If you have the next values:
30
24
21
26
23
19
17
11
6
10
How is it possible to do a peak2peak starting from 4 (thus value 26) untill the last value? So in this case the answer should be 20.
Thanks!
You can simply use only the values starting from 4th value of your vector:
a = [30,24,21,26,23,19,17,11,6,10];
peak2peak(a(4:end))
Related
I've made a calendar sheet and would like to fill it using an Arrayformula or some kind of Lookup.
The problem is, the code in each cell is different, do I need it all to be the same code or is it possible to do an Arrayformula that does a different formula for each line?
I spent ages getting the calendar code working but would now like to simplify the code and I'm not sure what my next step should be:
https://docs.google.com/spreadsheets/d/1u_J7bmOFyDlYXhcL5dW3CHFJ1esySAKK_yPc6nFTdLA/edit?usp=sharing
Any advice would be much appreciated.
I've added a new sheet in your file called 'Aresvik'.
The green cells have new formula.
Cell B3 can be =date(B1,1,1)
Then each successive month can be =eomonth(B3,0)+1, =eomonth(J3,0)+1 etc.
The date formula in cell B5 is:
=arrayformula(iferror(vlookup(sequence(7,7,1),{array_constrain(sequence(40,1),day(eomonth(B3,0))+weekday(B3,3),1),query({flatten(split(rept(",",day(eomonth(B3,0))-1),",",0,0));sequence(day(eomonth(B3,0)),1,1)},"offset "&day(eomonth(B3,0))-weekday(B3,3)&" ",0)},2,false),))
It can be copied to each other cell below Mo, so B5 will change to J5, R5, Z5 etc.
Notes
The concept revolves around using the SEQUENCE function to generate a grid of numbers, 6 rows, 7 columns:
sequence(6,7)
which looks like this:
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35
36 37 38 39 40 41 42
Then using these numbers in a VLOOKUP to get a corresponding date for the calendar. If the first of the month falls on a Thursday (April 2021), the vlookup range needs 3 gaps at the top of the list of dates. player0 has a more elegant solution than my original query using offset, so I've incorporated it below. Cell Z3 is the date 1/4/2021:
=arrayformula(
iferror(
vlookup(sequence(6,7),
{sequence(day(eomonth(Z3,0))+weekday(Z3,2),1,0),
{iferror(sequence(weekday(Z3,2),1)/0,);sequence(day(eomonth(Z3,0)),1,Z3)}},
2,false)
,))
The first column in the vlookup range is:
sequence(day(eomonth(Z3,0))+weekday(Z3,2),1,0)
which is an array of numbers from 0, corresponding with the number of days in the month plus the number of gaps before the 1st day.
The second column in the vlookup range is:
{iferror(sequence(weekday(Z3,2),1)/0,);sequence(day(eomonth(Z3,0)),1,Z3)}},
It is an array of 2 columns in this format: {x;y}, where y sits below x because of the ;.
These are the gaps: iferror(sequence(weekday(Z3,2),1)/0,), followed by the date numbers: sequence(day(eomonth(Z3,0)),1,Z3)
(Example below is April 2021):
0
1
2
3
4
5
6 44317
7 44318
8 44319
9 44320
10 44321
11 44322
12 44323
13 44324
14 44325
15 44326
16 44327
17 44328
18 44329
19 44330
20 44331
21 44332
22 44333
23 44334
24 44335
25 44336
26 44337
27 44338
28 44339
29 44340
30 44341
31 44342
32 44343
33 44344
34 44345
35 44346
36 44347
The vlookup takes each number in the initial sequence (6x7 layout), and brings back the corresponding date from col2 in the range, based on a match in col1.
When the first day of the month is a Monday, iferror(sequence(weekday(BB1,2),1)/0,) generates a gap in col2 of the vlookup range. This is why col1 in the vlookup range has to start with 0.
I've updated the sheet at https://docs.google.com/spreadsheets/d/1u_J7bmOFyDlYXhcL5dW3CHFJ1esySAKK_yPc6nFTdLA/edit#gid=68642071
Values on the calendar are dates so the formatting has to be d.
If you want numbers, then use:
=arrayformula(
iferror(
vlookup(sequence(6,7),
{sequence(day(eomonth(Z3,0))+weekday(Z3,2),1,0),
{iferror(sequence(weekday(Z3,2),1)/0,);sequence(day(eomonth(Z3,0)),1)}},
2,false)
,))
shorter solution:
=INDEX(IFNA(VLOOKUP(SEQUENCE(6, 7), {SEQUENCE(DAY(EOMONTH(B3, ))+WEEKDAY(B3, 2), 1, ),
{IFERROR(ROW(INDIRECT("1:"&WEEKDAY(B3, 2)))/0); SEQUENCE(DAY(EOMONTH(B3, )), 1, B3)}}, 2, )))
for a=1:50; %numbers 1 through 50
for b=1:50;
c=sqrt(a^2+b^2);
if c<=50&c(rem(c,1)==0);%if display only if c<=50 and c=c/1 has remainder of 0
pyth=[a,b,c];%pythagorean matrix
disp(pyth)
else c(rem(c,1)~=0);%if remainder doesn't equal to 0, omit output
end
end
end
answer=
3 4 5
4 3 5
5 12 13
6 8 10
7 24 25
8 6 10
8 15 17
9 12 15
9 40 41
10 24 26
12 5 13
12 9 15
12 16 20
12 35 37
14 48 50
15 8 17
15 20 25
15 36 39
16 12 20
16 30 34
18 24 30
20 15 25
20 21 29
21 20 29
21 28 35
24 7 25
24 10 26
24 18 30
24 32 40
27 36 45
28 21 35
30 16 34
30 40 50
32 24 40
35 12 37
36 15 39
36 27 45
40 9 41
40 30 50
48 14 50
This problem involves the Pythagorean theorem but we cannot use the built in function so I had to write one myself. The problem is for example columns 1 & 2 from the first two rows have the same numbers. How do I code it so it only deletes one of the rows if the columns 1 and 2 have the same number combination? I've tried unique function but it doesn't really delete the combinations. I have read about deleting duplicates from previous posts but those have confused me even more. Any help on how to go about this problem will help me immensely!
