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我有 tab delim 文件，其中包含以下信息

>fasta 
    >ss_23_122_0_1
    MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS
    >ss_23_167_0_1
    WEIURIOWERWKLEJDSAJFASDGASZDTTQZWTEZQWTEZUQWEZQWTEZQTWEZTQW
    >ss_23_167_0_1
    MAASDASDWEPWERIWERIWER
    >ss_23_167_0_1
    QWEKCKLSDOIEOWIOWEUWWEUWEZURZEWURZUWEUZUQZUWZUE
    >ss_45_201_0_1
    HZTMKSKDIUWZUWEZTZWERWUEOIRUOEROOWEWERSDFSDFRRRETERTER
    >ss_45_201_0_1
    ZTTRASOIIDIFOSDIOFISDOFSDFQAWTZETQWE
    >ss_89_10_0_2
    NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP

对于像这样的 idss_45_201_0_1并且ss_23_167_0_1有多个条目，我想只保留那些具有最大长度的条目。我想得到如下输出：

>fasta
    >ss_23_122_0_1
    MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS
    >ss_23_167_0_1
    WEIURIOWERWKLEJDSAJFASDGASZDTTQZWTEZQWTEZUQWEZQWTEZQTWEZTQW
    >ss_45_201_0_1
    HZTMKSKDIUWZUWEZTZWERWUEOIRUOEROOWEWERSDFSDFRRRETERTER
    >ss_89_10_0_2
    NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP

我在 R 中尝试了以下代码，但失败了

Unique(fasta)

谁能指导我。对于那些具有不同长度的多个条目的相同 ID，我如何才能获得最长的序列。

4

2 回答 2

2

这里有三个可供考虑的选项。

选项 1：基础 R

取消列出列表，使用该列表，nchar然后使用ave找出要保留的值。

x <- nchar(unlist(l))
l[as.logical(ave(x, names(x), FUN = function(x) x == max(x)))]
# $ss_23_122_0_1
# [1] "MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS"
# 
# $ss_23_167_0_1
# [1] "WEIURIOWERWKLEJDSAJFASDGASZDTTQZWTEZQWTEZUQWEZQWTEZQTWEZTQW"
# 
# $ss_45_201_0_1
# [1] "HZTMKSKDIUWZUWEZTZWERWUEOIRUOEROOWEWERSDFSDFRRRETERTER"
# 
# $ss_89_10_0_2
# [1] "NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP"

选项 2：“data.table”

使用meltfrom "reshape2" 创建一个data.frame. rank与子集一起使用nchar。（我使用 rank 而不是==这样我就不必使用nchar两次 - 没有检查比较效率。）

library(data.table)
library(reshape2)
as.data.table(melt(l))[, Rnk := rank(nchar(as.character(value))), 
                       by = L1][Rnk == 1]
#                                                 value            L1 Rnk
# 1: MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS ss_23_122_0_1   1
# 2:                             MAASDASDWEPWERIWERIWER ss_23_167_0_1   1
# 3:               ZTTRASOIIDIFOSDIOFISDOFSDFQAWTZETQWE ss_45_201_0_1   1
# 4:      NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP  ss_89_10_0_2   1

选项 3：“dplyr”

与“data.table”类似的方法。

library(dplyr)
library(reshape2)
melt(l) %>%
  group_by(L1) %>%
  mutate(Rnk = dense_rank(nchar(as.character(value)))) %>%
  filter(Rnk == 1)
# Source: local data frame [4 x 3]
# Groups: L1
# 
#                                                value            L1 Rnk
# 1 MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS ss_23_122_0_1   1
# 2                             MAASDASDWEPWERIWERIWER ss_23_167_0_1   1
# 3               ZTTRASOIIDIFOSDIOFISDOFSDFQAWTZETQWE ss_45_201_0_1   1
# 4      NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP  ss_89_10_0_2   1
于 2015-03-01T07:41:13.863 回答
1

也许有更优雅的方式......

l <-list(ss_23_122_0_1 = "MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS",
                           ss_23_167_0_1 = "WEIURIOWERWKLEJDSAJFASDGASZDTTQZWTEZQWTEZUQWEZQWTEZQTWEZTQW",
                           ss_23_167_0_1 = "MAASDASDWEPWERIWERIWER",
                           ss_23_167_0_1 = "QWEKCKLSDOIEOWIOWEUWWEUWEZURZEWURZUWEUZUQZUWZUE",
                           ss_45_201_0_1 = "HZTMKSKDIUWZUWEZTZWERWUEOIRUOEROOWEWERSDFSDFRRRETERTER",
                           ss_45_201_0_1 = "ZTTRASOIIDIFOSDIOFISDOFSDFQAWTZETQWE",
                           ss_89_10_0_2 = "NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP")

res <- split(l, names(l))
ind <- lapply(split(sapply(l, nchar), names(l)), which.max)
Map(function(x, y) x[y], res, ind)
$ss_23_122_0_1
$ss_23_122_0_1$ss_23_122_0_1
[1] "MJSDHWTEZTZEWUIASUDUAISDUASADIASDIAUSIDAUSIDCASDAS"


$ss_23_167_0_1
$ss_23_167_0_1$ss_23_167_0_1
[1] "WEIURIOWERWKLEJDSAJFASDGASZDTTQZWTEZQWTEZUQWEZQWTEZQTWEZTQW"


$ss_45_201_0_1
$ss_45_201_0_1$ss_45_201_0_1
[1] "HZTMKSKDIUWZUWEZTZWERWUEOIRUOEROOWEWERSDFSDFRRRETERTER"


$ss_89_10_0_2
$ss_89_10_0_2$ss_89_10_0_2
[1] "NJZTIWEIOIOIPIEPWIQPOEIQWIEPOQWIEPOQWIEPQIWEP"
于 2015-02-28T20:33:30.063 回答