If you have a list and another list with indices (limited number) in of the first list in ascending order.
How can you get a sum of elements in the first list between consecutive indices in the second list.
e.g:
list1: til 100;
idx: (1 20 50 70 100);
How can we get a list with sum of elements of list from elements 1:20, 20:50, 50:70, 70:100?
The obvious approach would be to use # and _ on elements of the idx but can we do that iteratively somehow without using first, first 1_idx etc.
Something like this would work:
q)sum each idx cut list1
190 1035 1190 2535 0
cut operates by cutting the second argument at the indices given in the first. Hence why you see the 0 at the end of the result, as it's cutting at the 100th element.
Related
I've come across the requirement to collect the percentiles from a list a few times:
Within what percentile is a certain number?
What is the nth percentile in a list?
I have written these methods to solve the issue:
/for 1:
percentileWithinThreshold:{[threshold;list] (100 * count where list <= threshold) % count list};
/for 2:
thresholdForPercentile:{[percentile;list] (asc list)[-1 + "j"$((percentile % 100) * count list)]};
They work well for both use cases, but I was thinking this is a too common use case, so probably Q offers already something out of the box that does the same. Any idea if there already exists something else?
'100 xrank' generates percentiles.
q) 100 xrank 1 2 3 4
q) 0 25 50 75
Solution for your second requirement:
q) f:{ y (100 xrank y:asc y) bin x}
Also, note that your second function result will not be always same as xrank. Reason for that is 'xrank' uses floor for fractional index output which is the normal scenario with calculating percentiles and your function round up the value and subtracts -1 which ensures that output will always be lesser-equal to input percentile. For example:
q) thresholdForPercentile[63;til 21] / output 12
q) f[63;til 21] / output 13
For first requirement, there is no inbuilt function. However you could improve your function if you keep your input list sorted because in that case you could use 'bin' function which runs faster on big lists.
q) percentileWithinThreshold:{[threshold;list] (100 * 1+list bin threshold) % count list};
Remember that 'bin' will throw type error if one argument is of float type and other is an integer. So make sure to cast them correctly inside the function.
qtln:{[x;y;z]cf:(0 1;1%2 2;0 0;1 1;1%3 3;3%8 8) z-4;n:count y:asc y;?[hf<1;first y;last y]^y[hf-1]+(h-hf)*y[hf]-y -1+hf:floor h:cf[0]+x*n+1f-sum cf}
qtl:qtln[;;8];
I am trying to realize my idea in matlab.
I consider two column A and B.
A=data(:,1)
B=data(:,5)
the data look like:
A B
1 1
2 1
3 1
... ...
100 20
... ...
150 30
151 1
... ...
The values in column A are timepoints.
I start with the first element in column A. It schould be A(1,1) and look on the first element in the column B B(1,1). If B(1,1)==1its true,if not its false. Then I increase consider the second raw of the column A and second raw of the column B and so on until the last raw of A and B.
How can I construck this loop??
You can just consider B likes the following:
result = (B == 1);
The result would be the same size of B such as you want. Nowm you can get the value of A on result likes the following:
valid_times = A(result);
I have a matrix (4096x4) containing all possible combinations of four values taken from a pool of 8 numbers.
...
3 63 39 3
3 63 39 19
3 63 39 23
3 63 39 39
...
I am only interested in the rows of the matrix that contain four unique values. In the above section, for example, the first and last row should be removed, giving us -
...
3 63 39 19
3 63 39 23
...
My current solution feels inelegant-- basically, I iterate across every row and add it to a result matrix if it contains four unique values:
result = [];
for row = 1:size(matrix,1)
if length(unique(matrix(row,:)))==4
result = cat(1,result,matrix(row,:));
end
end
Is there a better way ?
Approach #1
diff and sort based approach that must be pretty efficient -
sortedmatrix = sort(matrix,2)
result = matrix(all(diff(sortedmatrix,[],2)~=0,2),:)
Breaking it down to few steps for explanation
Sort along the columns, so that the duplicate values in each row end up next to each other. We used sort for this task.
