I have array X with some values
[[0.3,0.4,0.5],
[0.1,0.7,0.9],
.
.
.
[0.3,0.6,0.9]]
an I have array with indexes I =[0,2,1,2,0,..].
I would like to take value from array X for each row according to indexed in array I like, in array I first value is 0 so from first row in I will take value from column 0 which is 0.3 and so on.
Is there any possible to do this without loop?
My idea:
Y = X[:,I] has no sense.
You were almost there, what you need is some fancy indexing on top:
Y = X[np.arange(len(I)),I]
This kind of indexing tells numpy to select the entries (i, I(i)) in X.
Related
I have a 2D array PointAndTangent of dimension 8500 x 5. The data is row-wise with 8500 data rows and 5 data values for each row. I need to extract the row index of an element in 4th column when this condition is met, for any s:
abs(PointAndTangent[:,3] - s) <= 0.005
I just need the row index of the first match for the above condition. I tried using the following:
index = np.all([[abs(s - PointAndTangent[:, 3])<= 0.005], [abs(s - PointAndTangent[:, 3]) <= 0.005]], axis=0)
i = int(np.where(np.squeeze(index))[0])
which doesn't work. I get the follwing error:
i = int(np.where(np.squeeze(index))[0])
TypeError: only size-1 arrays can be converted to Python scalars
I am not so proficient with NumPy in Python. Any suggestions would be great. I am trying to avoid using for loop as this is small part of a huge simulation that I am trying.
Thanks!
Possible Solution
I used the following
idx = (np.abs(PointAndTangent[:,3] - s)).argmin()
It seems to work. It returns the row index of the nearest value to s in the 4th column.
You were almost there. np.where is one of the most abused functions in numpy. Half the time, you really want np.nonzero, and the other half, you want to use the boolean mask directly. In your case, you want np.flatnonzero or np.argmax:
mask = abs(PointAndTangent[:,3] - s) <= 0.005
mask is a 1D array with ones where the condition is met, and zeros elsewhere. You can get the indices of all the ones with flatnonzero and select the first one:
index = np.flatnonzero(mask)[0]
Alternatively, you can select the first one directly with argmax:
index = np.argmax(mask)
The solutions behave differently in the case when there are no rows meeting your condition. Three former does indexing, so will raise an error. The latter will return zero, which can also be a real result.
Both can be written as a one-liner by replacing mask with the expression that was assigned to it.
For example we have 5 values(a,b,c,e,d) in an array which may contain duplicate values.The values of a,b,c,d,e are calculated by another function and got assigned with some values.
let the user input value be x.
we need to find the closest value to x in the array and the output must be in a,b,c,e,d. if the closest number is one of the duplicates the alphabetical order must be considered.
for example:
Array: a,b,c,e,d
a=6,b=5,c=3,d=9,e=9 are the values assigned to them by a function.
for x : 5,
output : b
for x :11,
output : d
for x : 4,
output :c
Try and implement the following algorithm:
get the element at the middle of the array ;
if this element has the value x, the array contains it, you know what to print ;
if the element is larger, look in the first half of the array ;
otherwise look in the second half of the array ;
Once the array portion you search into has a size of 0, you have found the place where x would be inserted to preserve the ordering. The closest value is either the one of the left or the one on the right, if any. Compute the differences to determine what to print.
I'm quite new to MatLab and this problem really drives me insane:
I have a huge array of 2 column and about 31,000 rows. One of the two columns depicts a spatial coordinate on a grid the other one a dependent parameter. What I want to do is the following:
I. I need to split the array into smaller parts defined by the spatial column; let's say the spatial coordinate are ranging from 0 to 500 - I now want arrays that give me the two column values for spatial coordinate 0-10, then 10-20 and so on. This would result in 50 arrays of unequal size that cover a spatial range from 0 to 500.
II. Secondly, I would need to calculate the average values of the resulting columns of every single array so that I obtain per array one 2-dimensional point.
III. Thirdly, I could plot these points and I would be super happy.
Sadly, I'm super confused since I miserably fail at step I. - Maybe there is even an easier way than to split the giant array in so many small arrays - who knows..
I would be really really happy for any suggestion.
