1

I think I'm just too tired to see the mistake. I wrote a function to get the maximal value for two data sets from a 'for' loop: 

plot_zu <- function(x) {for (i in 1:x){
    z=data_raw[grep(a[i], data_raw$Gene.names),]
    b=data_raw_ace[grep(a[i], data_raw_ace$Gene.names),]
    p<-vector("numeric", length(1:length(a)))
    p[i]<-max(z$t_test_diff)
    return(p)} 
}

Imagine: a is a vector of names and the data set (data_raw(_ace)) is filtered by it. In the end, I would like to have all maxima values of column t_test_diff in a vector. After that I want to add the t_test_diff column values from data_raw_ace also.

So the problem is, that I get this:

[1] 1.210213 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
[8] 0.000000 0.000000

So there is a problem with brackets or something but I cannot see it ( first value fits). Sorry for no good example but i think it is understandable and an easy to solve question.

If need be, I can add another example.

Thanks a lot!!

gratefully,

Hendrik

4

3 回答 3

1

In the absence of data and even the call you make to this function, I'm going to offer an alternative based on what I think you are attempting. It appears you want to select only those rows of "data_raw" whose "Gene.names" column values are in the set defined by "a". If so, that is just:

 z <- data_raw[ data_raw$Gene.names %in% a , ]  # no loop needed

 b <- data_raw_ace[ data_raw_ace$Gene.names %in% a , ] # again no loop needed
 # Next step is unclear

If you want to use grep or grepl inside "[" then use sapply:

 z <- data_raw[ sapply(a, grep, x= data_raw$Gene.names), ]  # (still) no loop needed
 b <- data_raw_ace[ sapply(a, grep, x= data_raw_ace$Gene.names),  ]

When you do this, what is it that is desired? 

p<-vector("numeric", length(1:length(a)))
p[i]<-max(z$t_test_diff)

If you want the maximum value for an identically named column in the two subset of data, then do this:

p <- pmax( z$t_test_diff, b$t_test_diff )

Based on you further comments above I (now) think maybe:

p <- apply( cbind(z$t_test_diff), abs(b$t_test_diff), 1, function(x) x[which.max(abs(x))])
于 2012-12-02T19:14:07.403 回答
0

vapply does this, eliminating the for loop. Not tested.

vapply(1:x, FUN.VALUE=numeric(1), FUN=function(x) {
    z=data_raw[grep(a[i], data_raw$Gene.names),]
    b=data_raw_ace[grep(a[i], data_raw_ace$Gene.names),]  # Is this needed?

    return(max(z$t_test_diff))
})
于 2012-12-02T17:17:42.013 回答
0

It appears you are overwriting p in each new iteration by defining it inside the loop.

于 2012-12-02T17:28:20.153 回答