范围函数以 M 为单位?

Question

是否可以在M中创建数值范围？例如：

let
x = range(1,10) // {1,2,3,4,5,6,7,8,9,10}, from 1 to 10, increment by 1
x = range(1,10,2) // {1,3,5,7,9}, from 1 to 10, increment by 2

Answer 1

对于简单的方案，a..b可能是合适的。一些例子：

let
firstList = {1..10}, // {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
secondList = {1, 5, 12, 14..17, 18..20}, // {1, 5, 12, 14, 15, 16, 17, 18, 19, 20}
thirdList = {Character.ToNumber("a")..100, 110..112}, // {97, 98, 99, 100, 110, 111, 112}
fourthList = {95..(10 * 10)} // {95, 96, 97, 98, 99, 100}
in
fourthList

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否则，可以尝试在内部使用List.Generate的自定义函数：

let
range = (inclusiveStart as number, inclusiveEnd as number, optional step as number) as list => 
let
interval = if step = null then 1 else step,
conditionFunc = 
if (interval > 0) then each _ <= inclusiveEnd
else if (interval < 0) then each _ >= inclusiveEnd
else each false,
generated = List.Generate(() => inclusiveStart, conditionFunc, each _ + interval)
in generated,
firstList = range(1, 10), // {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
secondList = range(1, 10, 2), // {1, 3, 5, 7, 9}
thirdList = range(1, 10 , -1), // {} due to the combination of negative "step", but "inclusiveEnd" > "inclusiveStart"
fourthList = range(10, 1, 0), // {} current behaviour is to return an empty list if 0 is passed as "step" argument.
fifthList = range(10, 1, -1), // {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}
sixthList = range(10, 1, 1) // {} due to the combination of positive "step", but "inclusiveEnd" < "inclusiveStart"
in
sixthList

上面的range函数应该能够生成升序和降序序列(请参阅代码中的示例(。

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或者，您可以创建一个自定义函数，该函数通过首先计算count参数需要是什么来在内部使用List.Numbers。但我选择和List.Generate一起去.