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simple sort question

nosaj

I'm trying to sort a list of strings with sort alphalessp, can anyone explain why "l" in the list is dropped in the following example?

How may I sort the list without losing elments AND changing the original list?

I tried saving list a to a tmp list, that doesn't seem to help.  

Thanks in advance!

Commands :

a=list("res" "sumW" "w" "sumL" "l" "rs" "segSp" "srs")
length(a)
sort(a 'alphalessp)
length(a)
a

CIW Output :

a=list("res" "sumW" "w" "sumL" "l" "rs" "segSp" "srs")
("res" "sumW" "w" "sumL" "l"
    "rs" "segSp" "srs"
)
length(a)
8
sort(a 'alphalessp)
("l" "res" "rs" "segSp" "srs"
    "sumL" "sumW" "w"
)
length(a)
7
a
("res" "rs" "segSp" "srs" "sumL"
    "sumW" "w"
)

Parents

  • The sort() function is not dropping anything. You are supposed to use the return value of sort(), and not depend upon the side effects on the variable passed in.

    It's actually easy enough to understand what is happening without understanding the sort algorithm. The variable a points to a list cell containing the value "res" and the pointer for that list cell points to another list cell containing the value "sumW" and so on (before the sort).

    The sort will re-use the existing list cells, and just destructively modify the pointers - this avoids re-allocating any memory. So after the sort, the list cell containing "res" will point to the list cell containing "rs" (the next in the sorted sequence). In addition, the list cell containing "l" will point to the list cell containing "res" and the return value of the sort will be a pointer to the list cell containing "l" as that's the first in the list.

    The variable a is still pointing at the list cell containing "res" - the sort cannot change that pointer because it has no means of knowing what variables are pointing to the original list.

    So the variable a is pointing part way along the sorted list.

    What will happen is that when you call sort and do not re-assign the variable to the return value, the car() of the variable passed in will remain the same before and after the sort. You'll simply get the remainder of the sorted list from that point onwards. Since "res" is the second entry in the sorted list, you just get the list from that point onwards.

    Since this is a side effect, the only sensible thing to do is use the return value of the function, as is described in the documentation - which also indicates that sort() is a destructive function.

    Hope that clarifies things?

    Regards,

    Andrew.

Reply

  • The sort() function is not dropping anything. You are supposed to use the return value of sort(), and not depend upon the side effects on the variable passed in.

    It's actually easy enough to understand what is happening without understanding the sort algorithm. The variable a points to a list cell containing the value "res" and the pointer for that list cell points to another list cell containing the value "sumW" and so on (before the sort).

    The sort will re-use the existing list cells, and just destructively modify the pointers - this avoids re-allocating any memory. So after the sort, the list cell containing "res" will point to the list cell containing "rs" (the next in the sorted sequence). In addition, the list cell containing "l" will point to the list cell containing "res" and the return value of the sort will be a pointer to the list cell containing "l" as that's the first in the list.

    The variable a is still pointing at the list cell containing "res" - the sort cannot change that pointer because it has no means of knowing what variables are pointing to the original list.

    So the variable a is pointing part way along the sorted list.

    What will happen is that when you call sort and do not re-assign the variable to the return value, the car() of the variable passed in will remain the same before and after the sort. You'll simply get the remainder of the sorted list from that point onwards. Since "res" is the second entry in the sorted list, you just get the list from that point onwards.

    Since this is a side effect, the only sensible thing to do is use the return value of the function, as is described in the documentation - which also indicates that sort() is a destructive function.

    Hope that clarifies things?

    Regards,

    Andrew.

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