CODEWITHSUNDEEP
rdataframesubset
Sander Van der Zeeuw
Sander Van der Zeeuw

Reputation: 1092

Extract columns out of a data frame and order them

I have a data frame which looks like this:

structure(list(Mash_pear = c(0.192474082559755, 0.679726904159742, 
0.778564545349054, 0.573745352397321, 0.56633658385284, 0.472559997318901, 
0.462635414367878, 0.562128414492567, 0.354624921832056, 0.64532681437697
), tRap_pear = c(0.0350096175177328, 0.234255507711743, 0.23714999195134, 
0.185536020521134, 0.191585098617356, 0.201402054387186, 0.220911538536031, 
0.216072802572045, 0.132247101763063, 0.172753098431029), Beeml_pear = c(0.179209909971615, 
0.79129167285928, 0.856908302056589, 0.729078080521886, 0.709346164378725, 
0.669599784720647, 0.585348196746785, 0.639355942917055, 0.544909349368496, 
0.794652394149651), Mash_pear2080 = c(0.823944540480775, 0.816630852343513, 
0.81134728399675, 0.801065036203532, 0.799630945085954, 0.799195606444727, 
0.798637867344115, 0.798478922129054, 0.798090734787886, 0.797673368802285
)), .Names = c("Mash_pear", "tRap_pear", "Beeml_pear", "Mash_pear2080"
), row.names = c("Aft1", "Alx3_3418.2", "Alx4_1744.1", "Arid3a_3875.1_v1_primary", 
"Arid3a_3875.1_v2_primary", "Arid3a_3875.2_v1_primary", "Arid3a_3875.2_v2_primary", 
"Arid5a_3770.2_v1_primary", "Arid5a_3770.2_v2_primary", "Aro80"
), class = "data.frame")

Now i had the idea to rank those scores, but every columns should be ranked separately with keeping the row names intact. So i tried to extract all the columns 1 by 1 and order them. The problem i'm experiencing is occurring when trying to order 1 column. Namely my dataframe disappears and becomes a vector of numeric values and as i already pointed out, i need the data frame (rownames) to stay as is, only ordered. The code i'm working on right now is here:

rowname<-rownames(pearframe)
col1<-subset(pearframe, select=1)[order(pearframe),]
col2<-subset(pearframe, select=2)[order(pearframe),]
col3<-subset(pearframe, select=3)[order(pearframe),]
col4<-subset(pearframe, select=4)[order(pearframe),]

THIS removes my rownames and orginal data frame structure. Which makes it impossible to rank my data. So the actual question would be: How can i order/sort a data frame per column and create 4 new frames with 1 ordered column each. Eventually i want to have a table which exists of the rownames of every ranked frame and the scores.

Upvotes: 2

Views: 1239

Answers (2)

Simon O&#39;Hanlon
Simon O&#39;Hanlon

Reputation: 59970

Another way would be to take advantage of the fact that a data.frame is just a bunch of lists. You can use lapply and this will give you a list of data.frames. You can access each one by the column name and assign it to a new df if you wish:

ranks <- lapply( df , function(x) data.frame( rank = rownames(df)[ order( x ) ] , score = x[ order(x) ] ) )
names(ranks) <- names(df)

head(ranks[["Mash_pear"]])
#                     rank     score
#1                     Aft1 0.1924741
#2 Arid5a_3770.2_v2_primary 0.3546249
#3 Arid3a_3875.2_v2_primary 0.4626354
#4 Arid3a_3875.2_v1_primary 0.4725600
#5 Arid5a_3770.2_v1_primary 0.5621284
#6 Arid3a_3875.1_v2_primary 0.5663366

head(ranks[["tRap_pear"]])
#                     rank      score
#1                     Aft1 0.03500962
#2 Arid5a_3770.2_v2_primary 0.13224710
#3                    Aro80 0.17275310
#4 Arid3a_3875.1_v1_primary 0.18553602
#5 Arid3a_3875.1_v2_primary 0.19158510
#6 Arid3a_3875.2_v1_primary 0.20140205

Upvotes: 4

Frank
Frank

Reputation: 66819

You'll need to use drop=FALSE twice, I think:

subset(pearframe, select=1,drop=FALSE)[order(pearframe[,1]),,drop=FALSE]

The other cases look the same, incrementing 1 both places it appears.

EDIT: Also, this is more concise:

pearframe[order(pearframe[,1]),1,drop=FALSE]

EDIT2: And this is how to make your final data.frame with this approach:

col_list <- list()
for (i in 1:4){
    col_list[[i]] <- pearframe[order(pearframe[,i]),i,drop=FALSE]
    col_rnname <- paste(names(pearframe)[[i]],"rn",sep=".")
    col_list[[i]][[col_rnname]] <- rownames(col_list[[i]])
    rownames(col_list[[i]]) <- NULL
}
col_mat <- do.call(cbind,col_list)

Upvotes: 4

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