从R中的数据框列表中选择特定名称

Question

Sample data:

df <- data.frame(names=letters[1:10],name1=rnorm(10,1,1),name2=rexp(10,2))

list <- list(df,df)

vec_name <- c("f","i","c") # desired row names 

I would like to select per list rows given the vec_name names:

Desired outcome:

[[1]]
      names      value1    value2
   6   nd:f   -1.6323952 0.3117470
   9   nd:i    1.8270855 0.2475741
   3   nd:c    0.6978422 0.4695581   # the ordering does matter; must be as seen in vec_name

[[2]]
      names      value1    value2
   6   ad:f   -1.6323952 0.3117470
   9   ad:i    1.8270855 0.2475741
   3   ad:c    0.6978422 0.4695581

Desired output 2: Is in dataframe, which would be I believe just do.call(rbind,list) :

However the clean names from vec_names should be used instead.

      names      value1    value2
   1      f   -1.6323952 0.3117470
   2      i    1.8270855 0.2475741
   3      c    0.6978422 0.4695581 
   4      f   -1.6323952 0.3117470
   5      i    1.8270855 0.2475741
   6      c    0.6978422 0.4695581

I have tried sapply ; lapply ... for example:

lapply(list, function(x) x[grepl(vec_name,x$names),])

EDIT : PLEASE SEE THE EDITED QUESTION ABOVE.

Answer 1

You were almost there. The warning message was saying:

Warning messages:
1: In grepl(vec_name, x$names) :
   argument 'pattern' has length > 1 and only the first element will be used

Reason is that you provide a vector to grepl which is expecting a regex (see ?regex ). What you want to do is to match the contents:

lapply(list, function(x) x[match(vec_name,x$names),])

Which will give you a list of data.frame objects. If you want to combine them afterwards just use:

do.call(rbind, lapply(list, function(x) x[match(vec_name,x$names),]))

Or you use ldply from library(plyr) :

library(plyr)
ldply(list, function(x) x[match(vec_name,x$names),])
#   names       name1     name2
# 1     f  2.01421228 0.4489627
# 2     i  0.28899891 0.8323940
# 3     c -0.01746007 1.5309936
# 4     f  2.01421228 0.4489627
# 5     i  0.28899891 0.8323940
# 6     c -0.01746007 1.5309936

And as a remark: avoid to use protected names like list for your variables to avoid unwanted effects.

Update

Taking the comments into account ( vec_name does not match completely the names in the data.frame )you should clean first the names and then do the match . This is, however, assuming that your 'uncleaned' names contain the cleaned names with a pre-fix separated by a colon (':') (if this is not the case adapt the regex in the gsub statement):

ldply(list, function(x) x[match(vec_name, gsub(".*:(.*)", "\\1", x$names)),])

Answer 2

for the first output :

output1<-lapply(list,function(elt){
                       resmatch<-sapply(vec_name,function(x) regexpr(x,df$names))
                       elt<-elt[apply(resmatch,2,function(rg) which(rg>0)),]
                       colnames(elt)<-c("names","value1","value2")
                       return(elt)
                       })

>output1
[[1]]
  names     value1    value2
6  nd:f -0.2132962 0.7618105
9  nd:i -0.6580247 0.6010379
3  nd:c  0.9302625 0.1490061

[[2]]
  names     value1    value2
6  nd:f -0.2132962 0.7618105
9  nd:i -0.6580247 0.6010379
3  nd:c  0.9302625 0.1490061

For the second output, you can do what you wanted to :

output2<-do.call(rbind,output1)

> output2

   names     value1    value2
6   nd:f -0.2132962 0.7618105
9   nd:i -0.6580247 0.6010379
3   nd:c  0.9302625 0.1490061
61  nd:f -0.2132962 0.7618105
91  nd:i -0.6580247 0.6010379
31  nd:c  0.9302625 0.1490061