如何操作两根柱子？答案

【问题标题】：How to manipulate two pieces of a column?如何操作两根柱子？
【发布时间】：2018-12-11 02:01:38
【问题描述】：

我正在处理一些基因数据，但我的一个专栏不是我想要的格式。我不知道这里讨论了多少生物学，但我正在尝试修复我的氨基酸在数据中的显示方式。

氨基酸显然有名称，但它们也有 3 个字母的缩写和 1 个字母的缩写。我的数据包含 3 个字母形式的氨基酸，但我想将它们更改为 1 个字母的缩写。这是我的数据示例。

 chr location           effect   impact AA_change
   1    12543 missense_variant MODERATE  p.Ala12Val
   1    52367 missense_variant MODERATE  p.Leu54Pro
   1   752347 missense_variant MODERATE  p.Met99Ser
   1   984645 missense_variant MODERATE  p.Lys34Ile
   1   989845 missense_variant MODERATE  p.Arg4Cys
   1   999854 missense_variant MODERATE  p.His43Gly
   1   999855 missense_variant MODERATE  p.Glu14Phe

dat <- structure(list(chr = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), location = c(12543L, 
52367L, 752347L, 984645L, 989845L, 999854L, 999855L), effect = c("missense_variant", 
"missense_variant", "missense_variant", "missense_variant", "missense_variant", 
"missense_variant", "missense_variant"), impact = c("MODERATE", 
"MODERATE", "MODERATE", "MODERATE", "MODERATE", "MODERATE", "MODERATE"
), AA_change = c("Ala12Val", "Leu54Pro", "Met99Ser", "Lys34Ile", 
"Arg4Cys", "His43Gly", "Glu14Phe")), .Names = c("chr", "location", 
"effect", "impact", "AA_change"), row.names = c(NA, -7L), class = "data.frame")

这里列出了 3 个字母的氨基酸及其更好的缩写。

  Ala == A
  Arg == R
  Asn == N
  Asp == D
  Cys == C
  Glu == E
  Gln == Q
  Gly == G
  His == H
  Ile == I
  Leu == L
  Lys == K
  Met == M
  Phe == F
  Pro == P
  Ser == S
  Thr == T
  Trp == W
  Tyr == Y
  Val == V

我觉得有一个简单的功能可以做到这一点，但我正在努力解决如何做到这一点。我习惯于只更改一列的一部分，而不是一次更改两件事。所以我要问的是如何改变这个

Ala12Val
Leu54Pro
Met99Ser
Lys34Ile
Arg4Cys
His43Gly
Glu14Phe

到这里

A12V
L54P
M99S
K32I
R4C
E14F

这是可以做到的吗？

【问题讨论】：

是否总是采用“3 个字母，一些数字，3 个字母”的格式？
如果格式是固定的（根据 zx 的查询），您可以拆分 col 并匹配/合并以进行更新。如果它不固定，你可以使用正则表达式：stackoverflow.com/q/6954017
@zx8754 没错。它总是 3 个字母，一些数字，然后又是 3 个字母

标签： r regex bioinformatics genetics

【解决方案1】：

查找氨基酸，然后获取子字符串前 3 个字母和映射，提取数字，子字符串后 3 个字母和映射。然后将所有内容粘贴在一起。

# lookup map
AAmap <- setNames(c("A","R","N","D","C","E","Q","G","H","I","L","K","M","F","P","S","T","W","Y","V"),
                  c("Ala","Arg","Asn","Asp","Cys","Glu","Gln","Gly","His","Ile","Leu","Lys","Met","Phe","Pro","Ser","Thr","Trp","Tyr","Val"))

