Loop range with steps

Question

I'm trying to move from BATCH into PowerShell and I am trying to convert my own scripts.

My problem is with ranges in loop: my original BATCH script was like

   for /L %%U in (123,2,323) do ECHO %%U

and will print

123
125
127
...

with Powershell a range would be 123..323, thus

123..323 | % {Echo $_ }

Will give me

123
124
125
...

Is there a way to set a range with a step that is different than 1? All the examples I find either list all the numbers (I have hundreds...) or use the two points between the numbers.

Caramelize

#33892674 – Persephone

Thanks, I could not find this example while searching, now I will test all suggestions – Caramelize

Answer 1

Simple Google search should have found you this.

for ($i=123; $i -le 323; $i=$i+2 ) {Write-Host $i;}

(Initialize; condition to keep the loop running; iteration)

Seafowl

Thank you, I suspect I could not find this solution because I was searching for the wrong thing. Now I will study how to pass the variable onward. – Caramelize

@Squashman, you can also use +=2, (similarly to Set /A i+=2 with cmd.exe), e.g. For ($i=123; $i -LE 323; $i+=2) {$i}. – Housman

@Seafowl This solution gave me the desired control over the steps (and was easy to understand since it was like counters in php) and I could successfully do a conversion of a couple of my existing BATCH scripts. Powershell ones proved to be 2x faster, so it was a needed improvement for my scripting! – Caramelize

@Caramelize Powershell does do many tasks with less code then a batch-file, but there are still a lot of things that a batch-file can do more efficiently then PowerShell. There are things that I can write in a batch-file that has 10 times more lines of code but runs quicker then one line of code in Powershell. – Seafowl

actually, google led me here. thank you for adding to the SO docs – Chuff

Answer 2

Using the original approach (Step 2):

(123..323) | % {if( $_ -band 1 ) {$_ }}

Step 3 and greater:

 (123..323) | % {if( !($i++ % 3) ) {$_ }} 

Lentha

I hardly see a resemblance with an iterating for with start, step and end values, even if your approach has the same effect. – Aboutship

Ok, in this case (123..323) | % {if( !($i++ % 3) ) {$_ }} is working better, but I like the approach with the fast binary operator. – Lentha

I have problem in understanding how to change the step from 2 into another given number. – Caramelize

See comment before. – Lentha

Answer 3

A more concise way to create a range with a step value of n assuming variable $n contains the step value would be:

1..25 | ? { !($_ % $n) } 

The ? is an alias for the Where-Object cmdlet. And the $_ % $n can be written more explicitly as $_ % $n -eq 0 and hence the above can also be written as:

1..25 | Where-Object { $_ % $n -eq 0 }

Where-Object acts as a filter and only includes the values whose modulus with the step equals 0: $_ % $n -eq 0. The reason this can be written as !($_ % $n) is that PowerShell treats 0 as false and other numbers as true. So, negating the 0 with ! makes it evaluate to true and then causes the number to be included in the range.

So, for a step of 2 you would write:

123..323 | ? { !($_ % 2) }  

This includes only even numbers (124, 126, ...). To include only odd numbers, (starting from 123), a simple trick is:

123..323 | ? { $_ % 2 }

When the modulus evaluates to 1, which happens for an odd number, it is included in the final range.

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