开发者

grouping dataframe rows by factor and by function - output complete original dataframe row

开发者 https://www.devze.com 2023-04-12 14:45 出处:网络
My first post and I\'m very new to R so this may be a lob.I have search all over for a solution though, so I\'m finally posting for help.Let me know if I need to clarify or provide more information.

My first post and I'm very new to R so this may be a lob. I have search all over for a solution though, so I'm finally posting for help. Let me know if I need to clarify or provide more information.

I have a large dataframe that looks like the following:

numReads length    name2
0        7384      Ssxb2
7904     93237     St5
3438     12969     Taf9b
0        996       Tas2r138
0        882       Tas2r143
0        960       Tas2r144
0        6761      Tbx10
8125     43804     Tdrd1
8124     43738     Tdrd1
8102     39301     Tdrd1
1227     9286      Thnsl1

How can I group the data by the third column (name2), find the max() value for numReads, and maintain the associated length value?

My ideal output would be the above data with the two lines associated with "Tdrd1" that DO NOT contain the max value for that factor level (the lines with the 8124 and 8102 values).

I have tried tapply(), by(), and aggregate(). None of them can provide me with the proper output.

Thanks in advance.

Edit after comments that came FAR faster than expected. Thank you!

Ideal example results would look like the following

numReads  length  name2
0        7384      Ssxb2
7904     93237     St5
3438     12969     Taf9b
0        996       Tas2r138
0        882       Tas2r143
0        960       Tas2r144
0        6761      Tbx10
8125     43804     Tdrd1
1227     9286      Thnsl1

So it does seem like I have two questions here. The first is to group the data based on a factor. The second is how to calculate a function on the group, but output the entire row after calculating the chosen function.

I like the idea of an aggregate() followed by a merge(). But how will the merge() function know WHICH row of the original rows from which to grab the 'length' value based on a common factor level?

The data is a snapshot of gene expression data based on transcript annotations. I am trying to select the highest expressed transcript ( in terms of numReads) for an associated 'name2.' I need the length data for downstream normalization.

EDIT after trying to use the very helpful suggestion by ROLO. Thanks again!

also thank you Chase and daroczig for help as well

So I am trying to use the ddply() approach to split my dataframe by 'name2', sort by the number of reads in decreasing order, and selecting the top row. This effectively gives me the max 'name2' value of each group and maintains all my original information, especially the length.

Unfortunately, I'm trying to d开发者_运维百科o this on a dataframe with >34,000 rows. It works fine for ~1000 rows, and even ~5000 rows, but crashes when I give it my whole dataset.

I've trying to use the .parallel option but it fails with the following error:

Loading required package: foreach
Error: foreach package required for parallel plyr operation

I've also tried to monitor operation with the .progressbar option as well. the progress bar makes it to 100%, but the operation never finishes.

Any ideas on how to apply this operation to my complete dataset?


Use plyr to split on name2, then reverse sort numReads and select the first row:

require(plyr)
ddply(df, "name2", function(dat) {
    dat[order(dat$numReads, decreasing=TRUE), ][1,]
})

  numReads length    name2
1        0   7384    Ssxb2
2     7904  93237      St5
3     3438  12969    Taf9b
4        0    996 Tas2r138
5        0    882 Tas2r143
6        0    960 Tas2r144
7        0   6761    Tbx10
8     8125  43804    Tdrd1
9     1227   9286   Thnsl1


I might not get what you are after exactly, but I think you want to get the rows from the database which have the highest value in numReadsper level of name2. This could be done easily eg. with aggregate and later merge.

Your demo dataset:

df  <- structure(list(numReads = c(0L, 7904L, 3438L, 0L, 0L, 0L, 0L, 
8125L, 8124L, 8102L, 1227L), length = c(7384L, 93237L, 12969L, 
996L, 882L, 960L, 6761L, 43804L, 43738L, 39301L, 9286L), name2 = structure(c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, 8L, 8L, 8L, 9L), .Label = c("Ssxb2", 
"St5", "Taf9b", "Tas2r138", "Tas2r143", "Tas2r144", "Tbx10", 
"Tdrd1", "Thnsl1"), class = "factor")), .Names = c("numReads", 
"length", "name2"), class = "data.frame", row.names = c(NA, -11L
))

Let us aggregate the data frame by name2 with max function:

> df.a <- aggregate(numReads ~ name2, df, max)
> df.a
     name2 numReads
1    Ssxb2        0
2      St5     7904
3    Taf9b     3438
4 Tas2r138        0
5 Tas2r143        0
6 Tas2r144        0
7    Tbx10        0
8    Tdrd1     8125
9   Thnsl1     1227

And merge the original values of length to the data frame (df.a):

> merge(df.a, df)
     name2 numReads length
1    Ssxb2        0   7384
2      St5     7904  93237
3    Taf9b     3438  12969
4 Tas2r138        0    996
5 Tas2r143        0    882
6 Tas2r144        0    960
7    Tbx10        0   6761
8    Tdrd1     8125  43804
9   Thnsl1     1227   9286

I hope I did not misunderstood your question!


There are seemingly two different questions here. The first can be solved with the plyr package:

library(plyr)
txt <- "numReads length    name2

0   7384    Ssxb2
7904  93237      St5
3438  12969    Taf9b
0    996 Tas2r138
0    882 Tas2r143
0    960 Tas2r144
0   6761    Tbx10
8125  43804    Tdrd1
8124  43738    Tdrd1
8102  39301    Tdrd1
1227   9286   Thnsl1
"

dat <- read.table(textConnection(txt), header = TRUE)

ddply(dat, "name2", summarize, max = max(numReads))

Gives you:

     name2  max
1    Ssxb2    0
2      St5 7904
3    Taf9b 3438
4 Tas2r138    0
5 Tas2r143    0
6 Tas2r144    0
7    Tbx10    0
8    Tdrd1 8125
9   Thnsl1 1227

The second question can seemingly be answered with:

dat[dat$name2 == "Tdrd1" & dat$numReads != max(dat$numReads[dat$name2 == "Tdrd1"]),]

   numReads length name2
9      8124  43738 Tdrd1
10     8102  39301 Tdrd1

Provide some more context on what you're trying to do and I'll elaborate further.

0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号