比较两列“按序列”并制作新列-Java 学习之路

问题很难解释，但让我告诉你我想从这些数据中得到什么 . 所以，我有一个包含20个不同列的数据，其中有两个已在此处显示 .

Sequence             modifications
AAAAGAAAVANQGKK     [14] Acetyl (K)|[15] Acetyl (K)
AAAAGAAAVANQGKK     [14] Acetyl (K)|[15] Acetyl (K)
AAIKFIKFINPKINDGE   [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)
AAIKFIKFINPKINDGE   [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)
AAIKFIKFINPKINDGE   [7] Acetyl (K)|[12] Acetyl (K)
AAIKFIKFINPKINDGE   [4] Acetyl (K)|[7] Acetyl (K)
AAIYKLLKSHFRNE      [5] Biotin (K)|[8] Acetyl (K)
AAKKFEE             [3] Acetyl (K)|[4] Acetyl (K)

正如您在相同的序列中看到的那样，可以有不同的修改 . 有时可能有3x乙酰基，simetimes 2x乙酰基，有时只有一次，在其他情况下不会有任何修饰 . 我对“生物素和乙酰基”感兴趣只有2个修改，其他修改并不重要 . 修饰的数量取决于序列中“K”的数量 . 例如，如果序列中有3个“K”，则可能的修改数量为0 0,1,2,3且不超过3.因此，我想根据“K”的数量对这些序列（1000行）进行分组 . “在顺序和修改的数量和类型，它没有粉碎其他列 .

我希望通过R从这些数据中得到它，它是具有指定修改的不同序列组 . 例如：

First Group: (number of "K" in the sequence = 2, and both modified by acetyl)

Sequence             modifications
AAAAGAAAVANQGKK      [14] Acetyl (K)|[15] Acetyl (K)
AAIYKLLKSHFRNE       [5] Acetyl (K)|[8] Acetyl (K)

Second Group: (number of "K" in the sequence = 2, and one modified by acetyl, second nothing)

Third Group: (number of "K" in the sequence = 3, and one modified by acetyl, second acetyl, and last is biotin)

我必须包括所有可能性 . 这就是我认为在我试图编写的脚本的这个“部分”上最好的东西 . 也许你有任何其他建议如何插入这些数据 .

第二个问题是：我计算了3个不同列中的值的平均值，我想将结果放在相同的数据中但在另一列中 . 怎么做？

tbl_imp$mean <- rowMeans(subset(tbl_imp, select = c("x", "y", "w")), na.rm = TRUE)
tbl_imp$mean <- data.frame(tbl_imp$mean)

我用来计算行的平均值的代码 . 我只是不知道如何在我拥有的数据中创建一个新列，并将我的结果放在那里 . 我应该使用转换功能吗？

2 回答

我将您的数据加载为对象 aa .

mydata <-  data.frame(seqs = aa$Sequence, mods = aa$modifications) # subset of aa with sequences and modifications

    ##to find number of "K"s
    spl_seqs <- strsplit(as.character(mydata$seqs), split = "")  # split all sequences (use "as.character" because they are turned into factor)
    where_K <- lapply(spl_seqs, grep, pattern = "K") # find positions of "K"s in each sequence
    No_K <- lapply(where_K, length) # count "K"s in each sequence

    mydata$No_Ks <- No_K #add a column that informs about the number of "K"s in each sequence
    ##

我想所有看似“修改”列的大写字母都是指正在进行的修改或是“K” . 我想不出任何其他方法来简化“修改”列以便操纵它们 . 所以我只是保留不是“K”的大写字母：

names(LETTERS) <- LETTERS  # DWin's idea in this http://stackoverflow.com/questions/4423460/is-there-a-function-to-find-all-lower-case-letters-in-a-character-vector 

    spl_mods <- strsplit(as.character(mydata$mods), split = "")  # split the characters in each modification row

