Question

I have a data frame with the following structure:

> dftest
       element seqnames     start       end  width strand tx_id   tx_name
1            1    chr19  58858172  58864865   6694      - 36769 NM_130786
2           10     chr8  18248755  18258723   9969      + 16614 NM_000015
3          100    chr20  43248163  43280376  32214      - 37719 NM_000022
4         1000    chr18  25530930  25757445 226516      - 33839 NM_001792
5        10000     chr1 243651535 244006584 355050      -  4182 NM_181690
6        10000     chr1 243663021 244006584 343564      -  4183 NM_005465
1316 100302285    chr12  12264886  12264967     82      + 24050 NR_036052
1317 100302285    chr12   9392066   9392147     82      - 25034 NR_036052
1318 100302285     chr2 232578024 232578105     82      +  5491 NR_036052
1319 100302285     chr5 118310281 118310362     82      + 11128 NR_036052

As an intermediate step I am trying to get rid of the elements, such as “100302285”, that are present more than once, but with different “seqnames”. Element “10000” would be kept because all “seqnames” are the same. Elements that are present only once are also kept. This is the desired output:

> dftest
       element seqnames     start       end  width strand tx_id   tx_name
1            1    chr19  58858172  58864865   6694      - 36769 NM_130786
2           10     chr8  18248755  18258723   9969      + 16614 NM_000015
3          100    chr20  43248163  43280376  32214      - 37719 NM_000022
4         1000    chr18  25530930  25757445 226516      - 33839 NM_001792
5        10000     chr1 243651535 244006584 355050      -  4182 NM_181690
6        10000     chr1 243663021 244006584 343564      -  4183 NM_005465

So far I’ve played with ddply and custom function to include duplicates:

subChr <- function(df)
{
    df[duplicated(df$seqnames),]
}

ddply(df, .(element), subChr)

But the result is far from the intended – silly me, it could have not been that simple:

    element seqnames     start       end  width strand tx_id   tx_name
1     10000     chr1 243663021 244006584 343564      -  4183 NM_005465
2 100302285    chr12   9392066   9392147     82      - 25034 NR_036052

Since this is a step before another ddply, I would be happy with an alternative solution that does this:

ddply(df, .(element), summarize, chromosome=seqnames[1], gene_start=min(start), gene_end=max(end), strand=strand[1])

    element chromosome gene_start  gene_end strand
1         1      chr19   58858172  58864865      -
2        10       chr8   18248755  18258723      +
3       100      chr20   43248163  43280376      -
4      1000      chr18   25530930  25757445      -
5     10000       chr1  243651535 244006584      -
6 100302285      chr12    9392066 232578105      +

but summarizes element “100302285” for each “seqnames”:

 element chromosome gene_start  gene_end strand
1         1      chr19   58858172  58864865      -
2        10       chr8   18248755  18258723      +
3       100      chr20   43248163  43280376      -
4      1000      chr18   25530930  25757445      -
5     10000       chr1  243651535 244006584      -
6 100302285      chr12    9392066  12264967      +
7 100302285       chr2  232578024 232578105      +
8 100302285       chr5  118310281 118310362      +

Basically summarizing by .element and .seqname, if that makes sense. I have been searching for an answer for sometime now but did not progress much.

Test data:

dftest <- structure(list(element = c("1", "10", "100", "1000", "10000", 
"10000", "100302285", "100302285", "100302285", "100302285"), 
    seqnames = c("chr19", "chr8", "chr20", "chr18", "chr1", "chr1", 
    "chr12", "chr12", "chr2", "chr5"), start = c(58858172L, 18248755L, 
    43248163L, 25530930L, 243651535L, 243663021L, 12264886L, 
    9392066L, 232578024L, 118310281L), end = c(58864865L, 18258723L, 
    43280376L, 25757445L, 244006584L, 244006584L, 12264967L, 
    9392147L, 232578105L, 118310362L), width = c(6694L, 9969L, 
    32214L, 226516L, 355050L, 343564L, 82L, 82L, 82L, 82L), strand = c("-", 
    "+", "-", "-", "-", "-", "+", "-", "+", "+"), tx_id = c(36769L, 
    16614L, 37719L, 33839L, 4182L, 4183L, 24050L, 25034L, 5491L, 
    11128L), tx_name = c("NM_130786", "NM_000015", "NM_000022", 
    "NM_001792", "NM_181690", "NM_005465", "NR_036052", "NR_036052", 
    "NR_036052", "NR_036052")), .Names = c("element", "seqnames", 
"start", "end", "width", "strand", "tx_id", "tx_name"), class = "data.frame", row.names = c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L))

