如何解决 lambda 函数中的这个 TypeError？ [复制]

Question

在字符串格式化期间出现并非所有参数都转换的错误

def even_odd_lambda(list_object):
  
    #odd_ctr = list(filter(lambda x: (x%2 != 0) , list_1))
    #even_ctr = list(filter(lambda x: (x%2 == 0) , list_1))
    
    odd_ctr = len(list(filter(lambda x: (x%2 != 0) , list_object)))
    even_ctr = len(list(filter(lambda x: (x%2 == 0) , list_object)))

    return [odd_ctr, even_ctr] 


if __name__ == '__main__':
  
    a=input("Add list elements seperated by space ").split(' ')

    
    
    output = even_odd_lambda(a)
    print("\nNumber of even numbers in the above array: ", output[1])
    print("\nNumber of odd numbers in the above array: ", output[0])  

Answer 1

把你的输入变成整数：

a = [*map(int, input("Add list elements seperated by space ").split())]

另外，您应该避免仅仅为了测量它们的长度而构建列表。我也会避免使用map/filter 和lambda；它总是比生成器或理解器更长且可读性更低。比较：

list(filter(lambda x: (x%2 != 0) , list_object))
[x for x in list_object if x%2 != 0]

例如，您可以将sum 与生成器一起使用，以减少内存占用并提高可读性：

def even_odd_lambda(list_object):
    odd_ctr = sum(x%2 for x in list_object)
    even_ctr = sum(not x%2 for x in list_object) 
    return [odd_ctr, even_ctr]