Question

我是R的新手，无法搜索我遇到的具体问题的答案。

如果我的数据框如下所示：

d <- data.frame(Name = c("Jon", "Jon", "Jon", "Kel", "Kel", "Kel", "Don", "Don", "Don"),
             No1 = c(1,2,3,1,1,1,3,3,3),
             No2 = c(1,1,1,2,2,2,3,3,3))

Name No1 No2
Jon   1   1
Jon   2   1
Jon   3   1
Kel   1   2
Kel   1   2
Kel   1   2
Don   3   3
Don   3   3
Don   3   3 
...

我如何添加能够向数据框添加新列，其中列将指示列No1和No2中的唯一值：这将是（1,2,3），（1,2），（3）分别为John，Kelly，Don，

因此，如果新列名为ID#，则所需结果应为

d2 <- data.frame(Name = c("Jon", "Jon", "Jon", "Kel", "Kel", "Kel", "Don", "Don", "Don"),
          No1 = c(1,2,3,1,1,1,3,3,3),
          No2 = c(1,1,1,2,2,2,3,3,3),
          ID1 = c(1,1,1,1,1,1,3,3,3),
          ID2 = c(2,2,2,2,2,2,NA,NA,NA),
          ID3 = c(3,3,3,NA,NA,NA,NA,NA,NA))

Name No1 No2 ID1 ID2 ID3
Jon   1   1   1   2   3 
Jon   2   1   1   2   3
Jon   3   1   1   2   3 
Kel   1   2   1   2   NA
Kel   1   2   1   2   NA
Kel   1   2   1   2   NA
Don   3   3   3   NA  NA
Don   3   3   3   NA  NA
Don   3   3   3   NA  NA

Answer 1

declare var __moduleName: any;
@Component({
    moduleId: __moduleName,
    selector: 'dashboard',
    templateUrl: 'dashboard.html',
    styleUrls: ['dashboard.css']
})

Answer 2

整齐的方法：

library(dplyr)
library(tidyr)

      # evaluate separately for each name
d %>% group_by(Name) %>% 
    # add a column of the unique values pasted together into a string
    mutate(ID = paste(unique(c(No1, No2)), collapse = ' ')) %>% 
    # separate the string into individual columns, filling with NA and converting to numbers
    separate(ID, into = paste0('ID', 1:3), fill = 'right', convert = TRUE)

## Source: local data frame [9 x 6]
## Groups: Name [3]
## 
##     Name   No1   No2   ID1   ID2   ID3
## * <fctr> <dbl> <dbl> <int> <int> <int>
## 1    Jon     1     1     1     2     3
## 2    Jon     2     1     1     2     3
## 3    Jon     3     1     1     2     3
## 4    Kel     1     2     1     2    NA
## 5    Kel     1     2     1     2    NA
## 6    Kel     1     2     1     2    NA
## 7    Don     3     3     3    NA    NA
## 8    Don     3     3     3    NA    NA
## 9    Don     3     3     3    NA    NA

这是一个很好的基础版本，采用基本的split-apply-combine方法：

# store distinct values in No1 and No2
cols <- unique(unlist(d[,-1]))
                           # split No1 and No2 by Name,
ids <- data.frame(t(sapply(split(d[,-1], d$Name), 
                           # find unique values for each split,
                           function(x){y <- unique(unlist(x))
                                       # pad with NAs,
                                       c(y, rep(NA, length(cols) - length(y)))
                           # and return a data.frame
                           }))) 
# fix column names
names(ids) <- paste0('ID', cols)
# turn rownames into column
ids$Name <- rownames(ids)
# join two data.frames on Name columns
merge(d, ids, sort = FALSE)

##   Name No1 No2 ID1 ID2 ID3
## 1  Jon   1   1   1   2   3
## 2  Jon   2   1   1   2   3
## 3  Jon   3   1   1   2   3
## 4  Kel   1   2   1   2  NA
## 5  Kel   1   2   1   2  NA
## 6  Kel   1   2   1   2  NA
## 7  Don   3   3   3  NA  NA
## 8  Don   3   3   3  NA  NA
## 9  Don   3   3   3  NA  NA

只是为了踢，这里是一个创造性的备用基础版本，它利用table而不是分割/分组：

# copy d so as not to distort original with factor columns
d_f <- d
# make No* columns factors to ensure similar table structure
d_f[, -1] <- lapply(d[,-1], factor, levels = unique(unlist(d[, -1])))
# make tables of cols, sum to aggregate occurrences, and set as boolean mask for > 0
tab <- Reduce(`+`, lapply(d_f[, -1], table, d_f$Name)) > 0
# replace all TRUE values with values they tabulated
tab <- tab * matrix(as.integer(rownames(tab)), nrow = nrow(tab), ncol = ncol(tab))
# replace 0s with NAs
tab[tab == 0] <- NA
# store column names
cols <- paste0('ID', rownames(tab))
# sort each row, keeping NAs
tab <- data.frame(t(apply(tab, 2, sort, na.last = T)))
# apply stored column names
names(tab) <- cols
# turn rownames into column
tab$Name <- rownames(tab)
# join two data.frames on Name columns
merge(d, tab, sort = FALSE)

结果完全相同。

Answer 3

我们可以使用单个外部包，即data.table并获取输出。转换＆＃39; data.frame＆＃39;到＆＃39; data.table＆＃39; （setDT(d)），按姓名＆＃39;分组，我们unlist .SDcols中提到的列，获取唯一值，{＆＃39}}来自＆＃ 39;长＆＃39;广泛的＆＃39;格式，与原始数据集dcast进行联接＆＃34;名称＆＃34;列。

on

或者这可以通过library(data.table) dcast(setDT(d)[, unique(unlist(.SD)) , Name, .SDcols = No1:No2], Name~paste0("ID", rowid(Name)), value.var="V1")[d, on = "Name"] # Name ID1 ID2 ID3 No1 No2 #1: Jon 1 2 3 1 1 #2: Jon 1 2 3 2 1 #3: Jon 1 2 3 3 1 #4: Kel 1 2 NA 1 2 #5: Kel 1 2 NA 1 2 #6: Kel 1 2 NA 1 2 #7: Don 3 NA NA 3 3 #8: Don 3 NA NA 3 3 #9: Don 3 NA NA 3 3 {1}}元素中的paste元素一行完成。和＆＃39; No2＆＃39;，按名称＆＃39;分组，然后使用unique中的split将cSplit分为三列。

splitstackshape

或仅使用library(splitstackshape) cSplit(setDT(d)[, ID:= paste(unique(c(No1, No2)), collapse=" ") , Name], "ID", " ") # Name No1 No2 ID_1 ID_2 ID_3 #1: Jon 1 1 1 2 3 #2: Jon 2 1 1 2 3 #3: Jon 3 1 1 2 3 #4: Kel 1 2 1 2 NA #5: Kel 1 2 1 2 NA #6: Kel 1 2 1 2 NA #7: Don 3 3 3 NA NA #8: Don 3 3 3 NA NA #9: Don 3 3 3 NA NA来踢球

baseVerse

注意：没有使用任何包，也没有太多精力进行拆分。

使用现有列中的（提取唯一值）向数据框添加列

3 个答案: