rso*_*ren 6 r dataframe diagonal
我正在开发一个模型,预测一个年龄组的生育能力.我目前有一个这样的数据框,其中行是年龄,列是年.每个细胞的价值是该年度的特定年龄生育率:
> df1
iso3 sex age fert1953 fert1954 fert1955
14 AUS female 13 0.000 0.00000 0.00000
15 AUS female 14 0.000 0.00000 0.00000
16 AUS female 15 13.108 13.42733 13.74667
17 AUS female 16 26.216 26.85467 27.49333
18 AUS female 17 39.324 40.28200 41.24000
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但是,我想要的是每一行都是一个队列.因为行和列表示各个年份,所以可以通过获得对角线来获得群组数据.我正在寻找这样的结果:
> df2
iso3 sex ageIn1953 fert1953 fert1954 fert1955
14 AUS female 13 0.000 0.00000 13.74667
15 AUS female 14 0.000 13.42733 27.49333
16 AUS female 15 13.108 26.85467 41.24000
17 AUS female 16 26.216 40.28200 [data..]
18 AUS female 17 39.324 [data..] [data..]
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这是df1数据框:
df1 <- structure(list(iso3 = c("AUS", "AUS", "AUS", "AUS", "AUS"), sex = c("female",
"female", "female", "female", "female"), age = c(13, 14, 15,
16, 17), fert1953 = c(0, 0, 13.108, 26.216, 39.324), fert1954 = c(0,
0, 13.4273333333333, 26.8546666666667, 40.282), fert1955 = c(0,
0, 13.7466666666667, 27.4933333333333, 41.24)), .Names = c("iso3",
"sex", "age", "fert1953", "fert1954", "fert1955"), class = "data.frame", row.names = 14:18)
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编辑:
这是我最终使用的解决方案.这是基于大卫的答案,但我需要为每个级别做这个iso3.
df.ls <- lapply(split(f3, f = f3$iso3), FUN = function(df1) {
n <- ncol(df1) - 4
temp <- mapply(function(x, y) lead(x, n = y), df1[, -seq_len(4)], seq_len(n))
return(cbind(df1[seq_len(4)], temp))
})
f4 <- do.call("rbind", df.ls)
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我还没有测试速度,但是data.table v1.9.5,最近实现了一个新的(用 C 编写的)超前/滞后函数,称为shift
因此,对于您想要移动的列,您可以将其与 结合使用mapply,例如
library(data.table)
n <- ncol(df1) - 4 # the number of years - 1
temp <- mapply(function(x, y) shift(x, n = y, type = "lead"), df1[, -seq_len(4)], seq_len(n))
cbind(df1[seq_len(4)], temp) # combining back with the unchanged columns
# iso3 sex age fert1953 fert1954 fert1955
# 14 AUS female 13 0.000 0.00000 13.74667
# 15 AUS female 14 0.000 13.42733 27.49333
# 16 AUS female 15 13.108 26.85467 41.24000
# 17 AUS female 16 26.216 40.28200 NA
# 18 AUS female 17 39.324 NA NA
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data.table编辑:您可以使用以下命令轻松安装来自 GitHub的开发版本
library(devtools)
install_github("Rdatatable/data.table", build_vignettes = FALSE)
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不管怎样,如果你愿意dplyr,这里就是
library(dplyr)
n <- ncol(df1) - 4 # the number of years - 1
temp <- mapply(function(x, y) lead(x, n = y), df1[, -seq_len(4)], seq_len(n))
cbind(df1[seq_len(4)], temp)
# iso3 sex age fert1953 fert1954 fert1955
# 14 AUS female 13 0.000 0.00000 13.74667
# 15 AUS female 14 0.000 13.42733 27.49333
# 16 AUS female 15 13.108 26.85467 41.24000
# 17 AUS female 16 26.216 40.28200 NA
# 18 AUS female 17 39.324 NA NA
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