Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
242 views
in Technique[技术] by (71.8m points)

r - Nested if else statements over a number of columns

I have a large data.frame where the first three columns contain information about a marker. The remaining columns are of numeric type for that marker in each individual. Each individual has three columns. The dataset looks as follows:

                      marker alleleA alleleB   X818 X818.1 X818.2   X345 X345.1 X345.2   X346 X346.1 X346.2
1   kgp5209280_chr3_21902067       T       A 0.0000 1.0000 0.0000 1.0000 0.0000 0.0000 0.0000 1.0000 0.0000
2 chr3_21902130_21902131_A_T       A       T 0.8626 0.1356 0.0018 0.7676 0.2170 0.0154 0.8626 0.1356 0.0018
3 chr3_21902134_21902135_T_C       T       C 0.6982 0.2854 0.0164 0.5617 0.3749 0.0634 0.6982 0.2854 0.0164

That is, for each marker (row), each individual has three values, one in each column.

I want to create a new data.frame which has all the same rows as in the original, but only one column per individual. In the one column for each individual I want the value out of the three for each individual which is greater than 0.8. If no value is greater than 0.8 then I want to print NA. For instance, in the data set I have given for the first row I would want the second value for 818 (1.0000), and the first value for 345 (1.0000). In the second row, I want the first value for 818 (0.8626), and for 345 none of the values are above 0.8 so I want NA to be printed and so on. The new data set would therefore look like this:

                     marker alleleA alleleB   X818 X345
1   kgp5209280_chr3_21902067       T       A 1.0000    1
2 chr3_21902130_21902131_A_T       A       T 0.8626   NA

I have been trying to use if/else statements, along the lines of if [, 4] > 0.8 then [, 4], else... however it doesn't seem to give me what I want, and I would also have to loop this command so it doesn't just do it for one individual in the first three columns but for all columns.

Any help would be appreciated! Thanks in advance.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Edit: Updated solution using the fast melt/dcast methods implemented in data.table versions >= 1.9.0. Go here for more info.

require(data.table)
require(reshape2)
dt <- as.data.table(df)

# melt data.table
dt.m <- melt(dt, id=c("marker", "alleleA", "alleleB"), 
                 variable.name="id", value.name="val")
dt.m[, id := gsub("\.[0-9]+$", "", id)] # replace `.[0-9]` with nothing
# aggregation
dt.m <- dt.m[, list(alleleA = alleleA[1], 
         alleleB = alleleB[1], val = max(val)), 
        keyby=list(marker, id)][val <= 0.8, val := NA]
# casting back
dt.c <- dcast.data.table(dt.m, marker + alleleA + alleleB ~ id)
#                        marker alleleA alleleB X345   X346   X818
# 1: chr3_21902130_21902131_A_T       A       T   NA 0.8626 0.8626
# 2: chr3_21902134_21902135_T_C       T       C   NA     NA     NA
# 3:   kgp5209280_chr3_21902067       T       A    1 1.0000 1.0000

Solution 1: Probably not the best way, but this is what I could think of at the moment:

mm <- t(apply(df[-(1:3)], 1, function(x) tapply(x, gl(3,3), max)))
mode(mm) <- "numeric"
mm[mm < 0.8] <- NA 
# you can set the column names of mm here if necessary
out <- cbind(df[, 1:3], mm)

#                       marker alleleA alleleB      1  2      3
# 1   kgp5209280_chr3_21902067       T       A 1.0000  1 1.0000
# 2 chr3_21902130_21902131_A_T       A       T 0.8626 NA 0.8626
# 3 chr3_21902134_21902135_T_C       T       C     NA NA     NA

gl(3,3) gives a factor with values 1,1,1,2,2,2,3,3,3 with levels 1,2,3. That is, tapply will take the values x 3 at a time and get their max (first 3, next 3 and the last 3). And apply sends each row one by one.


Solution 2: A data.table solution with melt and cast within data.table without using reshape or reshape2:

require(data.table)
dt <- data.table(df)
# melt your data.table to long format
dt.melt <- dt[, list(id = names(.SD), val = unlist(.SD)), 
                  by=list(marker, alleleA, alleleB)]
# replace `.[0-9]` with nothing
dt.melt[, id := gsub("\.[0-9]+$", "", id)]
# get max value grouping by marker and id
dt.melt <- dt.melt[, list(alleleA = alleleA[1], 
                      alleleB = alleleB[1], 
                      val = max(val)), 
        keyby=list(marker, id)][val <= 0.8, val := NA]
# edit mnel (use setattr(,'names') to avoid copy by `names<-` within `setNames`
dt.cast <- dt.melt[, as.list(setattr(val,'names', id)), 
                   by=list(marker, alleleA, alleleB)]

#                        marker alleleA alleleB X345   X346   X818
# 1: chr3_21902130_21902131_A_T       A       T   NA 0.8626 0.8626
# 2: chr3_21902134_21902135_T_C       T       C   NA     NA     NA
# 3:   kgp5209280_chr3_21902067       T       A    1 1.0000 1.0000

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...