Thank you,
Arne
First of all, since you wish a data structure of array of different size you will need to place them in a cell array so you could try something like this:
res = arrayfun(#(x)arr(arr(:,1)==x,:), unique(arr(:,1)), 'UniformOutput', 0);
The previous code return a cell array with the array splitted according its first column with #(x)arr(arr(:,1)==x,:) you are doing a function on x and arrayfun(function, ..., 'UniformOutput', 0) applies function to each element in the following arguments (taken a single value of each argument to evaluate the function) but you must notice that arr must be numeric so if not you should map your values to numeric values or use another way to select this values.
In the same way you could do
uo = 'UniformOutput';
res = arrayfun(#(x){arr(arr(:,1)==x,:), mean(arr(arr(:,1)==x,2))), unique(arr(:,1)), uo, 0);
You will probably want to flat the returning value, check the function cat, you could do:
res = cat(1,res{:})
Plot your data depends on their format, so I can't help if i don't know how the data are, but you could try to plot inside a loop over your 'res' variable or something similar.
Step I indeed comes with some difficulties. Once these are solved, I guess steps II and III can easily be solved. Let me make some suggestions for step I:
You first define the maximum value (maxValue = 500;) and the step size (stepSize = 10;). Now it is possible to iterate through all steps and create your new vectors.
for k=1:maxValue/stepSize
...
end
As every resulting array will have different dimensions, I suggest you save the vectors in a cell array:
Y = cell(maxValue/stepSize,1);
Use the find function to find the rows of the entries for each matrix. At each step k, the range of values of interest will be (k-1)*stepSize to k*stepSize.
row = find( (k-1)*stepSize <= X(:,1) & X(:,1) < k*stepSize );
You can now create the matrix for a stepk by
Y{k,1} = X(row,:);
Putting everything together you should be able to create the cell array Y containing your matrices and continue with the other tasks. You could also save the average of each value range in a second column of the cell array Y:
Y{k,2} = mean( Y{k,1}(:,2) );
I hope this helps you with your task. Note that these are only suggestions and there may be different (maybe more appropriate) ways to handle this.
I have a matrix having values between [0,1]. I want to find and mark the locations of those elements which have values <0.1 and >0.9.
So I use the matlab function find; but it returns me two vectors: a row and column vector distinctly which is difficult for analysis. So is there a way by which I can see the location of which elements meet the conditions without losing the original matrix structure?
I used the below line of code:
[r,c,v]= find(X<0.1 | X>0.9); % X is my 512*512 matrix of values
Thanks!
See if this works out for you -
%// cell array with each cell housing the matching indices for each row
out = cellfun(#find,mat2cell(X<0.1 | X>0.9,ones(1,size(X,1)),size(X,2)),'uni',0)
Browse through the values of out using - celldisp(out)
Just by using condition like this :
mask = (X < 0.1 | X > 0.9)
Will return an array logical with 1 where condition is respected.
My array contains a string in the first row
how can I sum the array from the 2nd row to the Nth/1442th row (as in my example) disregarding the negative signs present in the column?
for example, my code for an array called data2 is:
S = sum(data2(2,15):data2(1442,15));
so sum all of the elements from row 2 to row 1442 in column 15.
This doesn't work but it also does not have anything to deal with the absolute value of whatever row its checking
data is from a .csv:
You should do something like this:
sum(abs(data(2:1442,15)));
The abs function will find the absolute value of each value in the array (i.e. disregard the negative sign). data(2:1442,15) will grab rows 2-1442 of the 15th column, as you wanted.
EDIT: apparently data is a cell array, so you could do the following, I think:
sum(abs([data{2:1442,15}]));
Ok so it looks like you have a constant column so
data2(2,15) = -0.02
and further down
data2(1442,15) = -0.02 %(I would assume)
So when you form:
data2(2,15):data2(1442,15)
this is essential like trying to create an array but of a single value since:
-0.02:-0.02
ans =
-0.0200
which of course gives:
>> sum(-0.02:-0.02)
ans =
-0.0200
What you want should be more like:
sum(data2(2:1442,15))
That way, the index: 2:1442, forms a vector of all the row references for you.
To disregard the negative values:
your answer = sum(abs(data2(2:1442,15)))
EDIT: For a cell array this works:
sum(abs(cell2mat(data2(2:1442,15))))