我试图获得类似于expand.grid的函数,并在data.frame上工作.
我在Alternative to expand.grid for data.frames找到了解决方案
使用合并功能来实现这一点.
由于与dplyr替代full_join相比合并相当慢,所以我尝试使用full_join来实现此功能,但我无法正确完成它.这是我失败的一个例子：
df <- data.frame(attribute = paste0('attr', rep(1:5, each=2)),
value = paste0(rep(1:5, each=2), rep(c('A','B'), 2)),
score = runif(10))
df
attribute value score
1 attr1 1A 0.75600171
2 attr1 1B 0.07086242
3 attr2 2A 0.92403325
4 attr2 2B 0.63414169
5 attr3 3A 0.78763834
6 attr3 3B 0.88576568
7 attr4 4A 0.75998967
8 attr4 4B 0.25205845
9 attr5 5A 0.99304728
10 attr5 5B 0.70389605
我尝试按属性拆分df并将得分列表加在一起：
dfList <- df %>%
mutate(attribute=1) %>%
split(df$attribute)
我通过以下方式将所有这5张桌子“expand.grid”放在一起：
Reduce(function(x, y) {full_join(x, y, by=c('attribute'='attribute'))}, dfList)
然而,结果很奇怪：
attribute value.x score.x value.y score.y value.x score.x value.y score.y value score
1 1 1A 0.75600171 2A 0.9240333 1A 0.75600171 2A 0.9240333 5A 0.9930473
2 1 1A 0.75600171 2A 0.9240333 1A 0.75600171 2A 0.9240333 5B 0.7038961
3 1 1A 0.75600171 2A 0.9240333 1A 0.75600171 2A 0.9240333 5A 0.9930473
4 1 1A 0.75600171 2A 0.9240333 1A 0.75600171 2A 0.9240333 5B 0.7038961
...
前两个表显示两次,这是不希望的.但是当我在前4个表上尝试这个时,它完美地工作：
Reduce(function(x, y) {full_join(x, y, by=c('attribute'='attribute'))}, dfList[1:4])
attribute value.x score.x value.y score.y value.x score.x value.y score.y
1 1 1A 0.75600171 2A 0.9240333 3A 0.7876383 4A 0.7599897
2 1 1A 0.75600171 2A 0.9240333 3A 0.7876383 4B 0.2520584
3 1 1A 0.75600171 2A 0.9240333 3B 0.8857657 4A 0.7599897
4 1 1A 0.75600171 2A 0.9240333 3B 0.8857657 4B 0.2520584
...
我哪里做错了？
我在Ubuntu 14.04上使用dplyr 0.4.3和R版本3.2.4
最佳答案 我可以在我的机器上重现你的dfList的损坏结果.在我看来,我已经发现它为什么会发生.
require(dplyr)
adf <- data.frame(c1 = 7, c1 = 8, jv = 1, check.names = F)
bdf <- data.frame(d1 = 1:3, d2 = letters[1:3], jv = 1)
cdf <- data.frame(v1.x = 1:3, v2 = letters[1:3], jv = 1)
ddf <- data.frame(v2 = 4:5, v2.x = letters[4:5], jv = 1)
full_join(adf, bdf, by = "jv")
c1 c1 jv d1 d2
1 7 7 1 1 a
2 7 7 1 2 b
3 7 7 1 3 c
我们可以注意到,在adf中使用重复的列名会导致错误的加入结果.当我们在Reduce的帮助下应用多个连接链时,会发生重复列名的自动重命名(默认添加.x和.y).这可能会导致产生另一个重复的名称(与其意图避免的事物相反).
full_join(cdf, ddf, by = "jv")
v1.x v2.x jv v2.y v2.x
1 1 a 1 4 d
2 1 a 1 5 e
3 2 b 1 4 d
4 2 b 1 5 e
5 3 c 1 4 d
6 3 c 1 5 e
这里我们在不同的data.frames中有一个重复的名称 – 列v2,在应用后缀 – v2.x后被另一个重复替换.
因此,为了使事情顺利进行,我们应该关注我们加入的data.frames中列的唯一名称.
我已经尝试了几种方法来获得理想的结果,并希望展示它们是什么.
>基础R解决方案使用合并,它是为速度比较.