简化修改列，仅保留每个修改的第一个字母：

mods_ls <- vector("list", length = nrow(mydata))  #list to fill with simplified modifications
    for(i in 1:length(spl_mods))
     {
      res <- as.character(na.omit(LETTERS[strsplit(as.character(mydata$mods), split = "")[[i]]])) #keep only upper-case letters

      res <- as.character(na.omit(gsub("K", NA, res)))  # exclude "K"s 
      res <- as.character(na.omit(gsub("M", NA, res)))  # and "M"s I guessed

      mods_ls[[i]] <- res
     }
    mydata$simplified_mods <- unlist(lapply(mods_ls, paste, collapse = " ; "))

到目前为止我们得到了什么：

mydata[1:10,]
    #                seqs                                          mods No_Ks simplified_mods
    #1    AAAAGAAAVANQGKK               [14] Acetyl (K)|[15] Acetyl (K)     2           A ; A
    #2    AAAAGAAAVANQGKK               [14] Acetyl (K)|[15] Acetyl (K)     2           A ; A
    #3      AAFTKLDQVWGSE                                [5] Acetyl (K)     1               A
    #4  AAIKFIKFINPKINDGE [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)     3       A ; A ; A
    #5  AAIKFIKFINPKINDGE [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)     3       A ; A ; A
    #6  AAIKFIKFINPKINDGE                [7] Acetyl (K)|[12] Acetyl (K)     3           A ; A
    #7  AAIKFIKFINPKINDGE                 [4] Acetyl (K)|[7] Acetyl (K)     3           A ; A
    #8     AAIYKLLKSHFRNE                 [5] Biotin (K)|[8] Acetyl (K)     2           B ; A
    #9            AAKKFEE                 [3] Acetyl (K)|[4] Acetyl (K)     2           A ; A
    #10           AAKYFRE                                [3] Acetyl (K)     1               A

然后，您可以对“K”的数量和所需的特定修改进行子集化 . 例如 . ：

how_many_K <- 2 
    what_mods <- "A ; A"    #separated by [space];[space]

    show_rows <- which(mydata$No_Ks == how_many_K & mydata$simplified_mods == what_mods)  
    mydata[show_rows,]
    #                             seqs                            mods No_Ks simplified_mods
    #1                 AAAAGAAAVANQGKK [14] Acetyl (K)|[15] Acetyl (K)     2           A ; A
    #2                 AAAAGAAAVANQGKK [14] Acetyl (K)|[15] Acetyl (K)     2           A ; A
    #9                         AAKKFEE   [3] Acetyl (K)|[4] Acetyl (K)     2           A ; A
    #11                     AANVKKTLVE   [5] Acetyl (K)|[6] Acetyl (K)     2           A ; A
    #14  AARDSKSPIILQTSNGGAAYFAGKGISNE  [6] Acetyl (K)|[24] Acetyl (K)     2           A ; A
    #20                        AEKLKAE   [3] Acetyl (K)|[5] Acetyl (K)     2           A ; A
    #21
    #....

编辑：这一切都可以在像 fun 这样的函数中完成 . x 是您的 data.frame （与 structure 上传的"for Henrik"） . noK 是您想要的"K"的数量 . mod 是您希望用[空格]分隔的修改; [空格]（例如"B ; A ; O"）：

fun <- function(x, noK, no_modK = NULL, mod = NULL) #EDIT_1e: update arguments; made optional
    {
     mydata <- data.frame(seqs = x$Sequence, mods = x$modifications) 

     spl_seqs <- strsplit(as.character(mydata$seqs), split = "")  
     where_K <- lapply(spl_seqs, grep, pattern = "K") 
     No_K <- lapply(where_K, length)

     mydata$No_Ks <- No_K 

     names(LETTERS) <- LETTERS  

     spl_mods <- strsplit(as.character(mydata$mods), split = "")  

     mods_ls <- vector("list", length = nrow(mydata))  
     for(i in 1:length(spl_mods))
      {
       res <- as.character(na.omit(LETTERS[strsplit(as.character(mydata$mods), split = "")[[i]]])) 

       no_modedK <- length(grep("K", res))   #EDIT_1a: how many "K"s are modified?

       res <- as.character(na.omit(gsub("K", NA, res)))   
       res <- as.character(na.omit(gsub("M", NA, res)))  

       mods_ls[[i]] <- list(mods = res, modified_K = no_modedK) #EDIT_1b: catch number of "K"s modified (along with the actual modifications) 
      }

     mydata$no_modK <- unlist(lapply(lapply(lapply(mods_ls, `[`, 2), unlist), paste, collapse = " ; ")) #EDIT_1d: insert number of modified "K"s in "mydata"   
     mydata$simplified_mods <- unlist(lapply(lapply(lapply(mods_ls, `[`, 1), unlist), paste, collapse = " ; ")) #EDIT_1c: insert mods in "mydata"  

     if(!is.null(no_modK) & !is.null(mod)) #EDIT_1f: update "return"
      {
       show_rows <- which(mydata$No_Ks == noK & mydata$no_modK == no_modK & mydata$simplified_mods == mod) 
      }
     if(is.null(no_modK) & !is.null(mod))
      {
       show_rows <- which(mydata$No_Ks == noK & mydata$simplified_mods == mod) 
      } 
     if(is.null(mod) & !is.null(no_modK)) 
      {
       show_rows <- which(mydata$No_Ks == noK & mydata$no_modK == no_modK)
      }

     if(is.null(no_modK) & is.null(mod)) 
      {
       show_rows <- which(mydata$No_Ks == noK) 
      } 

     return(mydata[show_rows,])
    }

例如 . ：

fun(aa, noK = 3) #aa is the the "for Henrik" loaded in `R` (aa <- structure( ... )
                                  seqs                                                             mods No_Ks no_modK simplified_mods
    4                AAIKFIKFINPKINDGE                    [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)     3       3       A ; A ; A
    5                AAIKFIKFINPKINDGE                    [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)     3       3       A ; A ; A
    6                AAIKFIKFINPKINDGE                                   [7] Acetyl (K)|[12] Acetyl (K)     3       2           A ; A
    #...
    fun(aa, noK = 3, no_modK = 2)
                             seqs                                             mods No_Ks no_modK simplified_mods
    6           AAIKFIKFINPKINDGE                   [7] Acetyl (K)|[12] Acetyl (K)     3       2           A ; A
    7           AAIKFIKFINPKINDGE                    [4] Acetyl (K)|[7] Acetyl (K)     3       2           A ; A
    #...

    fun(aa, noK = 2, mod = "A ; B")
                  seqs                           mods No_Ks no_modK simplified_mods
    200    ISAMVLTKMKE [8] Acetyl (K)|[10] Biotin (K)     2       2           A ; B
    441 NLKPSKPSYYLDPE  [3] Acetyl (K)|[6] Biotin (K)     2       2           A ; B
    #...

    fun(aa, noK = 2, no_modK = 1, mod =  "A")
                                     seqs            mods No_Ks no_modK simplified_mods
    15      AARDSKSPIILQTSNGGAAYFAGKGISNE [24] Acetyl (K)     2       1               A
    27                     AKALVAQGVKFIAE  [2] Acetyl (K)     2       1               A
    #...

EDIT_1：更新了 fun 和示例 .

回复于 2024-05-03T02:14:52+08:00

这样的事情可能适用于你的第一部分 . 我现在无法下载文件，但是当我可以的时候，我也会尝试回复第二部分 .

library(data.table)
library(stringr)

# Slightly modified dataset
dataset <- data.table(
Sequence  = c(
'AAAAGAAAVANQGKK'    
,'AAAAGAAAVANQGKK'    
,'AAIKFIKFINPKINDGE'  
,'AAIKFIKFINPKINDGE'  
,'AAIKFIKFINPKINDGE'  
,'AAIKFIKFINPKINDGE'
,'AAIYKLLKSHFRNE'
,'AAKKFEE'
),
 modifications = c(
'[14] Acetyl (K)|[15] Acetyl (K)'
,'[14] Acetyl (K)|[15] Acetyl (K)'
,'[4] Acetyl (K)|[7] Something (K)|[12] Acetyl (K)'
,'[4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)'
,'[7] Acetyl (K)|[12] Acetyl (K)'
,'[4] Acetyl (K)|[7] Acetyl (K)'
,'[5] Biotin (K)|[8] Acetyl (K)'
,'[3] Acetyl (K)'
)
)

# get the 1st, 2nd, 3rd modifications in separate columns
dataset <- data.table(cbind(
   dataset,
   str_split_fixed(dataset[,modifications], pattern = "\\(K\\)",3)
))

dataset[,':='(
   V1 = as.character(V1),
   V2 = as.character(V2),
   V3 = as.character(V3)
)]

# Count of modifications    
dataset[, NoOfKs := 3]
dataset[V3 == "", NoOfKs := 2]
dataset[V2 == "", NoOfKs := 1]
dataset[V1 == "", NoOfKs := 0]

# Retaining Acetyl/Biotin or no modification only
dataset[, AB01 := TRUE]
dataset[, AB02 := TRUE]
dataset[, AB03 := TRUE]

dataset[V1 != "",  AB01 := grepl(V1, pattern = "Acetyl|Biotin")]
dataset[V2 != "",  AB02 := grepl(V2, pattern = "Acetyl|Biotin")]
dataset[V3 != "",  AB03 := grepl(V3, pattern = "Acetyl|Biotin")]

dataset <- dataset[AB01 & AB02 & AB03]

# Marking each modification as acetyl/biotin/none
dataset[V1 != " " & grepl(V1, pattern = "Acetyl"), AB1 := "Acetyl"]
dataset[V1 != " " & grepl(V1, pattern = "Biotin"), AB1 := "Biotin"]
dataset[V2 != " " & grepl(V2, pattern = "Acetyl"), AB2 := "Acetyl"]
dataset[V2 != " " & grepl(V2, pattern = "Biotin"), AB2 := "Biotin"]
dataset[V3 != " " & grepl(V3, pattern = "Acetyl"), AB3 := "Acetyl"]
dataset[V3 != " " & grepl(V3, pattern = "Biotin"), AB3 := "Biotin"]

dataset[
   ,
   list(
   Sequence = Sequence, 
   modifications = modifications, 
   GroupID = .GRP
   ),
   by = c('NoOfKs','AB1','AB2','AB3')
]

产量

NoOfKs    AB1    AB2    AB3          Sequence                                 modifications GroupID
1:      2 Acetyl Acetyl     NA   AAAAGAAAVANQGKK               [14] Acetyl (K)|[15] Acetyl (K)       1
2:      2 Acetyl Acetyl     NA   AAAAGAAAVANQGKK               [14] Acetyl (K)|[15] Acetyl (K)       1
3:      2 Acetyl Acetyl     NA AAIKFIKFINPKINDGE                [7] Acetyl (K)|[12] Acetyl (K)       1
4:      2 Acetyl Acetyl     NA AAIKFIKFINPKINDGE                 [4] Acetyl (K)|[7] Acetyl (K)       1
5:      3 Acetyl Acetyl Acetyl AAIKFIKFINPKINDGE [4] Acetyl (K)|[7] Acetyl (K)|[12] Acetyl (K)       2
6:      2 Biotin Acetyl     NA    AAIYKLLKSHFRNE                 [5] Biotin (K)|[8] Acetyl (K)       3
7:      1 Acetyl     NA     NA           AAKKFEE                                [3] Acetyl (K)       4

回复于 2024-05-03T02:14:52+08:00

比较两列“按序列”并制作新列

2 回答

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