Numpy计算可整除的整数和余数Python

Question

如何编写一个numpy函数，列出一个数字是另一个数字的因子的次数，如果有余数的话。所以数字1和2被10和CCD_ 1整除。因此，当number2除以divisor时，存在7个除数值的整数和2的余数，这在expected results中可见。我不知道我该怎么做这个函数？

def calculating(number, divisor):

number = 70
number2 = 72
number3 = 7
divisor = 10
calculating(number2, divisor)
calculating(number2, divisor)

预期结果：

number1= [10 10 10 10 10 10 10]
number2= [10 10 10 10 10 10 10 2]
number3= [7]

Answer 1

def calculating(number, divisor):
list1=[]
while number - divisor > 0:
list1.append(divisor)
number-=divisor
list1.append(number)
return list1
print(calculating(70,10))
#[10, 10, 10, 10, 10, 10, 10]
print(calculating(72,10))
#[10, 10, 10, 10, 10, 10, 10, 2]
print(calculating(7,10))
#[7]

递归

def calculating(number, divisor,list1=[]):
if number - divisor > 0:
list1.append(divisor)
number-=divisor
calculating(number, divisor,list1)
else:
list1.append(number)
return list1
print(calculating(72,10))
#[10, 10, 10, 10, 10, 10, 10, 2]

Answer 2

如果您不需要尽可能快的实现，以下内容简短易懂：

def calculating(number, divisor):
return np.diff(np.r_[:number:divisor, number])