# get first 3 map to AA, get digits, get last 3 map to AA
dat$AA_change_short <-
  paste0(AAmap[ substr(dat$AA_change, 1, 3) ],
         gsub("[^\\d]+", "", dat$AA_change, perl = TRUE),
         AAmap[ substr(dat$AA_change, nchar(dat$AA_change) - 2, nchar(dat$AA_change)) ])

dat
#   chr location           effect   impact AA_change AA_change_short
# 1   1    12543 missense_variant MODERATE  Ala12Val            A12V
# 2   1    52367 missense_variant MODERATE  Leu54Pro            L54P
# 3   1   752347 missense_variant MODERATE  Met99Ser            M99S
# 4   1   984645 missense_variant MODERATE  Lys34Ile            K34I
# 5   1   989845 missense_variant MODERATE   Arg4Cys             R4C
# 6   1   999854 missense_variant MODERATE  His43Gly            H43G
# 7   1   999855 missense_variant MODERATE  Glu14Phe            E14F

【讨论】：

开枪！我对结构实际上是错误的。它是 5 个字符，一些数字，然后是 3 个字符。你写的很好用！我只是得到第一个氨基酸的 NA，因为它有一个 p。前面是 3 个字母的缩写。
@Brian 然后删除前 2 个字符，然后按照上面的代码：dat$AA_change <- gsub("p.", "", dat$AA_change, fixed = TRUE)

【解决方案2】：

b=which(adist(dat2$V1,dat$AA_change,partial = T)==0,T)

dat$AA_change1=`regmatches<-`(dat$AA_change,gregexpr("\\D+",dat$AA_change),
                                 value=split(dat2$V3[b[,1]],b[,2]))

dat
  chr location           effect   impact AA_change AA_change1
1   1    12543 missense_variant MODERATE  Ala12Val       A12V
2   1    52367 missense_variant MODERATE  Leu54Pro       L54P
3   1   752347 missense_variant MODERATE  Met99Ser       M99S
4   1   984645 missense_variant MODERATE  Lys34Ile       I34K
5   1   989845 missense_variant MODERATE   Arg4Cys        R4C
6   1   999854 missense_variant MODERATE  His43Gly       G43H
7   1   999855 missense_variant MODERATE  Glu14Phe       E14F



dat2 = read.table(text="Ala == A
  Arg == R
  Asn == N
  Asp == D
  Cys == C
  Glu == E
  Gln == Q
  Gly == G
  His == H
  Ile == I
  Leu == L
  Lys == K
  Met == M
  Phe == F
  Pro == P
  Ser == S
  Thr == T
  Trp == W
  Tyr == Y
  Val == V")[-2]

【讨论】：

【解决方案3】：

如果它始终是 {acid, numbers, acid} 的形式，您可以将其拆分为三列，并使用 match 或连接进行替换。使用 data.table，这看起来像......

library(data.table)
setDT(dat)

# put your mapping into a nicer format
abbrDT = fread(header = FALSE,"
  Ala == A
  Arg == R
  Asn == N
  Asp == D
  Cys == C
  Glu == E
  Gln == Q
  Gly == G
  His == H
  Ile == I
  Leu == L
  Lys == K
  Met == M
  Phe == F
  Pro == P
  Ser == S
  Thr == T
  Trp == W
  Tyr == Y
  Val == V")[, .(abbr3 = V1, abbr1 = V3)] 

# split the column
patt = "(?<=\\d)(?=\\D)|(?<=\\D)(?=\\d)"
dat[, c("AA1", "num", "AA2") := tstrsplit(AA_change, patt, perl=TRUE)]

# substitute for each part
dat[abbrDT, on=.(AA1 = abbr3), AA1 := abbr1]
dat[abbrDT, on=.(AA2 = abbr3), AA2 := abbr1]

给了

   chr location           effect   impact AA_change AA1 num AA2
1:   1    12543 missense_variant MODERATE  Ala12Val   A  12   V
2:   1    52367 missense_variant MODERATE  Leu54Pro   L  54   P
3:   1   752347 missense_variant MODERATE  Met99Ser   M  99   S
4:   1   984645 missense_variant MODERATE  Lys34Ile   K  34   I
5:   1   989845 missense_variant MODERATE   Arg4Cys   R   4   C
6:   1   999854 missense_variant MODERATE  His43Gly   H  43   G
7:   1   999855 missense_variant MODERATE  Glu14Phe   E  14   F

或者，再次合并列并删除不需要的列：

dat[, AA_change := paste0(AA1, num, AA2)]

dat[, c("AA1", "num", "AA2") := NULL]

【讨论】：