如果在循環中滿足特定條件，如何返回參數？

Question

import random


def lottery(lucky_numbers, run):
    i = 0
    while i < run:
        x = random.uniform(0, 1) #prints out a random number between 0 and 1
        numbers = lucky_numbers
        NewNumbers = numbers[-1:] + numbers[:-1] #shifts each element in the to the right
        lucky_numbers = NewNumbers
        print(lucky_numbers, x)
        i += 1

lottery([1, 2, 0], 3)

這段代碼輸出如下內容：

>>>>>>>>>>
[0, 1, 2] 0.33016179294984127 
[2, 0, 1] 0.7797639530009745
[1, 2, 0] 0.6292245916315391
>>>>>>>>>>

x值將始終是不同的，因為它們是0到1之間的隨機數。

我試圖添加一個函數，說明如果x是循環中的最小值（最小值），則程序應打印該迭代的列表，例如，在這種情況下，此循環中x的最小值為0.33016179 ..，因此，程序應打印列表[0，1，2]

Answer 1

我只是將信息保存在變量中，並在循環結束后打印出來：

import random

def lottery(lucky_numbers, run):
    i = 0
    min_x = 1
    while i < run:
        x = random.uniform(0, 1) #prints out a random number between 0 and 1           
        numbers = lucky_numbers
        NewNumbers = numbers[-1:] + numbers[:-1] #shifts each element in the to the right
        lucky_numbers = NewNumbers
        if x < min_x:
            min_x = x
            min_lucky_numbers = lucky_numbers
        i += 1        
    print(min_lucky_numbers, min_x)

lottery([1, 2, 0], 3)

Answer 2

您可以創建一個存儲所有x值的“緩存”，然后調用最低值。

cache = []
for _ in range(3):
    x = random.uniform(0, 1)
    cache.append(x)
print min(cache)

Answer 3

要執行您想要的操作，只需將項目存儲在兩個不同的列表中，對它們進行排序並顯示每個項目的firt元素：

import random

luckiest_num = list()
luckiest_list = list()
def lottery(lucky_numbers, run):
    i = 0
    while i < run:
        x = random.uniform(0, 1) 
        numbers = lucky_numbers
        NewNumbers = numbers[-1:] + numbers[:-1] 
        print(NewNumbers, x)
        i += 1
        luckiest_num.append(x)
        luckiest_list.append(NewNumbers)

lottery([1, 2, 0], 3)
luckiest_num.sort()
luckiest_list.sort()

print ("The luckiest item is : ")
print(luckiest_num[0],luckiest_list[0])