100的随机数的奇数和偶数计数器

Question

我有以下分配：

在本章中，您看到了如何编写，确定一个数是否 为偶数或奇数的算法的例子。编写一个程序产生100个随机数，并保持 这些随机数有多少是偶数，有多少是奇数。

这是我已经能够得到多远，我可以得到100个数字，但我无法弄清楚如何总计奇数和平均数。这应该包括一个值返回布尔函数。

我们允许使用的所有东西是循环，if-elif-else，函数和其他基本的东西。

import random 

NUMBER_LIST = [random.randint(0,1000)] 

def main(): 

    for numbers in range(100): 

     number = print(NUMBER_LIST) 
     number 

    is_even(number) 

    print('The total amount of even numbers is', even_count) 
    print('The total amount of odd numbers is', 100 - even_count) 

def is_even(number): 

    even_count = 0 

    for number in NUMBERS_LIST: 
     if (number % 2): 
      even_count += 1 

    return even_count 

main() 

编辑： 我不应该用一个列表，所以如果那里有一个办法做到这一点没有，让我知道！

Answer 1

import random 

def main(): 
    numberList = [] # create an empty list, to add 100 random ints to 
    for i in range(100): 
     numberList.append(random.randint(1,1000)) # add a random int 

    # now, numberList has 100 random numbers in it 

    # keep track of how many odd numbers 
    oddCount = 0 

    # loop through numberList 
    for number in numberList: 
     if number%2 == 1: # number is odd 
      oddCount += 1 

    evenCount = 100 - oddCount # if a number is not odd, it is not even 

    print("There are", oddCount, "odd numbers, and", evenCount, "even numbers") 

好了，现在我们有一个硬编码的版本，让我们尝试，允许你指定尽可能多的东西可以更灵活的方式：

def main(numNumbers, smallestNumber, biggestNumber): 
    numberList = [] 
    for i in range(numNumbers): 
     numberList.append(random.randint(smallestNumber, biggestNumber)) 
    oddCount = 0 
    for number in numberList: 
     if number%2: # oh look, I didn't have to do the `== 1` part 
      oddCount += 1 
    evenCount = numNumbers - oddCount 
    print("There are", oddCount, "odd numbers, and", evenCount, "even numbers") 

Answer 2

import random 

NUMBER_LIST = [random.randint(0,1000)] 
even = 0; 
odd = 0; 
for numbers in range(100): 
    if (numbers%2 == 1): 
     odd = odd+1 
    if (numbers%2 == 0): 
     even = even+1 
print('number of evens is: ',even) 
print('number of odds is: ',odd) 

所以你可以做这种事情。

Answer 3

您可以在没有列表的情况下执行此操作，但由于您的问题可能需要，请执行此操作。

首先，请注意您的代码只是在其中创建一个带有一个随机数的列表。如果你想填充与100个随机号的列表，你必须做一些类似这样：

NUMBER_LIST = [] 
i = 0 
while i < 100: 
    number = random.randint(0, 1000) 
    NUMBER_LIST.append(number) 
    i += 1 

然后，检查数为偶数，与number % 2 == 0（也就是说，数的除法的余2 0这将返回true或false），并增加相应的计数器：

NUMBER_LIST = [] 
# initialize both counters 
evens = 0 
odds = 0 
i = 0 
while i < 100: 
    number = random.randint(0, 1000) 
    NUMBER_LIST.append(number) 
    if number % 2 == 0: 
     evens += 1 
    else: 
     odds += 1 
    i += 1 

然后，你只需要打印计数：

print("The number of even numbers is: " + evens) 
print("The number of odd numbers is: " + odds) 

的FUL那么L代码如下：

import random 
NUMBER_LIST = [] 
evens = 0 
odds = 0 
i = 0 
while i < 100: 
    number = random.randint(0, 1000) 
    NUMBER_LIST.append(number) 
    if number % 2 == 0: 
     evens += 1 
    else: 
     odds += 1 
    i += 1 
print("The numbers were: " + str(NUMBER_LIST)) 
print("The number of even numbers is: " + evens) 
print("The number of odd numbers is: " + odds) 

而且没有名单：

import random 
evens = 0 
odds = 0 
i = 0 
while i < 100: 
    number = random.randint(0, 1000) 
    if number % 2 == 0: 
     evens += 1 
    else: 
     odds += 1 
    i += 1 
print("The number of even numbers is: " + evens) 
print("The number of odd numbers is: " + odds) 

Answer 4

#!/usr/bin/env python3 

import random 
def main(n=100): 
    NUMBER_LIST = [random.randint(0,1000) for x in range(0,n)] 
    odds = len(list(filter(lambda x: x % 2, NUMBER_LIST))) 
    print("Odd Numbers: {}\nEven Numbers: {}".format(odds, n-odds)) 

if __name__ == "__main__": 
    main() 

Answer 5

from random import randrange 

even = 0 
for i in range(100): 
    num = randrange(1000) 
    if num % 2 == 0: 
     even += 1 

print('There were {0} even numbers and {1} odd numbers.'.format(even, 100-even)) 

Answer 6

我在同一个班！此代码为我工作。

import random 

def main(): 

    counter = 1 
    even_numbers = 0 
    odd_numbers = 0 
    while counter < 101: 

     a = random.randint(1,100) 


     if a % 2 == 0: 
      even_numbers += 1 
     else: 
      odd_numbers += 1 
     counter += 1 

    if counter == 101 : 
     reveal_total(even_numbers, odd_numbers) 

def reveal_total(even_numbers, odd_numbers): 
    print("This many evens : ", even_numbers) 
    print("This many odds : ", odd_numbers) 

main()