創建一個數字的倍數列表

Question

問題：

倍數列表
創建一個 Python 3 函數，該函數接受兩個數字（值、長度）作為參數並返回值的倍數列表，直到列表的大小達到長度為止。

例子
list_of_multiples(value=7, length=5) ➞ [7, 14, 21, 28, 35]

list_of_multiples(value=12, length=10) ➞ [12, 24, 36, 48, 60, 72, 84, 96, 108, 120]

list_of_multiples(value=17, length=6) ➞ [17, 34, 51, 68, 85, 102]

def multiples (value,length):
    """
    Value is number to be multiply
    length is maximum number of iteration up to
    which multiple required.
    """
    for i in range(length):
        out=i
    return i

Answer 1

最 Pythonic 的方式

def multiples(value, length):
    return [*range(value, length*value+1, value)]

print(multiples(7, 5))
# [7, 14, 21, 28, 35]
print(multiples(12, 10))
# [12, 24, 36, 48, 60, 72, 84, 96, 108, 120]
print(multiples(17, 6))
# [17, 34, 51, 68, 85, 102]

Answer 2

Pythonic方式：

def multiples(value, length):
    return [value * i for i in range(1, length + 1)]


print(multiples(7, 5))
# [7, 14, 21, 28, 35]
print(multiples(12, 10))
# [12, 24, 36, 48, 60, 72, 84, 96, 108, 120]
print(multiples(17, 6))
# [17, 34, 51, 68, 85, 102]

Answer 3

def multiples(value, length): list_multiples = [] i = 0 while i < length: list_multiples.append(value*(i+1)) i+=1 return list_multiples

Answer 4

Nite Block給出了小length值 (< 100) 的最佳答案。

但是，如果length變大，使用 numpy 比 python 循環要快得多：

numpy.arange(1, length+1) * value

長度為 1000 時，python 循環的時間幾乎是 numpy 的 4 倍。 見下面的代碼：

import timeit

testcode_numpy = ''' 
import numpy
def multiples_numpy(value, length):
    return numpy.arange(1, length+1) * value
multiples_numpy(5, 1000)
'''

testcode = ''' 
def multiples(value, length):
    return [*range(value, length*value+1, value)]
multiples(5, 1000)
'''

print(timeit.timeit(testcode_numpy))
print(timeit.timeit(testcode))

# Result:
# without numpy: 9.7 s
# with numpy: 2.4 s

Answer 5

簡單/不在線的方式是：

def multiples(value, length):
    l = []
    for i in range(1, length+1):
        l.append(value*i)
    return l