Question

我有一个类似下面的文件，并希望将其转换为输出

中给出的R数据框

        A    B    C     D   E   
2010    25   74   85    88  89
2011    27   86   97    99  
2012    37   115  131   
2013    47   146            
2014    56

输出：

Year  Year_No   Division   Amount 
2010    1       A          25
2010    1       B          74
2010    1       C          85
2010    1       D          88
2010    1       E          89
2011    2       A          27
2011    2       B          86
2011    2       C          97
2011    2       D          99
2012    3       A          37
2012    3       B          115
2012    3       C          131
2013    4       A          47
2013    4       B          146
2014    5       A          56

我真的很感激如果有人能帮助我解决这个问题

Answer 1

我们可以使用tidyverse。

中的函数

library(tidyverse)

dt2 <- dt %>%
  rownames_to_column("Year") %>%
  rowid_to_column("Year_No") %>%
  gather(Division, Amount, -Year, -Year_No, na.rm = TRUE) %>%
  arrange(Year_No, Division) %>% 
  select(Year_No, Year, Division, Amount)
dt2
   Year_No Year Division Amount
1        1 2010        A     25
2        1 2010        B     74
3        1 2010        C     85
4        1 2010        D     88
5        1 2010        E     89
6        2 2011        A     27
7        2 2011        B     86
8        2 2011        C     97
9        2 2011        D     99
10       3 2012        A     37
11       3 2012        B    115
12       3 2012        C    131
13       4 2013        A     47
14       4 2013        B    146
15       5 2014        A     56

数据

dt <- read.table(text = " A B C D E 2010 25 74 85 88 89 2011 27 86 97 99 NA 2012 37 115 131 NA NA 2013 47 146 NA NA NA 2014 56 NA NA NA NA", header = TRUE)

Answer 2

我们可以转换为matrix，然后转换为melt

library(reshape2)
library(data.table)
dt <- setDT(melt(as.matrix(df1), na.rm = TRUE))[ , YearNo := .GRP, Var1][order(Var1)]
setnames(dt, c("Var1", "Var2", "value"), c("Year", "Division", "Amount"))[]
#     Year Division Amount YearNo
# 1: 2010        A     25      1
# 2: 2010        B     74      1
# 3: 2010        C     85      1
# 4: 2010        D     88      1
# 5: 2010        E     89      1
# 6: 2011        A     27      2
# 7: 2011        B     86      2
# 8: 2011        C     97      2
# 9: 2011        D     99      2
#10: 2012        A     37      3
#11: 2012        B    115      3
#12: 2012        C    131      3
#13: 2013        A     47      4
#14: 2013        B    146      4
#15: 2014        A     56      5

注意：假设缺失值为NA，因为它似乎是数字列，即'A'到'E'

数据

df1 <- structure(list(A = c(25L, 27L, 37L, 47L, 56L), B = c(74L, 86L, 
115L, 146L, NA), C = c(85L, 97L, 131L, NA, NA), D = c(88L, 99L, 
NA, NA, NA), E = c(89L, NA, NA, NA, NA)), .Names = c("A", "B", 
"C", "D", "E"), class = "data.frame", row.names = c("2010", "2011", 
"2012", "2013", "2014"))

Answer 3

基础

中的简单解决方案

# create the data
mat <- matrix(1:30,6,5)
for(z in 1:nrow(mat)){
  a <- (1:(5-z+1))*-1
  mat[z,a] <- NA
}
rownames(mat) <- 2012:2017
colnames(mat) <- LETTERS[1:5]
mat

# start the task
col <- rep(colnames(mat), each=nrow(mat))
value <- as.vector(mat)
row <- rownames(mat)
table <- data.frame(col,row,value)
table <- table[!is.na(table$value),]
table

将宽数据转换为长格式

3 个答案:

数据