Thank you
welcome to StackOverflow.
The problem in your code seems to be, that pyth only contains 3 values, [a, b, c]. The unique() funcion used in the next line has no effect in that case, because only one row is contained in pyth. another issue is, that the values idx and out are calculated in each loop cycle. This should be placed after the loops. An example code could look like this:
pyth = zeros(0,3);
for a=1:50
for b=1:50
c = sqrt(a^2 + b^2);
if c<=50 && rem(c,1)==0
abc_sorted = sort([a,b,c]);
pyth = [pyth; abc_sorted];
end
end
end
% do final sorting outside of the loop
[~,idx] = unique(pyth, 'rows', 'stable');
out = pyth(idx,:);
disp(out)
a few other tips for writing MATLAB code:
You do not need to end for or if/else stements with a semicolon
else statements cover any other case not included before, so they do not need a condition.
Some performance reommendations:
Due to the symmetry of a and b (a^2 + b^2 = b^2 + a^2) the b loop could be constrained to for b=1:a, which would roughly save you half of the loop cycles.
if you use && for contencation of scalar values, the second part is not evaluated, if the first part already fails (source).
Regards,
Chris
You can also linearize your algorithm (but we're still using bruteforce):
[X,Y] = meshgrid(1:50,1:50); %generate all the combination
C = (X(:).^2+Y(:).^2).^0.5; %sums of two square for every combination
ind = find(rem(C,1)==0 & C<=50); %get the index
res = unique([sort([X(ind),Y(ind)],2),C(ind)],'rows'); %check for uniqueness
Now you could really optimized your algorithm using math, you should read this question. It will be useful if n>>50.
I got a dataset (azimuth vs time) with measure the compass of an object trough time. So I can see when the object is moving (the compass vary so much), and when it's static, without moving (compass do not vary). My question is how to program this in matlab in order to eliminate the data which show that the object is moving and just filter data that shows the object is static.
For example:
Azimuth (angle) | 30 30 30 15 10 16 19 24 24 24 17 14 12 15 16
Time (s) | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
The output would be:
Azimuth (angle) | 30 30 30 24 24 24
Time (s) | 1 2 3 8 9 10
s=diff(Azumuth)==0
%diff only would skip the values at t=1 and t=8. Modify to include them as well:
s=[s(1),s(2:end)|s(1:end-1),s(end)]
Azumuth(s)
Time(s)
I have a loop:
for i=1:size(A,1),
if A(i,4:6) == [0,0,3.4]
K = [K; A(i,:)];
end
end
and I would like to delete the last row in the matrix but I do not know what number row it will be. How do I delete the last row in the matrix in the loop? Or should I do it after the loop?
Why do you have loop? it is a one time action, not something you do several times.
check this out, I delete the last row:
>> a = magic(5);
>> a
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
>> a = a(1:end-1,:);
>> a
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
you can refer to last row by END keyword:
A= A(1:end-1, :)
I have a large number of entries arranged in three columns. Sample of the data is:
A=[1 3 2 3 5 4 1 5 ;
22 25 27 20 22 21 23 27;
17 15 15 17 12 19 11 18]'
I want the first column (hours) to control the entire matrix to create new matrix as follows:
Anew=[1 2 3 4 5 ; 22.5 27 22.5 21 24.5; 14 15 16 19 15]'
Where the 2nd column of Anew is the average value of each corresponding hour for example:
from matrix A:
at hour 1, we have 2 values in 2nd column correspond to hour 1
which are 22 and 23 so the average is 22.5
Also the 3rd column: at hour 1 we have 17 and 11 and the
average is 14 and this continues to the hour 5 I am using Matlab
You can use ACCUMARRAY for this:
Anew = [unique(A(:,1)),...
cell2mat(accumarray(A(:,1),1:size(A,1),[],#(x){mean(A(x,2:3),2)}))]
This uses the first column A(:,1) as indices (x) to pick the values in columns 2 and 3 for averaging (mean(A(x,2:3),1)). The curly brackets and the call to cell2mat allow you to work on both columns at once. Otherwise, you could do each column individually, like this
Anew = [unique(A(:,1)), ...
accumarray(A(:,1),A(:,2),[],#mean), ...
accumarray(A(:,1),A(:,3),[],#mean)]
which may actually be a bit more readable.
EDIT
The above assumes that there's no missing entry for any of the hours. It will result in an error otherwise. Thus, a more robust way to calculate Anew is to allow for missing values. For easy identification of the missing values, we use the fillval input argument to accumarray and set it to NaN.
Anew = [(1:max(A(:,1)))', ...
accumarray(A(:,1),A(:,2),[],#mean,NaN), ...
accumarray(A(:,1),A(:,3),[],#mean,NaN)]
You can use consolidator to do the work for you.
[Afinal(:,1),Afinal(:,2:3)] = consolidator(A(:,1),A(:,2:3),#mean);
Afinal
Afinal =
1 22.5 14
2 27 15
3 22.5 16
4 21 19
5 24.5 15