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问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . How is it possible that the binary executed 'by default' -- ie. the first one found in my path, as returned by the which command, is in fact not the binary actually being executed by default? (I'm running Mac OS X 10.5 on a 2008 MacBook Pro.) I just installed the latest version of git by running brew install git

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . How is it possible that the binary executed 'by default' -- ie. the first one found in my path, as returned by the which command, is in fact not the binary actually being executed by default? (I'm running Mac OS X 10.5 on a 2008 MacBook Pro.) I just installed the latest version of git by running brew install git

问题 I read the similar topic here. I think the question is different or at least .index() couldnot solve my problem. This is a simple code in R and its answer: x <- c(1:4, 0:5, 11) x #[1] 1 2 3 4 0 1 2 3 4 5 11 which(x==2) # [1] 2 7 min(which(x==2)) # [1] 2 which.min(x) #[1] 5 Which simply returns the index of the item which meets the condition. If x be the input for Python, how can I get the indeces for the elements which meet criteria x==2 and the one which is the smallest in the array which

问题 There are some answers on stack to the below type of question, but they are all inefficient and do not scale well. To reproduce it, suppose I have data that looks like this: tempmat=matrix(c(1,1,0,4,1,0,0,4,0,1,0,4, 0,1,1,4, 0,1,0,5),5,4,byrow=T) tempmat=rbind(rep(0,4),tempmat) tempmat=data.table(tempmat) names(tempmat)=paste0('prod1vint',1:4) This is what the data look like, although it is MUCH bigger, so the solution cannot be an "apply" or row-wise based approach. > tempmat prod1vint1

问题 Time Velocity 0 0 1.5 1.21 3 1.26 4.5 1.31 6 1.36 7.5 1.41 9 1.46 10.5 1.51 12 1.56 13 1.61 14 1.66 15 1.71 16 1.76 17 1.81 18 1.86 19 1.91 20 1.96 21 2.01 22.5 2.06 24 2.11 25.5 2.16 27 2.21 28.5 2.26 30 2.31 31.5 2.36 33 2.41 34.5 2.4223 36 2.4323 So I have data about Time and Velocity...I want to use the cut or the which function to separate my data into 6 min intervals...my Maximum Time usually goes up to 3000 mins So I would want the output to be similar to this... Time Velocity 0 0 1.5

问题 This question already has answers here : How to check if a program exists from a Bash script? (35 answers) Closed 3 years ago . I've read in some bash FAQ a while ago (that I don't remember), that which should be avoided and command -v preferred. Why is that so? What are the advantages, disadvantages of either one? 回答1: Well... command is likely built in to your shell, and with the -v option will tell you how your shell will invoke the command specified as its option. which is an external

问题 This question already has answers here : How to check if a program exists from a Bash script? (35 answers) Closed 3 years ago . I've read in some bash FAQ a while ago (that I don't remember), that which should be avoided and command -v preferred. Why is that so? What are the advantages, disadvantages of either one? 回答1: Well... command is likely built in to your shell, and with the -v option will tell you how your shell will invoke the command specified as its option. which is an external

问题 Just been playing with some basic functions and it seems rather strange how ifelse behaves if I use which() function as one of the arguments when the ifelse condition is true, e.g.: #I want to identify the location of all values above 6.5 #only if there are more than 90 values in the vector a: set.seed(100) a <- rnorm(100, mean=5, sd=1) ifelse(length(a)>90, which(a>6.5), NA) I get this output: [1] 4 When in fact it should be the following: [1] 4 15 25 40 44 47 65 How then can I make ifelse

问题 I have a vector of values, call it X, and a data frame, call it dat.fram. I want to run something like "grep" or "which" to find all the indices of dat.fram[,3] which match each of the elements of X. This is the very inefficient for loop I have below. Notice that there are many observations in X and each member of "match.ind" can have zero or more matches. Also, dat.fram has over 1 million observations. Is there any way to use a vector function in R to make this process more efficient?

问题 This question already has answers here : For each row return the column name of the largest value (7 answers) Closed last year . I am trying to extract the colnames of a data frame, based on the values in the cells. My data is a series of a couple hundred categories, with a simple binary 0 or 1 in the cells to indicate which column name I want in my new df. To illustrate my point: year cat1 cat2 cat3 ... catN 2000 0 0 1 0 2001 1 0 0 0 2002 0 0 0 1 .... 2018 0 1 0 0 I am trying to get a df