Question

我正在处理一些遗传数据，但我的一栏内容不是我想要的格式。我不知道在这里谈论了多少生物学，但我正在尝试固定氨基酸在数据中的显示方式。

氨基酸显然有一个名称，但也有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，但我想将其更改为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")

以下是3个字母的氨基酸及其更好的缩写的列表。

  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

我觉得可以做一个简单的功能来做到这一点，但是我正在努力做到这一点。我习惯于只更改列的一部分而不是一次更改两件事。所以我要问的是我该如何改变

Ala12Val
Leu54Pro
Met99Ser
Lys34Ile
Arg4Cys
His43Gly
Glu14Phe

对此

A12V
L54P
M99S
K32I
R4C
E14F

这可以做吗？

Answer 1

查找氨基酸，然后获取子字符串的前3个字母并映射，提取数字，子字符串的后3个字母并映射。然后将所有粘贴在一起。

scalaVersion := "2.11.12"
libraryDependencies += "org.apache.spark" %% "spark-sql" % "2.3.0"
libraryDependencies += "com.microsoft.sqlserver" % "mssql-jdbc" % "6.2.1.jre8"
libraryDependencies += "org.scalafx" %% "scalafx" % "8.0.144-R12"
libraryDependencies += "org.apache.ignite" % "ignite-core" % "2.5.0"
libraryDependencies += "org.apache.ignite" % "ignite-spring" % "2.5.0"
libraryDependencies += "org.apache.ignite" % "ignite-indexing" % "2.5.0"
libraryDependencies += "org.apache.spark" % "spark-streaming_2.11" % "2.3.0"
libraryDependencies += "org.apache.spark" % "spark-streaming-kafka-0-10_2.11" % "2.3.0"
libraryDependencies += "org.apache.spark" % "spark-sql-kafka-0-10_2.11" % "2.3.0"
libraryDependencies += "org.apache.kafka" % "kafka-streams" % "1.1.0"
libraryDependencies += "com.101tec" % "zkclient_2.11" % "0.10"
libraryDependencies += "org.apache.kafka" % "kafka_2.11" % "1.1.0"

Answer 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]

Answer 3

如果它始终为{acid，numbers，acid}形式，则可以将其分为三列，并用match或连接进行替换。使用data.table，看起来就像...

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]

给出

   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

（可选）再次合并各列，然后删除不需要的列：

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

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

如何操纵两根柱子？

3 个答案: