minimum

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 5 years ago . I don't know how to make python "scan" through the list for a candidate and then go back to the loop again to find another candidate for the min. candidate = 0 maximum = 0 a = [12, 10, 50, 100, 24] for i in len(s): for j in range(len(s)): 回答1: You can use the builtin min function to do this a = [12

问题 I have a question and I could not find the answer on the internet nor on this website. I am sure it is very easy though. Let's say I have a set of 20 numbers and I have them in a 5x4 matrix: numbers = np.arange(20).reshape(5,4) This yields the following matrix: [ 0, 1, 2, 3] [ 4, 5, 6, 7] [ 8, 9, 10, 11] [12, 13, 14, 15] [16, 17, 18, 19] Now I would like to have the minimum value of each row, in this case amounting to 0,4,8,12,16. However, I would like to add that for my problem the minimum

问题 I have a long list of longitude values (len(Lon) = 420481), and another one of latitude values. I want to find the corresponding latitude to the minimum of the longitude. I tried: SE_Lat = [Lat[x] for x,y in enumerate(Lon) if y == min(Lon)] but this takes ages to finish. Does anyone know a more efficient way? Maybe you also have a suggestions for this: I now try to find the closest corresponding latitude to a new longitude, which is not in the original longitude vector. I tried this: minDiff

问题 How do I add a Fetched Property in XCode's Data Model Editor for minimum value of one attribute?? My model: Model http://www.freeimagehosting.net/uploads/b48853070e.png Item (name, note, storedItem) StoredItem (price, item) Item 1 ---> N StoredITem (1->N Relationship) I want that Item has a fetched property named minPrice and its value is the minimum value setted for price in the storedItems. Example: Item1 (banana, storedItem1 ... storedItem4, 10) StoredItem1 (10,item1) StoredItem2 (15,item1

问题 I am trying to extract the daily minimum zenith angle in a dataset which consists of 24h values (1 zenith angle value every hour) over ~31 days for 12 months. It looks like this: JulianDay Azimuth Zenith Date (YYMMDD HH:MM:SS) 2455928 174.14066 70.04650 2012-01-01 13:00:00 2455928 188.80626 70.30747 2012-01-01 14:00:00 2455928 203.03458 73.12297 2012-01-01 15:00:00 2455928 216.28061 78.20131 2012-01-01 16:00:00 2455928 228.35929 85.10759 2012-01-01 17:00:00 .... 2456293 146.33844 77.03456

问题 I have some integer variables and I want to find smallest one. When I use: m1 = min(v1, v2, ...) I get the value of the smallest one, not its name. I want to know which one is smallest, not know it's value! How should I do this? 回答1: so you have 2 variables v1 and v2 and want to print v1 is small or v2: if( v1 > v2 ): print "v1 =": str(v1) #or print "v1 is smaller" else: print "v2 =": str(v2) if you have a lot of variables then storing them in a dictionary might be a better idea. 回答2: If the

问题 I have the original code: min = INT_MAX; for (i=0;i<N;i++) if (A[i]<min) min = A[i]; for (i=0;i<N;i++) A[i]=A[i]-min; I want to get the parallel version of this and I did this: min = INT_MAX; #pragma omp parallel private(i){ minl = INT_MAX; #pragma omp for for (i=0;i<N;i++) if (A[i]<minl) minl=A[i]; #pragma omp critical{ if (minl<min) min=minl; } #pragma omp for for (i=0;i<N;i++) A[i]=A[i]-min; } Is the parallel code right? I was wondering if it is necessary to write #pragma omp barrier

问题 I want to get the minimum value in an array and then get the index of that item, in one step without writing my own loop (if I have to please let me know). I know I can just do the $b = ($a | Measure -Minimum).Minimum But then I have to do [array]::IndexOf($a, $b) And while that is normally okay, I'm looking for a way to do it once because I'm running this MANY MANY times in a loop. Thanks! EDIT: One step meaning without looping through the array twice 回答1: Personally, I might consider a

问题 I want to extract the two points (i.e their values) which are marked with black outline in figure. These minima points are 2 and 5. Then after extraction these marked points coordinates I want to calculate the distance between them. The code that I am using to plot average values of image, calculate minimas and locations is I1=imread('open.jpg'); I2=rgb2gray(I1); figure, title('open'); plot(1:size(I2,1), mean(I2,2)); hold on horizontalAverages = mean(I2 , 2); plot(1:size(I2,1) ,

问题 I'm facing some difficulties in the last few days while trying to finish the following task, I hope you guys can assist : I'm given a single number N, and I'm allowed to perform any of the two operations on N in each move : One - If we take 2 integers where N = x * y , then we can change the value of N to the maximum between x and y. Two - Decrease the value of N by 1. I want to find the minimum number of steps to reduce N to zero. This is what I have so far, I'm not sure what is the best way