Question

我没有找到解决方案所以我想知道是否有人可以提供帮助。对于大多数人来说，这只是几秒钟的事情。

示例数据：

df <- data.frame(id=c(rep(1,4),rep(2,1),rep(3,2),rep(4,3)),rep("Name",10),rnorm(10,2,2),rbinom(10,2,0.7),rexp(10,2),rnorm(10,0,1),
             rgamma(10,2,7),rnorm(10,3,2),rexp(10,5))
colnames(df) <- c("id","Person","X2012","X2011","X2010","X2009","X2013","X2008","X2007")
df$X2007[1:4] <- NA;df$X2008[1:4] <- NA;df$X2011[1:4] <- NA;df$X2012[5:5] <- NA;df$X2013[5:5] <- NA
df$X2011[5:5] <- NA;df$X2007[5:5] <- NA;df$X2008[5:5] <- NA;df$X2009[5:5] <- NA;df$X2013[5:5] <- NA
df$X2012[6:7] <- NA;df$X2011[6:7] <- NA;df$X2013[6:7] <- NA;df$X2009[6:7] <- NA;df$X2010[6:7] <- NA
df$X2008[8:10] <- NA;df$X2007[8:10] <- NA;df$X2008[8:10] <- NA;df$X2013[8:10] <- NA;df$X2009[8:10] <- NA

预览上述数据：

> df
      id Person     X2012 X2011      X2010      X2009     X2013    X2008     X2007
  1   1   Name  1.0235950    NA 0.97326338 -0.3755807 0.7672341       NA        NA
  2   1   Name  0.1920131    NA 0.08904475 -0.5014424 0.2120472       NA        NA
  3   1   Name  1.2191632    NA 0.04186686  0.4966126 0.3395579       NA        NA
  4   1   Name  3.6281268    NA 1.63105614  1.5208809 0.2904623       NA        NA
  5   2   Name         NA    NA 0.34680106         NA        NA       NA        NA
  ....

预览我想要的内容：

> df.ok
    Year  id Person      X2012 X2011      X2010      X2009     X2013    X2008     X2007
1  X2009  1   Name  1.0235950    NA 0.97326338 -0.3755807 0.7672341       NA        NA
2  X2010  1   Name  0.1920131    NA 0.08904475 -0.5014424 0.2120472       NA        NA
3  X2012  1   Name  1.2191632    NA 0.04186686  0.4966126 0.3395579       NA        NA
4  X2013  1   Name  3.6281268    NA 1.63105614  1.5208809 0.2904623       NA        NA
5  X2010  2   Name         NA    NA 0.34680106         NA        NA       NA        NA
...

基本上，我在给出与列向量对应的数值后，在列名称中创建新的列向量。使用NA输入意味着年份不能用于给定的行。

编辑：数据集更新：

my.vec <- c(1,"Activa","","","","","","","")
df <- rbind(df[1:4, ], my.vec, df[5:nrow(df), ])

df[,3:ncol(df)] <- apply(df[, 3:ncol(df)],2, function(x) as.numeric(as.character(x)))
df[,2] <- rep("Name",nrow(df))

所以现在每个id之前还有1行。现在该如何处理？

Answer 1

这是一种可行的方法：

## Identify the non-`NA` values, make necessary adjustment
matches <- which(!is.na(df[-c(1, 2)]), arr.ind=TRUE)
matches[, "col"] <- matches[, "col"] + 2

## Create a `data.frame` from which we can aggregate the
##   sorted column names according to the "id" variable
matches <- data.frame(matches, 
                      Year = names(df)[matches[, 2]],
                      ID = df$id[matches[, 1]])

## Aggregate and retain only the unlisted aggregated values
Year <- unlist(aggregate(as.character(Year) ~ ID, matches, 
                         function(x) sort(unique(x)))[[2]], use.names=FALSE)

## Bind that back with the original data
cbind(Year, df)
#     Year id Person     X2012 X2011      X2010      X2009     X2013     X2008      X2007
# 1  X2009  1   Name  2.724841    NA 0.57090153  0.2346739 0.1861127        NA         NA
# 2  X2010  1   Name  1.604744    NA 0.54476273  1.4647495 0.1477001        NA         NA
# 3  X2012  1   Name  0.887420    NA 0.03375393 -0.7514787 0.1252799        NA         NA
# 4  X2013  1   Name  1.827136    NA 2.48897284  1.3697001 0.1026063        NA         NA
# 5  X2010  2   Name        NA    NA 0.70943724         NA        NA        NA         NA
# 6  X2007  3   Name        NA    NA         NA         NA        NA -4.196199 0.00789531
# 7  X2008  3   Name        NA    NA         NA         NA        NA  5.853701 0.16814434
# 8  X2010  4   Name  2.887311     1 0.64874998         NA        NA        NA         NA
# 9  X2011  4   Name -0.610759     2 0.74106815         NA        NA        NA         NA
# 10 X2012  4   Name  1.827282     2 0.25219805         NA        NA        NA         NA

Answer 2

Year = rep("", nrow(df))
cols = colnames(df[,-c(1:2)])
for (id in unique(df$id)) {
  # rows should match id AND should have at least a numeric value:
  rows = df$id == id & apply(!is.na(df[,cols]), 1, any) 
  isYearOk = !is.na(apply(df[rows, cols], 2, sum))
  Year[rows] = sort(cols[isYearOk])
}
cbind(Year, df) # This is equal to df.ok

# Test: 
all(Year == df.ok$Year) 
# TRUE

列名称为R中具有条件的行

2 个答案: