8:24 p.m.
> Hi all
> I am unable to sort a matrix and sorting the same colname.
>
>
> E.G.
>
> matrix M = {Class_A..Class_D}
>
> colnames(M, "Class_A Class_B Class_C Class_D")
>
> A = dsort(M)
>
> printing A I see the right numerbs but the wrong colnames and
> despite my
> research in the Gretl User?s Guide I was unable to find myself the
> solution.
Hi Guido,
what gretl version are you using? According to the dsort()
documentation
it doesn't apply to a matrix that is not a vector (more than one
column). And indeed here for me I get an (expected) error when I try
dsort(m) where m has two columns.
Also it is not clear to me what you want to achieve. Perhaps the
msortby() function is right for you.
cheers,
sven
Dear Sven
thanks for your quick answer.
My Gretl version is 2016-c but it does not matter.
You are right, I didnt tell all the story. But was not for cheating. I
tried to cut short the description of the problem.
I mentioned a matrix, but indeed I was working with a 1 row "matrix".
My real problem is ranking some time series, day by day;
And the trouble is the sorting mismatch the names of the series with
their values.
(E.G.) If I have A, B, C, D daily series and I want (recursively) rank
their values at day t, and then at day t+1, and so on, I am in
trouble,
because all my attempt of sorting does not keep the values with the
original series names, while in Excel is trivial.
I tried in many ways, unsuccessfully.
My last attempt was
a) limiting the whole dataset to only 1 day, (by
smpl)
b) declaring it as a one row matrix
c) giving it the right names
(via colnames)
d) ranking the values
but as before, I got the (values) rank but I lost the match with the names of time
series.
So any help, or reference to the User Guide, will be greatly appreciated.
Guido
9:54 p.m.
New subject: Gretl-users Digest, Vol 117, Issue 6
On Mon, 10 Oct 2016, guido giaume wrote:
Dear Sven
thanks for your quick answer.
My Gretl version is 2016-c but it does not matter.
You are right, I didnt tell all the story. But was not for cheating. I
tried to cut short the description of the problem.
I mentioned a matrix, but indeed I was working with a 1 row "matrix".
My real problem is ranking some time series, day by day;
And the trouble is the sorting mismatch the names of the series with
their values.
(E.G.) If I have A, B, C, D daily series and I want (recursively) rank
their values at day t, and then at day t+1, and so on, I am in
trouble,
because all my attempt of sorting does not keep the values with the
original series names, while in Excel is trivial.
I tried in many ways, unsuccessfully.
My last attempt was
a) limiting the whole dataset to only 1 day, (by
smpl)
b) declaring it as a one row matrix
c) giving it the right names
(via colnames)
d) ranking the values
but as before, I got the (values) rank but I lost the match with the names of time
series.
So any help, or reference to the User Guide, will be greatly appreciated.
I don't think this problem is considered in the User's Guide.
However, it's not beyond hansl's capacity. If I understand you
correctly, the following script should provide a template. We start
with a "raw" matrix M containing 4 "series" (columns) followed by
consective ID numbers of those series. After the loop we have a
matrix Ms where each row is sorted by value of the "series", and the
ID numbers are also reordered to show if it's series 1, 2, 3 or 4
that has the smallest values, the next smallest value, and so on.
<hansl>
scalar T = 100
# raw data
matrix M = mnormal(T,4) ~ mshape(seq(1,4), 4, T)'
matrix Ms = zeros(T,8)
loop i=1..T -q
matrix tmp = msortby((M[i,1:4] | M[i,5:])',1)
Ms[i,] = tmp[,1]' ~ tmp[,2]'
endloop
print M
# sorted data
print Ms
</hansl>
Allin Cottrell
10:51 p.m.
New subject: sorting/ranking (was Re: Gretl-users Digest, Vol 117, Issue 6)
On Mon, 10 Oct 2016, Allin Cottrell wrote:
If I understand you correctly, the following
script should provide a template. [...]
Ah, it struck me you may have something a bit different in mind.
Here's another interpretation: from a raw data matrix with 4 columns
in fixed positions, produce a matrix which shows the rank order of
the data values on each row.
<hansl>
# Example 2
scalar T = 100
matrix M = mnormal(T,4)
matrix Mr = zeros(T,4)
colnames(Mr, "A B C D")
loop i=1..T -q
Mr[i,] = ranking(M[i,])
endloop
print Mr
</hansl>
Allin Cottrell
11 a.m.
New subject: Gretl-users Digest, Vol 117, Issue 6
Am 10.10.2016 um 23:54 schrieb Allin Cottrell:
On Mon, 10 Oct 2016, guido giaume wrote:
> but as before, I got the (values) rank but I lost the match with
the
> names of time series.
I don't think this problem is considered in the User's Guide. However,
it's not beyond hansl's capacity.
Coming back to the problem of column label mismatch, here's a complete
code example:
<hansl>
matrix m = {3,4,1,2}
colnames(m, "a b c d")
print m
matrix m2 = sort(m) # still "a", "b", "c", "d"
print m2
m = sort(m) # still "a", "b", "c", "d"
print m
</hansl>
I'm not saying (yet) it's a bug, but is it really the desired behavior,
when the contents of the vector and the labels become totally
disconnected? Perhaps it would be better to simply delete the labels.
thanks,
sven
1:54 p.m.
New subject: Gretl-users Digest, Vol 117, Issue 6
On Tue, 11 Oct 2016, Sven Schreiber wrote:
Am 10.10.2016 um 23:54 schrieb Allin Cottrell:
> On Mon, 10 Oct 2016, guido giaume wrote:
>> but as before, I got the (values) rank but I lost the match with the
>> names of time series.
>
> I don't think this problem is considered in the User's Guide. However,
> it's not beyond hansl's capacity.
Coming back to the problem of column label mismatch, here's a complete code
example:
<hansl>
matrix m = {3,4,1,2}
colnames(m, "a b c d")
print m
matrix m2 = sort(m) # still "a", "b", "c", "d"
print m2
m = sort(m) # still "a", "b", "c", "d"
print m
</hansl>
I'm not saying (yet) it's a bug, but is it really the desired behavior, when
the contents of the vector and the labels become totally disconnected?
Perhaps it would be better to simply delete the labels.
Thanks for the example. Yes, this was an oversight. Now in git, when
sort() or dsort() is applied to a vector we keep any column names (or
row names, as the case may be) in alignment with the original data
values.
Allin
2757
days inactive
2758
days old
4 comments
3 participants
participants (3)
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Allin Cottrell -
guido giaume -
Sven Schreiber