Answer 1

Answering your first question: If you like, here’s a data.table solution:

require(data.table)
dt <- data.table(dftest, key="element")
dt.out <- dt[, .SD[length(table(seqnames)) == 1],by=c("element")]
> dt.out

#    element seqnames     start       end  width strand tx_id   tx_name
# 1:       1    chr19  58858172  58864865   6694      - 36769 NM_130786
# 2:      10     chr8  18248755  18258723   9969      + 16614 NM_000015
# 3:     100    chr20  43248163  43280376  32214      - 37719 NM_000022
# 4:    1000    chr18  25530930  25757445 226516      - 33839 NM_001792
# 5:   10000     chr1 243651535 244006584 355050      -  4182 NM_181690
# 6:   10000     chr1 243663021 244006584 343564      -  4183 NM_005465

And if you prefer the plyr solution:

require(plyr)
out <- ddply(dftest, .(element), function(x) {
    if( length(table(x$seqnames)) == 1) {
        x
    }
})

#   element seqnames     start       end  width strand tx_id   tx_name
# 1       1    chr19  58858172  58864865   6694      - 36769 NM_130786
# 2      10     chr8  18248755  18258723   9969      + 16614 NM_000015
# 3     100    chr20  43248163  43280376  32214      - 37719 NM_000022
# 4    1000    chr18  25530930  25757445 226516      - 33839 NM_001792
# 5   10000     chr1 243651535 244006584 355050      -  4182 NM_181690
# 6   10000     chr1 243663021 244006584 343564      -  4183 NM_005465

Edit: For your second question, basically, in addition to the old solution, you just want to return the first row when your first condition is not satisfied.

plyr solution: (without summarise)

out <- ddply(dftest, .(element), function(x) {
    if (length(table(x$seqnames)) == 1) {
        x
    } else {
        x[1, ]
    }
})

> out
#     element seqnames     start       end  width strand tx_id   tx_name
# 1         1    chr19  58858172  58864865   6694      - 36769 NM_130786
# 2        10     chr8  18248755  18258723   9969      + 16614 NM_000015
# 3       100    chr20  43248163  43280376  32214      - 37719 NM_000022
# 4      1000    chr18  25530930  25757445 226516      - 33839 NM_001792
# 5     10000     chr1 243651535 244006584 355050      -  4182 NM_181690
# 6     10000     chr1 243663021 244006584 343564      -  4183 NM_005465
# 7 100302285    chr12  12264886  12264967     82      + 24050 NR_036052

data.table solution.

dt <- data.table(dftest, key="element")
dt[, .SD[(if(length(table(seqnames)) == 1) seq_len(.N) else 1)], by = element]

> dt.out
#      element seqnames     start       end  width strand tx_id   tx_name
# 1:         1    chr19  58858172  58864865   6694      - 36769 NM_130786
# 2:        10     chr8  18248755  18258723   9969      + 16614 NM_000015
# 3:       100    chr20  43248163  43280376  32214      - 37719 NM_000022
# 4:      1000    chr18  25530930  25757445 226516      - 33839 NM_001792
# 5:     10000     chr1 243651535 244006584 355050      -  4182 NM_181690
# 6:     10000     chr1 243663021 244006584 343564      -  4183 NM_005465
# 7: 100302285    chr12  12264886  12264967     82      + 24050 NR_036052