As simple as in title. I have nx1 sized vector p. I'm interested in the maximum value of r = p/foo - floor(p/foo), with foo being a scalar, so I just call:
max_value = max(p/foo-floor(p/foo))
How can I get which value of p gave out max_value?
I thought about calling:
[max_value, max_index] = max(p/foo-floor(p/foo))
but soon I realised that max_index is pretty useless. I'm sorry asking this, real beginner here.
Having dropped the issue to pieces, I realized there's no unique corrispondence between values p and values in my related vector p/foo-floor(p/foo), so there's a logical issue rather than a language one.
However, given my input data, I know that the solution is unique. How can I fix this?
I ended up doing:
result = p(p/foo-floor(p/foo) == max(p/foo-floor(p/foo)))
Looks terrible, so if you know any other way...
Once you have the index, use it:
result = p(max_index)
You can create a new vector with your lets say "transformed" values:
p2 = (p/foo-floor(p/foo))
and then just use find to find the max values on p2:
max_index = find(p2 == max(p2))
that will return the index or indices of p2 with the max value of that operation, and finally just lookup the original value in p
p(max_index)
in 1 line, this is:
p(find((p/foo-floor(p/foo) == max((p/foo-floor(p/foo))))))
which is basically the same thing you did in the end :)
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.
Here's the thing: I want to modify (and then return) a matrix of integers that is given in the parameters of the function. The funcion average (of the class MatrixMotionBlur) gives the average between the own pixel, upper, down and left pixels. Follows the following formula:
result(x, y) = (M1(x, y)+M1(x-1, y)+M1(x, y-1)+M1(x, y+1)) / 4
This is the code i've implemented so far
MatrixMotionBlur - Average function
MotionBlurSingleThread - run
The objetive here is to apply "average" method to alter the matrix value and return that matrix. The thing is the program gives me error when I to insert the value on the matrix.
Any ideas how to do this ?
The functional way
val updatedData = data.map{ outter =>
outter(i).map{ inner =>
mx.average(i.j)
}
}
Pay attention that Seq is immutable collection type and you can't just modify it, you can create new, modified collection only.
By the way, why you iterate starting 1, but not 0. Are you sure you want it?
I have a set of IDs associated with costs which is just a double value. IDs are integers and unique. Two IDs may have same costs. I stored them as:-
a=containers.Map('KeyType','uint32','ValueType','double');
a(1)=7.3
a(2)=8.4
a(3)=7.3
Now i want to find the minimum cost.
b=[];
c=values(a);
b=[b,c{:}];
cost_min=min(b);
Now i want to find all IDs associated i.e. 1 and 3 with the minimum cost i.e. 7.3. I can collect all the keys into an array and then do a for loop over this array. Is there a better way to do this entire thing in Matlab so that for loops are not required?
sparse matrix can work as a hashmap, just do this:
a= sparse(1:3,1,[7.3 8.4 7.3])
find(a == min(nonzeros(a))
There are methods which can be used on maps for this kind of operations
http://se.mathworks.com/help/matlab/ref/containers.map-class.html
The approach finding minimum values and minimum keys can be done something like this,
a=containers.Map('KeyType','uint32','ValueType','double');
a(1)=7.3;
a(3)=8.4;
a(4)=7.3;
minval = inf;
minkeys = -1;
for k = keys(a)
val = a.values(k);
val = val{1};
if (val < minval(1))
minkeys = k;
minval = val;
elseif (val == minval(1))
minkeys = [minkeys,k];
end
end
disp(minval);
disp(minkeys);
This is not efficient though and value search is clumsy for maps. This is not what they are intended for. Maps is supposed to do efficient key lookup. In case you are going to do a lot of lookups and this is what takes time, then use a map. If you need to do a lot of value searches, I would recommend that you use a matrix (or two arrays) for this instead.
idx = [1;3;4];
val = [7.3,8.3,7.3];
minval = min(val);
minidx = idx(val==minval);
disp(minval);
disp(minidx);
There is also another post with an example where it is shown how a sparse matrix can be used as a hashmap. Let the index become the key. This will take about 3 times the memory as all non-zero elements an ordinary array, but a map uses more memory than an array as well.
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.
i have a cell array as below, which are dates. I am wondering how can i extract the year at the last 4 digits? Could anyone teach me how to locate the year in the string? Thank you!
'31.12.2001'
'31.12.2000'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2000'
'31.12.1999'
'31.12.1998'
'31.12.1997'
'31.12.2005'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2001'
'31.12.2000'
'31.12.1999'
'31.12.1998'
'31.12.2005'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2005'
Example cell array:
A = {'31.12.2001'; '31.12.2002'; '31.12.2003'};
Apply some regular expressions:
B = regexp(A, '\d\d\d\d', 'match')
B = [B{:}];
EDIT: I never realized that matlab will "nest" an extra layer of cells until I tested this. I don't like this solution as much now that I know the second line is necessary. Here is an alternative approach that gets you the years in numeric form:
C = datevec(A, 'dd.mm.yyyy');
C = C(:, 1);
SECOND EDIT: Suprisingly, if your cell array has less than 10000 elements, the regexp approach is faster on my machine. But the output of it is another cell array (which takes up much more memory than a numeric matrix). You can use B = cell2mat(B) to get a character array instead, but this brings the two approaches to approximately equal efficiency.
Just to add a fun answer, designed to take the OP to the stranger regions of Matlab:
C = char(C);
y = (D(:,7:end)-'0') * 10.^(3:-1:0).'
which is an order of magnitude faster than anything posted in the other answers :)
Or, to stay a bit closer to home,
y = cellfun(#(x)str2double(x(7:end)),C);
or, yet another regexp variation:
y = str2num(char(regexprep(C, '\d+\.\d+\.','')));
Assuming your matrix with dates is M or a cell array C:
In case your data is in a cell array start with
M = cell2mat(C)
Then get the relevant part
Y=M(:,end-4:end)
If required you can even make the year a number
Year = str2num(Y)
Using regexp this will works also with dates with slightly different formats, like 1.1.2000, which can mess with you offsets
res = regexp(dates, '(?<=\d+\.\d+\.)\d+', 'match')