让python代码更高效

Question

我有密码

num = 1
num2 = 1
num3 = 1
list = []
list2 = []
list3 = []

def numCheck1 (num):
    while num<1001:
        if (num%3==0):
            if (num%5==0):
                print num
                list.append(num)
                num+=1
                numCheck1(num)
                break
            else:
                print "error 1"
        else:
            print "error 2"
        num+=1

numCheck1(num)
total=sum(list)
print list
print total

def numCheck2 (num2):
    while num2<1001:
        if (num2%5==0):
            print num2
            list2.append(num2)
            num2+=1
            numCheck1(num2)
            break
        else:
            print "error"

numCheck2(num2)

def numCheck3 (num3):
    while num3<1001:
        if (num3%3==0):
            print num3
            list3.append(num3)
            num3+=1
            numCheck1(num3)
            break
        else:
            print "error"

numCheck3(num3)

total2 = sum(list2)
total3 = sum(list3)
overall = (total2 + total3) - total
print list2
print list3
print total2
print total3
print overall

作为我的代码的基本摘要，我有 3 个函数，以及每个函数对应的列表和变量。第一个函数检查小于等于 1000 的 3 和 5 的所有倍数。第二个函数检查小于等于 1000 的所有 5 的倍数。第三个函数检查小于等于 1000 的 3 的所有倍数。是倍数的数字被添加到相应的列表中，而相应的变量递增以允许函数检查所有数字。最后，程序计算 4 个总数：每个列表的总数，以及一个特殊总数，它将后两个总数相加并减去第一个总数，以防止过度计数。这只是程序的总体结构。

这个程序应该解决this problem（不是作业，只是好玩）。代码正在工作（据我所知；第一个函数肯定有效）但它不断使编译器崩溃（我正在使用在线编译器，repl。我想知道是否有任何方法可以使这段代码更高效。

谢谢！

Answer 1

我已经重写了您的代码，内嵌了 cmets。您可以在此基础上进行更多改进，但我认为看看您可以在哪些方面改进此代码会对您有所帮助。

# If you use this, it should be the first line
# in your code. If you're using Python2, you *should*
# use this, because it will make switching to Python3
# that much easier.
from __future__ import print_function

# No need for these global variables
#num = 1
#num2 = 1
#num3 = 1
#list = []
#list2 = []
#list3 = []
#

# Functions should be snake_cased, not camelCase
# also, the blank space between the function name
# and the paren is inconsistent with typical
# Python code.
def num_check_one(start):
    # We'll put the return list in here. Also
    # calling it `list = []` would have shadowed
    # a builtin function. Let's give it a better name:
    multiples = []

    # If you're iterating over known values in Python
    # use a for loop over a range instead.
    for num in range(start, 1001):

        # No need to nest your ifs.
        # Parenthesis are usually just extra noise,
        # But you might find it a little clearer with
        # them to clarify the grouping
        #if (num % 3 == 0) and (num % 5 == 0):
        # This could also be written as
        # if (not num % 3) and (not num % 5):
        # Or, using De Morgan's law:
        # if not (num % 3 or num % 5):
        if num % 3 == 0 and num % 5 == 0:
            print(num)
            multiples.append(num)
            # Uh, no need to recursively call
            # your function here.
            #numCheck1(num)
            #break
        # And finally, you don't need any of these
#            else:
#                print "error 1"
#        else:
#            print "error 2"
#        num+=1
    return multiples


# You can use the keyword in your function call
# which makes it clear what the value is for
multiples = num_check_one(start=1)
total=sum(multiples)
print('Multiples:', multiples)
print('Total:', total)

# Again, fixed casing/spacing, and gave the
# parameter a meaningful name
def num_check_two(start):
    multiples = []
    # Again, for loop
    for num in range(start, 10001):
        # 0 is Falsey, so we can
        # just treat it as a bool value
        if not num % 5:
            print(num)
            multiples.append(num)
            # Got rid of the recursive call again
        # You were never incrementing `num2` here,
        # which is why your code got into an infinite
        # loop. Also why you should use `for` loops by
        # default

num_check_two(1)

Answer 2

numCheck1：由于所有 3和5的倍数也是 15的倍数，您可以简单地打印出range(0, 1001, 15)中的每个项目（按15计数）

numCheck2:range(0, 1001, 5)（按 5 计数）应该已经是 5 的所有倍数小于或等于 1000。

numCheck3:range(0, 1001, 3)（按 3 计数）与上述相同，只是 3 的倍数。