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QUESTION: I have a color transformation that I want to apply to an RGB pixel in MedianImage = totalImage - minValue - maxValueĪpply a Pixel Transformation to an Entire Image The median value is just the total image value minus the minimum value and the maximum Next, total the images in the third dimension. First construct a three-dimensional image array from your three images.Ĭonsult the Array Concatenation Tutorial if you are Is there a way to do this without using a loop?ĪNSWER: Yes. I would like to create a fourth image with the median value Or, James Kuyper points out, you can avoid the problem like this:Ĭ = (A+B)/((A NE MISSING) + (B NE MISSING) > 1.0)Įxtendible to the case where you have N arrays with M values each.Ĭ = TOTAL(DATA,2)/(TOTAL(DATA NE MISSING,2) > 1.0)Ĭreate a Median Image from Three Original Images You'll get NaN whenever both values are missing. IF count GT 0 THEN C = Aīut Craig Markwardt provided the IDL solution.Ĭ = (A+B)/((A NE MISSING) + (B NE MISSING)) Indices = WHERE(A NE 0 AND B EQ 0, count) Indices = WHERE(A EQ 0 AND B NE 0, count) Third array to take the value of the array that has good data. The two arrays if there are two good values. I want to create a third array that has the average of Each array has floating point values and missingĭata ( value=0). This turned out to be what the original requester wanted, and he reported it Indoffset=n_elements(listlowestind)-countindok Indok=where(listlowestind GT -1L,countindok) Find the lowest index for each bin element. Listlowestind = Replicate(-1L, N_Elements(h)) H = Histogram(array, Reverse_Indices=ri, Min=1L) Sum 1-min(a) to the array before performing the operation. I assumed that the arrayĪ has integer elements starting from 1, if not the case, just Since we have sorted theĪrray first, we know which rank to assign to each element. Sort the list, such that we know which element occurs first,Īnd then with a second loop we fill the output array with Of each integer number and store it in listlowestind. The idea here is to use Histogram to scout for the first occurrence Provided that we just do simple operations in the loop. a loop, but this should not be tooīad as long as you don't have more than a few million elements, Paolo Grigis proposed an alternative solution, based on this one. Who posed the original question did not want to skip rankings like this? (He didn't, There is no "ranking" of 3 or 4 in this output. Reminder to people to check their algorithms with more than one data set!) This solution brought objections from some folks on the newsgroup, whoĪrgued that with some data sets "rankings" could be skipped. H = Histogram(array, Reverse_Indices=ri, Min=0)ī = Indgen(N_Elements(h) > N_Elements(array)) + 1Ĭ = Intarr(N_Elements(h) > N_Elements(array)) IF N_Elements(array) EQ 0 THEN Message, 'Must pass an array of integer values.' That uses the reverse indices of the Histogram command to set the proper values In IDL to find something which could help.ĪNSWER: In situations like this, you always look to the Would like to process large arrays ( of over 60000 elements), I'm I guess this process is similar to a lookup table. Multiple valuesĪre mapped to the value in which they first appear. The valueĦ is the second value in the array and is mapped to 2, and do on. Here the value 3 is the first value in the array and is mapped to 1.
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I would like to obtain an array like this: QUESTION: I would like to change the values in an array according to the order IDL> j = REBIN(TRANSPOSE(LINDGEN(n)), n, n)Ĭhange Values in Array According to Appearance Order Size in which all the values on the diagonal on above it are ones and all make all entries below the diagonal) zero ?ĪNSWER: You need to multiply your array times a mask of the same QUESTION: Is there an easy way to make an upper triangular Reading the the Dimensional Juggling Tutorial. Moment I am using a FOR loop and it is extremely slow.ĪNSWER: You can find the answer to this question in a careful Times, so my final array is 2000 elements long with alternating 2s and 3s. QUESTION: Suppose I have an array, say a = ,Īnd i wish to repeat these two elements say 1000 Increment an Array Subset by Another Array Subset.Determining the duration of events in a time series.Pixel color transformation applied to entire image.Change values in array according to appearance order.Replicate two-element array 1000 times.Examples To Illuminate the IDL Way Contents