重新排列纵向数据

Question

我有一个大致如下结构的数据集：

case Year      2001 2002 2003 2004
1    2003      0    0    0    3
2    2002      0    5    3    2
3    2001      3    3    2    2

我正在尝试对其进行重组，以便每一列代表从“年份”变量计数的第一年、第二年（等等），即：

case Year      yr1  yr2  yr3 yr4
1    2003      0    3    0    0 
2    2002      5    3    2    0
3    2001      3    3    2    2

此代码下载数据集并尝试@akrun 建议的解决方案，但失败了。

library("devtools")
df1 <- source_gist("b4c44aa67bfbcd6b72b9")

df1[-(1:2)] <- do.call(rbind,lapply(seq_len(nrow(df1)), function(i) {x <- df1[i, ]; x1 <- unlist(x[-(1:2)]); indx <- which(!is.na(x1))[1]; i <- as.numeric(names(indx))-x[,2]+1; x2 <- x1[!is.na(x1)]; x3 <- rep(NA, length(x1)); x3[i:(i+length(x2)-1)]<- x2; x3}))

这会生成：

Error in i:(i + length(x2) - 1) : NA/NaN argument
In addition: Warning message:
In FUN(1:234[[1L]], ...) : NAs introduced by coercion

如何转换数据，以便每一列代表第一年、第二年（等等），从每行的“年份”变量中的值开始计算？

Answer 1

这是一种可能性：

library(dplyr)
library(reshape2)

df %>%
  melt(id.vars = c("case", "Year")) %>%
  mutate(variable = as.numeric(as.character(variable)),
         yr = variable - Year + 1) %>%
  filter(variable >= Year) %>%
  dcast(case + Year ~ yr, fill = 0)

#   case Year 1 2 3 4
# 1    1 2003 0 3 0 0
# 2    2 2002 5 3 2 0
# 3    3 2001 3 3 2 2

数据：

df <- structure(list(case = 1:3, Year = c(2003L, 2002L, 2001L), `2001` = c(0L, 
0L, 3L), `2002` = c(0L, 5L, 3L), `2003` = c(0L, 3L, 2L), `2004` = c(3L, 
2L, 2L)), .Names = c("case", "Year", "2001", "2002", "2003", 
"2004"), class = "data.frame", row.names = c(NA, -3L))

Answer 2

这应该会创建您正在寻找的操作。

library("devtools")
df1 <- source_gist("b4c44aa67bfbcd6b72b9")
temp <- df1[[1]]

library(dplyr); library(tidyr); library(stringi) 

temp <- temp %>% 
  gather(new.Years, X, -Year) %>%  # convert rows to one column
  mutate(Year.temp=paste0(rownames(temp), "-", Year)) %>% # concatenate the Year with row number to make them unique
  mutate(new.Years = as.numeric(gsub("X", "", new.Years)), diff = new.Years-Year+1) %>% # calculate the difference to get the yr0 yr1 and so on
  mutate(diff=paste0("yr", stri_sub(paste0("0", (ifelse(diff>0, diff, 0))), -2, -1))) %>% # convert the differences in Yr01 ...
  select(-new.Years) %>% filter(diff != "yr00") %>% # drop new.Years column
  spread(diff, X) %>%  # convert column to rows
  select(-Year.temp) # Drop Year.temp column

temp[is.na(temp)] <- 0 # replace NA with 0

temp %>% View

请注意，这将工作长达 99 年。

Answer 3

这是data.table 解决方案：

require(data.table)
require(reshape2)
dt.m = melt(dt, id = 1:2, variable.factor = FALSE)
dt.m[, variable := as.integer(variable)-Year+1L]
dcast.data.table(dt.m, case + Year ~ variable, fill=0L, 
      value.var = "value", subset = (variable > 0L))
#    case Year 1 2 3 4
# 1:    1 2003 0 3 0 0
# 2:    2 2002 5 3 2 0
# 3:    3 2001 3 3 2 2

Answer 4

library("devtools")
df1 <- source_gist("b4c44aa67bfbcd6b72b9")$value

我在列名中有一个 X 并将其删除：

colnames(df1) <- gsub("X", "", colnames(df1))

我有一个没有任何额外软件包的解决方案：

startYear <- as.numeric(colnames(df1)[2])
shifts <- df1$Year - startYear
n <- ncol(df1)

df2 <- df1
colnames(df2)[-1] <- 1:(n-1) 
df2[,2:n]  <- NA

for(row in 1:nrow(df1)){
    if(shifts[row]>=0){
        df2[row,2:(n-shifts[row])] <- df1[row, (shifts[row]+2):n]
        #df2[row,2:(n-shifts[row])] <- colnames(df1)[(shifts[row]+2):n]
    }else{
        df2[row, (-shifts[row]+2):n] <- df1[row, 2:(n+shifts[row])]
        #df2[row, (-shifts[row]+2):n] <- colnames(df1)[2:(n+shifts[row])]
    }
}

您可以用0 代替NA 预填充df2。删除 ifelse 条件中的第二行并注释第一行以验证排列。

希望它能如你所愿。