使用字符数组按行填充列值

Question

一列big.data（year.60）按行描述所需值的列名称，如下所示：

big.data= data.frame(ID= c(1,2), Y.1990= c(100, 120), 
Y.1991= c(NA, 125), Y.1992= c(115, 130), year.60= c("Y.1990", 
"Y.1991"), Y.60= c(NA, NA) )

big.data$year.60 = as.character(big.data$year.60)

big.data
#  ID   Y.1990 Y.1991 Y.1992  year.60 Y.60
#1  1      100     NA    115   Y.1990   NA
#2  2      120    125    130   Y.1991   NA

我怎样才能让Y.60等于row1中的big.data $ Y.1990和row2中的big.data $ Y.1991等10,000行（即Y.60= year.60 by row)？< / H2>

big.data$Y.60= big.data[[year.60]]
Error in (function(x, i, exact) if (is.matrix(i)) 
  as.matrix(x)[[i]] else .subset2(x,  : 
    object 'year.60' not found

真实数据无法按预期方式运行

我希望skinny.data$Y.60在skinny.data$y.1970 skinny.data$Y.60[1,]，skinny.data$y.1953 skinny.data$Y.60[2,]，skinny.data$y.1963 skinny.data$Y.60[3,] skinny.data$y.1993，skinny.data$Y.60[4,] 1}}在str(skinny.data) 'data.frame': 42001 obs. of 39 variables: $ Y.60 : chr NA NA NA NA ... $ year.60 : chr "y.1970" "y.1953" "y.1963" "y.1993" ... $ y.1968 : num 10006 19467 19467 19467 19467 ... $ y.1969 : num NA 18994 18994 18994 18994 ... $ y.1970 : num NA 23150 23150 23150 23150 ... $ y.1971 : num NA 15041 15041 15041 25773 ... $ y.1972 : num NA 17183 17183 NA 17183 ... $ y.1973 : num NA 14354 14354 NA 14354 ... $ y.1974 : num NA 6829 6829 NA 6829 ... $ y.1975 : num NA 9444 9444 NA 9444 ... $ y.1976 : num NA 3717 3717 NA 9294 ... $ y.1977 : num NA 0 0 NA 2636 ... $ y.1978 : num NA 0 0 NA 4125 ... $ y.1979 : num NA 3394 3394 NA 12577 ... $ y.1980 : num NA 0 0 NA 4821 ... $ y.1981 : num NA 0 0 NA 7257 ... $ y.1982 : num NA 4778 4778 NA 8009 ... # ... # etc. skinny.data.mini= skinny.data[1:5, ] ## mapply skinny.data.mini$Y.60 <- mapply(getElement, name=skinny.data.mini$year.60, data.frame(t(skinny.data.mini))) # Error in `[[.default`(object, name, exact = TRUE) : # subscript out of bounds ## quick vectorized approach skinny.data.mini$Y.60 <- skinny.data[cbind(seq_len(nrow(skinny.data.mini)), match(skinny.data.mini$year.60, names(skinny.data.mini)))] describe(skinny.data.mini$Y.60) # Error in x - mx : non-numeric argument to binary operator # In addition: Warning messages: # 1: In mean.default(x, na.rm = na.rm) : # argument is not numeric or logical: returning NA # 2: In mean.default(x, na.rm = na.rm, trim = trim) : # argument is not numeric or logical: returning NA # 3: In mean.default(x) : argument is not numeric or logical: returning NA ## getval getval <- function(byrow) skinny.data.mini[, match(skinny.data.mini$year.60[byrow], names(skinny.data.mini))][byrow] skinny.data.mini$Y.60 <- sapply(1:nrow(skinny.data.mini), getval) # Show Traceback # Rerun with Debug # Error in `[.data.frame`(skinny.data.mini, , # match(skinny.data.mini$forty.fam.head.laby.year[byrow], : # undefined columns selected 等等。但即使我将此限制为5行，以下所有方法都会失败。

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Answer 1

这对你有用吗？

getval <- function(byrow) 
  big.data[byrow, match(big.data$year.60[byrow], names(big.data))]

big.data$Y.60 <- sapply(1:nrow(big.data), getval)

或者，如果你真的想这么做：

big.data$Y.60 <- mapply(getElement, name=big.data$year.60, data.frame(t(big.data)))

Answer 2

这是一种快速矢量化方法

big.data$Y.60 <- big.data[cbind(seq_len(nrow(big.data)), 
                          match(big.data$year.60, names(big.data)))]
big.data
#   ID Y.1990 Y.1991 Y.1992 year.60 Y.60
# 1  1    100     NA    115  Y.1990  100
# 2  2    120    125    130  Y.1991  125