sample

问题 Number of Threads (users) : 10 Ramp-up Period (in seconds) : 1 Loop Count : 2 Result - When I ran the test it shows 40 samples rather expected count was 20 . I want to ask what could be the reason behind 40 samples . Number of Threads (users): 10 Ramp-up Period (in seconds): 1 Loop Count : 1 Result - When I ran the test it shows 20 samples rather expected count was 10 . Error - I tried to compute but I can't understand how it is doubling the user count every time 回答1: It just means that every

问题 I'll preface this with: I have very little database experience. I understand the concept of a database and how to use a database within Java, but I have no idea how to create one and my knowledge of SQL is essentially zilch. I am creating a sample project to show off the features of IntelliJ to my company, in the hopes they will convert from Eclipse. One of the big features for our company will be database integration, so I would like to show off the Database tools. To do this, I need a

问题 Hi I need to read SampleRate, SignalFrequency and Amplitude from .wave file. How can I do that using JavaSound? 回答1: You can get the sampling rate by getting a handle on the AudioFormat object: AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(new File("test.wav")); AudioFormat audioFormat = audioInputStream.getFormat(); Once you have that, you can get the sample rate as follows: float sampleRate = audioFormat.getSampleRate(); As for the amplitude, that is basically the raw

问题 I have a data.frame like this: DqStr <- "Group q Dq SD.Dq 1 -3.0 0.7351 0.0067 1 -2.5 0.6995 0.0078 1 -2.0 0.6538 0.0093 2 -3.0 0.7203 0.0081 2 -2.5 0.6829 0.0094 2 -2.0 0.6350 0.0112" Dq1 <- read.table(textConnection(DqStr), header=TRUE) I would like to randomize group membership but only for rows with the same value of Dq1$q g <-unique(Dq1$q) Dq2<- data.frame() for(n in g) { Dqq <- Dq1[Dq1$q==n,] Dqq$Group <-sample(Dqq$Group) Dq2 <- rbind(Dq2,Dqq) } That could also be done with plyr library

问题 I am trying to randomly select n samples from a graph. In order to do so I create a list called X using the random.sample function like the following: X= random.sample(range(graph.ecount()), numPosSamples) The problem is that when numPosSamples is equal to graph.ecount() I receive the following error: ValueError: Sample larger than population Any help will be much appreciated. Thanks 回答1: I'm not sure how numPosSamples is getting its value, but because random.sample does sampling without

问题 I am trying to sample between a range of values as part of a larger loop in R. As the loop progresses to each row j , I want to sample a number between the value given in the start column and the value given in the end column, placing that value in the sampled column for that row. The results should look something like this: ID start end sampled a 25 67 44 b 36 97 67 c 23 85 77 d 15 67 52 e 21 52 41 f 43 72 66 g 39 55 49 h 27 62 35 i 11 99 17 j 21 89 66 k 28 65 48 l 44 58 48 m 16 77 22 n 25

问题 I am trying to sample between a range of values as part of a larger loop in R. As the loop progresses to each row j , I want to sample a number between the value given in the start column and the value given in the end column, placing that value in the sampled column for that row. The results should look something like this: ID start end sampled a 25 67 44 b 36 97 67 c 23 85 77 d 15 67 52 e 21 52 41 f 43 72 66 g 39 55 49 h 27 62 35 i 11 99 17 j 21 89 66 k 28 65 48 l 44 58 48 m 16 77 22 n 25

问题 I want create a distribution using the sample() function with the probability of each value defined by a data.frame() column. When I try the code below however it produces the error: Error in sample.int(length(x), size, replace, prob) : incorrect number of probabilities I'm guessing this is because mutate is passing the entire column and not just one integer. Can anyone help me with this? library(dplyr) input = data.frame(input = 1:100) output = input %>% mutate(output = sum( sample(0:1, 10,

问题 We know that prob argument in sample is used to assign a probability of weights. For example, table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.4, 0.3, 0.1)))/1e6 # 1 2 3 4 #0.2 0.4 0.3 0.1 table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.4, 0.3, 0.1)))/1e6 # 1 2 3 4 #0.200 0.400 0.299 0.100 In this example, the sum of probability is exactly 1 (0.2 + 0.4 + 0.3 + 0.1), hence it gives the expected ratio but what if the probability does not sum to 1? What output would it give? I

问题 We know that prob argument in sample is used to assign a probability of weights. For example, table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.4, 0.3, 0.1)))/1e6 # 1 2 3 4 #0.2 0.4 0.3 0.1 table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.4, 0.3, 0.1)))/1e6 # 1 2 3 4 #0.200 0.400 0.299 0.100 In this example, the sum of probability is exactly 1 (0.2 + 0.4 + 0.3 + 0.1), hence it gives the expected ratio but what if the probability does not sum to 1? What output would it give? I