Shifts the values of the record to the right by one

[aborruso]

Hi,
I'm sure it's very easy to do it.

I have this no header CSV

a,2,3,
2,3,,
b,4,5,

And I want to transform it, shifting all values to the right by one, from this (I am adding the header also to make my target more readable)

1	2	3	4
a	2	3
2	3
b	4	5

to this one

1	2	3	4
	a	2	3
	2	3
	b	4	5

I thought of using positional field names in this way

mlr --csv --implicit-csv-header put 'for (k in $*){$[ $[[k+1]] ]=$[ $[[k]] ]}' input.csv

but I get back

Record/map indices must be string, int, or array thereof; got error

And I'm sure there will be a very easy array way to do it.
I'm not able to do it.

Thank you

[aborruso]

And my way has no sense, also correcting it.

Running

mlr --csv --implicit-csv-header put  'for (k in $*){$[[[int(k)+1]]]=$[[[int(k)]]]}' input.csv

I have all equal to the first field

1	2	3	4
a	a	a	a
2	2	2	2
b	b	b	b

[johnkerl]

@aborruso how about:

$ cat shifter.mlr
keys   = get_keys($*);
values = get_values($*);

newrec = {};
newrec[keys[1]] = "";
for (i = 1; i <= length(keys); i += 1) {
    newrec[keys[i+1]] = values[i];
}

$* = newrec;

$ mlr --implicit-csv-header --c2p --from input.csv cat
1 2 3 4
a 2 3 -
2 3 - -
b 4 5 -

$ mlr --implicit-csv-header --c2p --from input.csv put -f shifter.mlr
1 2 3 4
- a 2 3
- 2 3 -
- b 4 5

[johnkerl]

Or this:

keys   = get_keys($*);
values = get_values($*);

newrec = {};
# newrec[keys[1]] = "";
newrec[keys[1]] = values[length(keys)]; # <----- here
for (i = 1; i <= length(keys); i += 1) {
    newrec[keys[i+1]] = values[i];
}

$* = newrec;

will "rotate" the last column's values into the first column

[aborruso]

Thank you very much, great.

shift could be a new verb :)