Question

 List=([1+3j,12j,-1j,12+3j,13,12j,12-3j,-45j,-45-5j,0,-323j,-2+3j])

这是我的功能，我称之为：

kernel void sum_float (global float* sum, global float* values)
{
  float sum_val = 0.0f;

  for (size_t i = 0u; i < get_num_groups(0); ++i)
  {
    size_t index = get_local_id(0) + i * get_enqueued_local_size(0);
    float value = work_group_reduce_add(values[index]);
    sum_val += work_group_broadcast(value, 0u);
  }

  sum[0] = sum_val;
}

该列表定义为：

kernel void sum_float2 (global float* sum, global float* values)
{
      float sum_partial = work_group_reduce_add(values[get_global_id(0)]);
      if(get_local_id(0)==0)
        sum[get_group_id(0)] = sum_partial; 
}

为什么它打印的所有数字不仅是模数<10

的数字

Answer 1

你可以这样做：

op = {'<': '__lt__', '>': '__gt__', '<=': '__le__', '>=': '__ge__'}

def print_numbers_modulo(List,Operator,ModuloNumber):
    CheckIfPrintedNumber=False
    for i in range(0,len(List)):
        if getattr(abs(List[i]), op[Operator])(ModuloNumber):
            print(List[i])
            CheckIfPrintedNumber = True
    if (CheckIfPrintedNumber == False):
        print("There are no numbers having modulo",Operator,ModuloNumber)
List=([1+3j,12j,-1j,12+3j,13,12j,12-3j,-45j,-45-5j,0,-323j,-2+3j])

print_numbers_modulo(List,'<',10)

op是一个字典，它将您的运算符“转换”为float类的相应魔术方法，作为字符串。然后，您可以使用getattr在abs(List[i])上调用该方法，并将ModuloNumber作为参数：

getattr(abs(List[i]), op[Operator])(ModuloNumber)

例如，如果Operator = "<"，则相当于：

abs(List[i]).__lt__(ModuloNumber)

Python Modulo复数

1 个答案: