R - 生成动态列数和子串列值-Java 学习之路

在R中寻找数据操作的帮助 . 我有以下格式的数据;

ID  L1  L2  L3
1   BBCBCACCBCB CBCBBBB BEBBBAAB
2   BBCBCCCCBCB CBCCCBC BBAACCCB
3   BBCBCACCBCB CBCBBBB BEBBBAAB
4   BBCBCACCBCB CBCBBBB BEBBBAAB
5   BBCBACBCCCB BBCCCBC BBCBAAAAB
6   BBCBBCCBBCB BBCBCEB BBBBCAACB
7   BBCBBCCBBCB BBCBCEB BBBBCAACB
8           
9   BBCBCACCBCB CBCBBBB BEBBBAAB
10  BBCBBCCBBCB BBCBCEB BBBBCAACB
11  BBCBBCCBBCB BBCBCEB BBBBCAACB

每列中的值将是不同长度的字符串 . 我想要一个R函数，对于上面的每一列，都会

1）基于列中任何字符串的最大长度生成动态数量的列，例如， L1最大长度= 11，因此11个新列各自标记为L1_1：L1_11

2）然后将字符串分成三元组，例如

ID  L1  L2  L3  L1_1    L1_2    L1_3    L1_4    L1_5    L1_6    L1_7    L1_8    L1_9
1   BBCBCACCBCB CBCBBBB BEBBBAAB    BBC BCB CBC BCA CAC ACC CCB CBC BCB

3）对三元组中的三元组进行计算，即（'a'* 1的数量）（'b'* 3的数量）（'c'* 7的数量） .

4）在新列中返回此计算的值 .

我发现建议的代码完全符合我在运行列L1，L2时所需的功能，但不适用于L3 . 我收到的错误是'as.data.frame.matrix中的错误（passed.args [[i]]，stringsAsFactors = st：缺少值，其中需要TRUE / FALSE'

有任何想法吗？非常感谢 .

编辑

dput（DF）：

structure(list(ID = 1:11, L1 = structure(c(4L, 5L, 4L, 4L, 2L, 3L, 3L, 1L, 4L, 3L, 3L), .Label = c("", "BBCBACBCCCB","BBCBBCCBBCB","BBCBCACCBCB", "BBCBCCCCBCB"), class = "factor"), L2 = structure(c(4L, 5L, 4L, 4L, 3L, 2L, 2L, 1L, 4L, 2L, 2L), .Label = c("","BBCBCEB","BBCCCBC", "CBCBBBB", "CBCCCBC"), class = "factor"), L3 = structure(c(5L,2L, 5L, 5L, 4L, 3L, 3L, 1L, 5L, 3L, 3L), .Label = c("", "BBAACCCB", "BBBBCAACB", "BBCBAAAAB", "BEBBBAAB"), class = "factor")), .Names = c("ID", "L1", "L2", "L3"), class = "data.frame", row.names = c(NA,-11L))

结构（列表（ID = 1：11，L1 =结构（c（4L，5L，4L，4L，2L，3L，3L，1L，4L，3L，3L）,. Label = c（“”，“BBCBACBCCCB” ，“BBCBBCCBBCB”，“BBCBCACCBCB”，“BBCBCCCCBCB”），类=“因子”），L2 =结构（c（4L，5L，4L，4L，3L，2L，2L，1L，4L，2L，2L）， .Label = c（“”，“BBCBCEB”，“BBCCCBC”，“CBCBBBB”，“CBCCCBC”），class =“factor”），L3 =结构（c（5L，2L，5L，5L，4L，3L， 3L，1L，5L，3L，3L）， . Label = c（“”，“BBAACCCB”，“BBBBCAACB”，“BBCBAAAAB”，“BEBBBAAB”），class =“factor”））， . Name = c（“ ID“，”L1“，”L2“，”L3“），class =”data.frame“，row.names = c（NA，-11L））

2 回答

#DATA
df = structure(list(ID = 1:4, L1 = c("abbbcc", "aabacd", "abbda", 
"bbad")), .Names = c("ID", "L1"), class = "data.frame", row.names = c(NA, 
-4L))

#Go through the strings and split into subgroups of 3 characters.
#Put the substrings in a list
temp = lapply(df$L1, function(x) sapply(3:nchar(x), function(i) substr(x, i-2, i)))

#Obtain the length of the subgroup with the most triplets
temp_l = max(lengths(temp))

#Subset the subgroups from 1 to temp_l so that remianing values are NA
cbind(df, setNames(data.frame(do.call(rbind, lapply(temp, function(a)
    a[1:temp_l]))), nm = paste0("L1_",1:temp_l)))
#  ID     L1 L1_1 L1_2 L1_3 L1_4
#1  1 abbbcc  abb  bbb  bbc  bcc
#2  2 aabacd  aab  aba  bac  acd
#3  3  abbda  abb  bbd  bda <NA>
#4  4   bbad  bba  bad <NA> <NA>

如果要基于三元组进行计算，请在执行 cbind 步骤之前运行以下命令

temp_L1 = lapply(df$L1, function(x) sapply(3:nchar(x), function(i) substr(x, i-2, i)))
temp_L1_length = max(lengths(temp_L1))
temp_L1 = lapply(temp_L1, function(x)
             sapply(x, function(y){
                     num_a = unlist(gregexpr(pattern = "a", text = y))
                     num_a = sum(num_a > 0)  #length of positive match
                     num_b = unlist(gregexpr(pattern = "b", text = y))
                     num_b = sum(num_b > 0)
                     num_c = unlist(gregexpr(pattern = "c", text = y))
                     num_c = sum(num_c > 0)
                     num_a * 1 + num_b * 3 + num_c * 7
                 })
         )
temp_L1 = setNames(data.frame(do.call(rbind, lapply(temp_L1, function(a)
              a[1:temp_L1_length]))), nm = paste0("L1_",1:temp_L1_length))

#REPEAT FOR L2, L3, ...

cbind(df, temp_L1)   #Run cbind(df, temp_L1, temp_L2, ...)
#  ID     L1 L1_1 L1_2 L1_3 L1_4
#1  1 abbbcc    7    9   13   17
#2  2 aabacd    5    5   11    8
#3  3  abbda    7    6    4   NA
#4  4   bbad    7    4   NA   NA

UPDATE

您可以创建一个函数并使用它，如下所示

#FUNCTION
foo = function(data, column){
    temp_L1 = lapply(as.character(data[[column]]), function(x) sapply(3:nchar(x), function(i) substr(x, i-2, i)))
    temp_L1_length = max(lengths(temp_L1))
    temp_L1 = lapply(temp_L1, function(x)
        sapply(x, function(y){
            num_a = unlist(gregexpr(pattern = "a", text = y, ignore.case = TRUE))
            num_a = sum(num_a > 0)  #length of positive match
            num_b = unlist(gregexpr(pattern = "b", text = y, ignore.case = TRUE))
            num_b = sum(num_b > 0)
            num_c = unlist(gregexpr(pattern = "c", text = y, ignore.case = TRUE))
            num_c = sum(num_c > 0)
            num_a * 1 + num_b * 3 + num_c * 7
        })
    )
    temp_L1 = setNames(data.frame(do.call(rbind, lapply(temp_L1, function(a)
        a[1:temp_L1_length]))), nm = paste0(column,"_",1:temp_L1_length))
    return(temp_L1)
}

#USING ON NEW DATA
cbind(df, do.call(cbind, lapply(colnames(df)[-1], function(x) foo(data = df, column = x))))

回复于 2024-05-08T01:19:37+08:00

如果你想使用 tidyverse 动词

library(tidyverse)
df1 <- df %>%
      mutate(L2=L1) %>%              # copies L1
      nest(L2) %>%                   # nest L1
      mutate(data=map(data,~sapply(1:(nchar(.x)-2), function(y) substr(.x, y, y+2)))) %>%       # makes triplets
      unnest(data) %>%        # unnest triplets
      group_by(ID) %>%        # perform next operations group wise
      mutate(rn=letters[row_number()]) %>%        # make future column names
      spread(rn,data)         # spread long format into wide format (columns)

     ID     L1     a     b     c     d
1     1 abbbcc   abb   bbb   bbc   bcc
2     2 aabacd   aab   aba   bac   acd
3     3  abbda   abb   bbd   bda  <NA>
4     4   bbad   bba   bad  <NA>  <NA>

回复于 2024-05-08T01:19:37+08:00

R - 生成动态列数和子串列值

2 回答

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