Question

我有包含以下信息的tab delim文件文件

>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ss_45_201_0_1和ss_23_167_0_1这样的ID，有多个条目，我想只保留那些最大长度为全部的条目。我希望获得如下输出：

>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，但它失败了

Unique(fasta)

任何人都可以指导我。如何才能获得具有不同长度的多个条目的相同ID的最长序列。

Answer 1

以下三个选项需要考虑。

选项1：基础R

取消列表，在其上使用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：＆＃34; data.table＆＃34;

使用来自＆＃34; reshape2＆＃34;的melt创建data.frame。将rank与nchar一起使用到子集。（我使用了排名而不是==，因此我没有必要使用nchar两次 - 还没有检查比较效率。）

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：＆＃34; dplyr＆＃34;

＆＃34; data.table＆＃34;。

的类似方法

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

Answer 2

也许有一种更优雅的方式...

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"

从制表符分隔文件中提取最长的序列

2 个答案:

选项1：基础R

选项2：＆＃34; data.table＆＃34;

选项3：＆＃34; dplyr＆＃34;