使用 ddply 排除某些列中的重复值

我有一个具有以下结构的数据框：

> 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

作为中间步骤，我试图删除多次出现但具有不同“seqnames”的元素，例如“100302285”。元素“10000”将被保留，因为所有“seqnames”都是相同的。仅出现一次的元素也会被保留。这是所需的输出：

> 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

到目前为止，我已经使用 ddply 和自定义函数来包含重复项：

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

ddply(df, .(element), subChr)

但结果与预期相去甚远——愚蠢的我，事情不可能那么简单：

    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

由于这是另一个 ddply 之前的一步，我会很高兴替代解决方案这样做是这样的：

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      +

但总结了每个“seqnames”的元素“100302285”：

 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      +

基本上通过 .element 和 .seqname 进行总结，如果这有意义的话。我已经寻找答案有一段时间了，但没有取得多大进展。

测试数据：

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))

回答你的第一个问题：如果你愿意，这里有一个data.table解决方案：

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

如果您更喜欢plyr解决方案：

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:对于你的第二个问题，基本上，除了旧的解决方案之外，你只想在第一个条件不满足时返回第一行。

plyr解决方案：（无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解决方案。

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