Question

我有兴趣弄清楚每个数字在5位数的数字范围内出现的次数，如01234,01243,01324等.... 我知道发生的时间是120次目前我已经用很多for循环和很多ands编程了它。正如你在代码中看到的那样

number = 5
for a in range(number):
    for b in range(number):
        for c in range(number):
            for d in range(number):
                for e in range(number):
                    if (a != b and a != c and a != d and a != e and
                            b != c and b != d and b != e and c != d and 
                            c != e and d != e):
                        print ('{}{}{}{}{}'.format(a, b, c, d, e))

是否有一种不同的，更好的方式来编写上面的代码？

映入眼帘，

超级苍蝇

Answer 1

有人建议使用itertools。以下是要使用的代码：

from itertools import permutations
#generator
factorials = permutations(set(range(5)))
for fact in factorials:
    print(fact)

如果您不想使用itertools，但想生成每个字符串，可以使用列表推导将所有字符串放在一行中。

rn = range(number)
number= 5

sequences = [(a,b,c,d,e) for a in rn for b in rn for c in rn for d in rn for e in rn if a != b and a!= c and a!= d and a!= e and b!= c and b!=d and b!=e and c!=d and c!= e and d!= e]

或者，您可以定义一种递归创建新列表和追加项目的方法

def rGenFactorials(L, leng=5, collected = [], restrictions=set()):

    # base case
    if len(L) == leng:
        collected.append(L)
        return

    #induction
    options = set(range(leng))
    options  -= restrictions

    for op in options:

        # copy restrictions
        r = set(restrictions)
        # add new restriction
        r.add(op)

        # copy list
        l = L[:]
        # append unused int
        l.append(op)

        # recursive invocation
        rGenFactorials(L=l,collected = collected, restrictions=r)


C = []
rGenFactorials([], collected = C)
for c in C:
    print(c)

数字范围内的唯一出现次数

1 个答案: