I know how to 'select' a range in LO (7.2.4.1) Calc BASIC ....
ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
But how to write a value, e.g. "1", into that range using BASIC?
myRange = ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
myRange.Value = 1
Gives an "property or method not found" error. But I can't find any properties or values to go after Range to allow me to do what I want. Flailing around and trying
myRange.setValue = 1
myRange.writeValue = 1
myRange.setString = "1"
and numerous other variants don't work either.
Would really appreciate the solution. Thanks.
You can edit the value of an individual cell, but not the entire range. You will have to iterate over all the cells in the range one at a time, changing the value of each of them.
Sub Set1ToD1H6
myRange = ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
For i = 0 To myRange.getRows().getCount()-1
For j = 0 To myRange.getColumns().getCount()-1
myRange.getCellByPosition(j, i).setValue(1)
Next j
Next i
End Sub
But since the read-write operation to a cell is comparable in time to the read-write operation to a whole range, it is preferable to use another method - to prepare data in an array and write from it to a range in one operation:
Sub Set1ToRange
myRange = ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
dataOfRange = myRange.getData()
For i = LBound(dataOfRange) To UBound(dataOfRange)
For j = LBound(dataOfRange(i)) To UBound(dataOfRange(i))
dataOfRange(i)(j) = 1
Next j
Next i
myRange.setData(dataOfRange)
End Sub
(For your example, this will be approximately 30 times faster, for a larger range the time winnings will be even more significant)
The .getData() and .setData() methods work on numeric range values. To work with text strings (and numbers), use .getDataArray() and .setDataArray(), for working with cell formulas use .getFormulaArray() and .setFormulaArray()
Related
I have an array of 1 x 400, where all element values are above 1500. However, I have some elements that have values<50 which are wrong measures and I would like to have the mean of the elements before and after the wrong measured data points and replace it in the main array.
For instance, element number 17 is below 50 so I want to take the mean of elements 16 and 18 and replace element 17 with the new mean.
Can someone help me, please? many thanks in advance.
No language is specified in the question, but for Python you could work with List Comprehension:
# array with 400 values, some of which are incorrect
arr = [...]
arr = [arr[i] if arr[i] >= 50 else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
That is, if arr[i] is less than 50, it'll be replaced by the average value of the element before and after it. There are two issues with this approach.
If i is the first or last element, then one of the two values will be undefined, and no mean can be obtained. This can be fixed by just using the value of the available neighbour, as specified below
If two values in a row are very low, the leftmost one will use the rightmost one to calculate its value, which will result in a very low value. This is a problem that may not occur for you in practice, but it is an inherent result of the way you wish to recalculate values, and you might want to keep it in mind.
Improved version, keeping in mind the edge cases:
# don't alter the first and last item, even if they're low
arr = [arr[i] if arr[i] >= 50 or i == 0 or i+1 == len(arr) else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
# replace the first and last element if needed
if arr[0] < 50:
arr[0] = arr[1]
if arr[len(arr)-1] < 50:
arr[len(arr)-1] = arr[len(arr)-2]
I hope this answer was useful for you, even if you intend to use another language or framework than python.
Say ...
you have about 20 Thing
very often, you do a complex calculation running through a loop of say 1000 items. The end result is a varying number around 20 each time
you don't know how many there will be until you run through the whole loop
you then want to quickly (and of course elegantly!) access the result set in many places
for performance reasons you don't want to just make a new array each time. note that unfortunately there's a differing amount so you can't just reuse the same array trivially.
What about ...
var thingsBacking = [Thing](repeating: Thing(), count: 100) // hard limit!
var things: ArraySlice<Thing> = []
func fatCalculation() {
var pin: Int = 0
// happily, no need to clean-out thingsBacking
for c in .. some huge loop {
... only some of the items (roughly 20 say) become the result
x = .. one of the result items
thingsBacking[pin] = Thing(... x, y, z )
pin += 1
}
// and then, magic of slices ...
things = thingsBacking[0..<pin]
(Then, you can do this anywhere... for t in things { .. } )
What I am wondering, is there a way you can call to an ArraySlice<Thing> to do that in one step - to "append to" an ArraySlice and avoid having to bother setting the length at the end?
So, something like this ..
things = ... set it to zero length
things.quasiAppend(x)
things.quasiAppend(x2)
things.quasiAppend(x3)
With no further effort, things now has a length of three and indeed the three items are already in the backing array.
I'm particularly interested in performance here (unusually!)
Another approach,
var thingsBacking = [Thing?](repeating: Thing(), count: 100) // hard limit!
and just set the first one after your data to nil as an end-marker. Again, you don't have to waste time zeroing. But the end marker is a nuisance.
Is there a more better way to solve this particular type of array-performance problem?
Based on MartinR's comments, it would seem that for the problem
the data points are incoming and
you don't know how many there will be until the last one (always less than a limit) and
you're having to redo the whole thing at high Hz
It would seem to be best to just:
(1) set up the array
var ra = [Thing](repeating: Thing(), count: 100) // hard limit!
(2) at the start of each run,
.removeAll(keepingCapacity: true)
(3) just go ahead and .append each one.
(4) you don't have to especially mark the end or set a length once finished.
It seems it will indeed then use the same array backing. And it of course "increases the length" as it were each time you append - and you can iterate happily at any time.
Slices - get lost!
I have a data (matrix) with 3 columns : DATA=[ID , DATE, Value]
I want to filter my data by ID for example DATAid1= DATA where ID==1 and so on ..
for that I write this code in MATLAB
load calibrage_capteur.mat
data = [ID ,DATE , Valeur]
minid = min(data(:,1));
maxid = max(data(:,1));
for i=minid:maxid
ind=find(data(:,1) == i)
dataID = [ID(ind) ,DATE(ind) , Valeur(ind)]
end
As a result he register the last value in this example the max ID=31 so he register dataId31. Now I need how to save the variable each iteration. How can I do this?
You will want to use a cell array to hold your data rather than saving them as independent variables that are named based upon the ID.
data_by_ID = cell();
ids = minid:maxid;
for k = 1:numel(ids)
data_by_ID{k} = data(data(:,1) == ids(k),:);
end
Really though, depending on what you're doing with it, you can use data all of the time since all operations are going to be faster on a numeric matrix than they are on a cell array.
%// Do stuff with data ID = 10
do_stuff(data(data(:,1) == 10, :));
Update
If you absolutely must name your variables you could do the following (but please don't do this and use one of the methods above).
for k = 1:numel(ids)
eval(['dataId', num2str(ids(k)), '= data(k,:);']);
end
Your question is a bit unclear but it sounds like you simply want to save the result at each iteration of the for loop.
I'm assuming min and max id are arbitrary and not necessarily the variable you are trying to index on.
kk = min_id:max_id;
dataID=nan(size(kk));
for ii = 1:numel(kk)
ind=find(data(:,1) == kk(ii))
dataID(kk) = [ID(ind) ,DATE(ind) , Valeur(ind)]
end
This is better than indexing by min_id or max_id since it isn't clear that min_id starts at at 1 (maybe it starts at 0, or something else.)
include "globals.mzn";
%Data
time_ID = [11,12,13,14,15];
eventId = [0011, 0012, 0013, 0021, 0022, 0031, 0041, 0051, 0061, 0071];
int:ntime = 5;
int:nevent = 10;
set of int: events =1..nevent;
set of int: time = 1..ntime;
array[1..nevent] of int:eventId;
array[1..nevent] of var time:event_time;
array[1..ntime] of int:time_ID;
solve satisfy;
constraint
forall(event in eventId)(
exists(t in time_ID)(
event_time[event] = t ));
output[ show(event_time) ];
I'm trying to assign times to an event using the code above.
But rather than randomly assign times to the events, it returns an error " array access out of bounds"
How can I make it select randomly from the time array?
Thank you
The error was because you tried to assign the index 11 (the first element in eventId array) in "event_time" array.
The assigment of just 1's is correct since you haven't done any other constraints on the "event_time" array. If you set the number of solutions to - say - 3 you will see other solutions. And, in fact, the constraint as it stand now is not really meaningful since it just ensures that there is some assignment to the elements in "event_time", but this constraint is handled by the domain of "event_time" (i.e. that all indices are in the range 1..ntime).
We've got an array of values, and we would like to create another array whose values are not in the first one.
Example:
load('internet.mat')
The first column contains the values in MBs, we have thought in something like:
MB_no = setdiff(v, internet(:,1))
where v is a 0 vector whose length equals to the number of rows in internet.mat. But it just doesn't work.
So, how do we do this?
You need to specify the range of possible values to define what values are not in internet . Say the range is v = 1:10 then setdiff(v,internet(:,1)) will give you the values in 1:10 that are not in the first column of internet.
It seems as if you don't want the first column.
You can simply do:
MB_no=internet(:,2:end);
assuming internet(:,1) has only positive integers and you wish to find which are the integers in [1,...,max( internet(:,1) )] that do not appear in that range you can simply do
app = [];
app( internet(:,1) ) = 1;
MB_no = find( app == 0 );
This is somewhat like bucket sort.