Find the difference between consecutive elements, which will catch those duplicate after sorting. diff was the tool for this purpose.
For any row with at least one zero indicates rows with duplicate rows. To put it other way, any row with no zero would indicate rows with no duplicate rows, which we are looking to have in the output. all got us the job done here to get a logical array of such matches.
Finally, we have used matrix indexing to select those rows from matrix to get the expected output.
Approach #2
This could be an experimental bsxfun based approach as it won't be memory-efficient -
matches = bsxfun(#eq,matrix,permute(matrix,[1 3 2]))
result = matrix(all(all(sum(matches,2)==1,2),3),:)
Breaking it down to few steps for explanation
Find a logical array of matches for every element against all others in the same row with bsxfun.
Look for "non-duplicity" by summing those matches along dim-2 of matches and then finding all ones elements along dim-2 and dim-3 getting us the same indexing array as had with our previous diff + sort based approach.
Use the binary indexing array to select the appropriate rows from matrix for the final output.
Approach #3
Taking help from MATLAB File-exchange's post combinator
and assuming you have the pool of 8 values in an array named pool8, you can directly get result like so -
result = pool8(combinator(8,4,'p'))
combinator(8,4,'p') basically gets us the indices for 8 elements taken 4 at once and without repetitions. We use these indices to index into the pool and get the expected output.
For a pool of a finite number this will work. Create is unique array, go through each number in pool, count the number of times it comes up in the row, and only keep IsUnique to 1 if there are either one or zero numbers found. Next, find positions where the IsUnique is still 1, extract those rows and we finish.
matrix = [3,63,39,3;3,63,39,19;3,63,39,23;3,63,39,39;3,63,39,39;3,63,39,39];
IsUnique = ones(size(matrix,1),1);
pool = [3,63,39,19,23,6,7,8];
for NumberInPool = 1:8
Temp = sum((matrix == pool(NumberInPool))')';
IsUnique = IsUnique .* (Temp<2);
end
UniquePositions = find(IsUnique==1);
result = matrix(UniquePositions,:)
I have a question : I know merge two list in SML but i can not do the total number of elements of the first list and the second list is less than n, append them fully and return the resulting list appended with 0’s, totaling n elements.
Sample :
f2([1,4,5],[3,6],7);
val it = [1,4,5,3,6,0,0] : int list // 7 elements
Thank you in advance..
Get the length of the two lists with LIST.length and compare the sum with the third argument.
I am not sure what to do if this sum is larger than the third argument but you get the idea.
if sum < n then list1 # list2 # 0.....
So, presume a matrix like so:
20 2
20 2
30 2
30 1
40 1
40 1
I want to count the number of times 1 occurs for each unique value of column 1. I could do this the long way by [sum(x(1:2,2)==1)] for each value, but I think this would be the perfect use for the UNIQUE function. How could I fix it so that I could get an output like this:
20 0
30 1
40 2
Sorry if the solution seems obvious, my grasp of loops is very poor.
Indeed unique is a good option:
u=unique(x(:,1))
res=arrayfun(#(y)length(x(x(:,1)==y & x(:,2)==1)),u)
Taking apart that last line:
arrayfun(fun,array) applies fun to each element in the array, and puts it in a new array, which it returns.
This function is the function #(y)length(x(x(:,1)==y & x(:,2)==1)) which finds the length of the portion of x where the condition x(:,1)==y & x(:,2)==1) holds (called logical indexing). So for each of the unique elements, it finds the row in X where the first is the unique element, and the second is one.
Try this (as specified in this answer):
>>> [c,~,d] = unique(a(a(:,2)==1))
c =
30
40
d =
1
3
>>> counts = accumarray(d(:),1,[],#sum)
counts =
1
2
>>> res = [c,counts]
Consider you have an array of various integers in 'array'
the tabulate function will sort the unique values and count the occurances.
table = tabulate(array)
look for your unique counts in col 2 of table.