>使用dplyr包中的full_join的方法
>使用dts顺序合并的data.table解决方案
>基于tidyr不需要的功能
>另一个data.table解决方案,它首先生成目标结果长度的关键表(在CJ的帮助下)然后进行几个左连接
>与之前相同,但使用on参数进行连接而不是设置键
require(data.table)
require(dplyr)
require(tidyr)
require(stringi)
require(microbenchmark)
expand.grid.df_base <- function(...) {
dfList <- list(...)
if (length(dfList) == 1) dfList <- dfList[[1]]
if (is.null(names(dfList))) names(dfList) <- paste0("df", 1:length(dfList))
lapply(1:length(dfList), function(i)
data.frame(dfN = i, colN = 1:length(dfList[[i]]),
dfname = names(dfList)[i], colname = names(dfList[[i]]),
stringsAsFactors = F)) %>% bind_rows %>%
mutate(dum_names = stri_rand_strings(nrow(.), 12)) %>% rowwise %>%
mutate(out_names = paste(dfname, colname, sep = ".")) %>% ungroup -> manage_names
for (i in 1:nrow(manage_names)) names(dfList[[manage_names$dfN[i]]])[manage_names$colN[i]] <- manage_names$dum_names[i]
Reduce(function(x, y) merge(x, y, by = NULL), dfList) %>% setNames(manage_names$out_names)
}
expand.grid.df_dplyr <- function(...) {
dfList <- list(...)
if (length(dfList) == 1) dfList <- dfList[[1]]
if (is.null(names(dfList))) names(dfList) <- paste0("df", 1:length(dfList))
lapply(1:length(dfList), function(i)
data.frame(dfN = i, colN = 1:length(dfList[[i]]),
dfname = names(dfList)[i], colname = names(dfList[[i]]),
stringsAsFactors = F)) %>% bind_rows %>%
mutate(dum_names = stri_rand_strings(nrow(.), 12)) %>% rowwise %>%
mutate(out_names = paste(dfname, colname, sep = ".")) %>% ungroup -> manage_names
for (i in 1:nrow(manage_names)) names(dfList[[manage_names$dfN[i]]])[manage_names$colN[i]] <- manage_names$dum_names[i]
joinvar <- stri_rand_strings(1, 12)
Reduce(function(x, y) {
mutate_def <- list(1L)
names(mutate_def) <- joinvar
full_join(x %>% mutate_(.dots = mutate_def), y %>% mutate_(.dots = mutate_def), by = joinvar)
}, dfList) %>% select(-contains(joinvar)) %>% setNames(manage_names$out_names) %>% tbl_df
}
expand.grid.dt <- function(...) {
dtList <- list(...)
if (length(dtList) == 1) dtList <- dtList[[1]]
if (!all(sapply(dtList, is.data.table))) dtList <- lapply(dtList, as.data.table)
if (is.null(names(dtList))) setnames(dtList, paste0("dt", 1:length(dtList)))
lapply(1:length(dtList), function(i)
data.frame(dfN = i, colN = 1:length(dtList[[i]]),
dfname = names(dtList)[i], colname = names(dtList[[i]]),
stringsAsFactors = F)) %>% bind_rows %>%
mutate(dum_names = stri_rand_strings(nrow(.), 12)) %>% rowwise %>%
mutate(out_names = paste(dfname, colname, sep = ".")) %>% ungroup -> manage_names
for (i in 1:nrow(manage_names)) setnames(dtList[[manage_names$dfN[i]]], old = manage_names$colN[i], new = manage_names$dum_names[i])
joinvar <- stri_rand_strings(1, 12)
setnames(Reduce(function(x, y) merge(copy(x)[,(joinvar) := 1], copy(y)[,(joinvar) := 1],
by = joinvar, all = T, allow.cartesian = T), dtList)[,(joinvar) := NULL],
manage_names$out_names)[]
}
expand.grid.df_tidyr <- function(...) {
dfList <- list(...)
if (length(dfList) == 1) dfList <- dfList[[1]]
if (is.null(names(dfList))) names(dfList) <- paste0("df", 1:length(dfList))
lapply(1:length(dfList), function(i)
data.frame(dfN = i, colN = 1:length(dfList[[i]]),
dfname = names(dfList)[i], colname = names(dfList[[i]]),
stringsAsFactors = F)) %>% bind_rows %>%
mutate(dum_names = stri_rand_strings(nrow(.), 12)) %>% rowwise %>%
mutate(out_names = paste(dfname, colname, sep = ".")) %>% ungroup -> manage_names
for (i in 1:nrow(manage_names)) names(dfList[[manage_names$dfN[i]]])[manage_names$colN[i]] <- manage_names$dum_names[i]
Reduce(function(x, y) x %>% rowwise %>% mutate(dfcol = list(y)) %>% ungroup %>% unnest(dfcol), dfList) %>%
setNames(manage_names$out_names) %>% tbl_df
}
expand.grid.dt2 <- function(...) {
dtList <- list(...)
if (length(dtList) == 1) dtList <- dtList[[1]]
dum_names <- stri_rand_strings(length(dtList), 12)
dtList <- lapply(1:length(dtList), function(i)
setkeyv(as.data.table(dtList[[i]])[, (dum_names[i]) := .I], dum_names[i]))
Reduce(function(result, dt) setkeyv(result, names(result)[1])[dt][, (names(result)[1]) := NULL],
dtList,
setnames(do.call(CJ, c(sapply(dtList, function(df) seq_len(nrow(df))), list(sorted = F))), dum_names))[]
}
expand.grid.dt3 <- function(...) {
dtList <- list(...)
if (length(dtList) == 1) dtList <- dtList[[1]]
dum_names <- stri_rand_strings(length(dtList), 12)
dtList <- lapply(1:length(dtList), function(i) as.data.table(dtList[[i]])[, (dum_names[i]) := .I])
Reduce(function(result, dt) result[dt, on = names(result)[1]][, (names(result)[1]) := NULL],
dtList,
setnames(do.call(CJ, c(sapply(dtList, function(df) seq_len(nrow(df))), list(sorted = F))), dum_names))[]
}
现在让我们创建data.frames列表来测试这个函数.
set.seed(1)
bigdfList <- data.frame(type = sample(letters[1:10], 50, T),
categ = sample(LETTERS[1:10], 50, T),
num = sample(100L:500L, 50, T),
val = rnorm(50)) %>% split(., .$type)
smalldfList <- data.frame(type = sample(letters[1:5], 50, T),
categ = sample(LETTERS[1:5], 50, T),
num = sample(100L:500L, 50, T),
val = rnorm(50)) %>% split(., .$type)
smalldfList的扩展连接产生一个维度[60,480 x 20]和bigdfList – [6,451,200 x 40]的表,占用1230.5 MB的RAM.
从smalldfList开始.
microbenchmark(expand.grid.df_base(smalldfList), expand.grid.df_dplyr(smalldfList),
expand.grid.dt(smalldfList), expand.grid.df_tidyr(smalldfList),
expand.grid.dt2(smalldfList), expand.grid.dt3(smalldfList), times = 10)
Unit: milliseconds
expr min lq mean median uq max neval cld
expand.grid.df_base(smalldfList) 178.36192 188.54955 201.28729 198.79644 209.86934 229.85360 10 b
expand.grid.df_dplyr(smalldfList) 16.04555 16.91327 18.91094 17.64907 18.45307 29.58192 10 a
expand.grid.dt(smalldfList) 20.33188 21.42275 26.30034 23.22873 31.66666 39.37922 10 a
expand.grid.df_tidyr(smalldfList) 722.06572 738.02188 801.41820 792.23725 859.96186 905.99190 10 c
expand.grid.dt2(smalldfList) 32.22650 33.68353 36.89386 36.39713 37.39182 48.93550 10 a
expand.grid.dt3(smalldfList) 29.13399 30.69299 34.51265 34.03198 37.48651 41.73543 10 a
所以,tidyr解决方案根本不是一个选项,基本合并也很慢. bigdfList上的其他4个函数显示效率.
microbenchmark(expand.grid.df_dplyr(bigdfList), expand.grid.dt(bigdfList),
expand.grid.dt2(bigdfList), expand.grid.dt3(bigdfList), times = 10)
Unit: seconds
expr min lq mean median uq max neval cld
expand.grid.df_dplyr(bigdfList) 1.326336 1.354706 1.456805 1.449781 1.481836 1.703158 10 a
expand.grid.dt(bigdfList) 1.763174 1.820004 1.894813 1.893910 1.939879 2.127097 10 b
expand.grid.dt2(bigdfList) 14.164731 14.332872 14.452933 14.452221 14.551982 14.740852 10 d
expand.grid.dt3(bigdfList) 10.589517 10.828548 11.104010 11.021519 11.368172 11.976976 10 c
而dplyr :: full_join解决方案效果最好！
也许,这是dplyr比data.table真的更好的选择之一,也许是我缺乏data.table知识,这使我无法实现真正快速的功能:-)
