯一标识为您的for循环。

就像

return number

Answer 3

只需对现有代码稍作更改，返回0即可明确添加。

def getMax4(list):
    highest = 0
    if len(list) == 0:
        return -999
    else:
        for number in list:
            if number % 4 == 0:
                if number < highest:
                    highest = number

                else:
                    highest = 0               
                return number

    return 0

列表和最大数量，不使用内置函数

3 个答案: