假设我有一个带有 id
Alibrary(data.table)
library(dplyr)
mydf <- data.table(
A = as.factor(c("A", "B", "C", "D","E")),
V1 = as.factor(c("x", "x", "y", "x", "y")),
V2 = as.factor(c("u", "v", "w", "v", "u"))
)
在一个真实的示例中，可以有更多列（比如 1000），并且数据集有几百万行，所以我真的很关心性能。
我想将其转换成以下data.frame：
A V1_x V1_y V2_u V2_v V2_w
1 A 1 0 1 0 0
2 B 1 0 0 1 0
3 C 0 1 0 0 1
4 D 1 0 0 1 0
5 E 0 1 1 0 0
实现这种转换的有效方法是什么？
我现在的做法是分别对每一列多次应用 dcast 。并相应地重命名列。
f <- function(x) as.integer(length(x) > 0)
mydf2 =
mydf %>%
dcast(... ~ V1, fun.aggregate = f, value.var = "V1") %>%
dcast(... ~ V2, fun.aggregate = f, value.var = "V2")
colnames(mydf2) = c("A", "V1_x", "V1_y", "V2_u", "V2_v", "V2_w")
如果我们在循环中运行它，这对于许多变量来说似乎效率不高。此外，如果因子数量因变量而异，则在每个步骤重命名变量并不是很稳健。
同一解决方案的循环版本：
mydf2 = mydf
for(tmp_var in c("V1","V2")){
ncol_before = ncol(mydf2)
mydf2 =
mydf2 %>%
dcast( as.formula( sprintf("... ~ %s", tmp_var) ),
fun.aggregate = f, value.var = tmp_var)
ncol_after = ncol(mydf2)
# assign correct names to created vars
colnames(mydf2)[ncol_before:ncol_after ] = paste0(tmp_var, "_", colnames(mydf2)[ncol_before:ncol_after ])
}
mydf2
这是一个带有
cSplit_elibrary(splitstackshape)
cSplit_e(mydf, 'V1', type = 'character', fill = '0') %>%
cSplit_e('V2', type = 'character', fill = '0')
# A V1 V2 V1_x V1_y V2_u V2_v V2_w
#1: A x u 1 0 1 0 0
#2: B x v 1 0 0 1 0
#3: C y w 0 1 0 0 1
#4: D x v 1 0 0 1 0
#5: E y u 0 1 1 0 0
或与
tablebase R do.call(cbind, lapply(2:3, function(i) table(mydf$A, mydf[[i]])))
或者在
data.tablenm1 <- names(mydf)[-1]
out <- mydf[, lapply(.SD, function(x)
as.data.frame.matrix(table(A, x))), .SDcols = nm1]
mydf[, names(out) := out][]
# A V1 V2 V1.x V1.y V2.u V2.v V2.w
#1: A x u 1 0 1 0 0
#2: B x v 1 0 0 1 0
#3: C y w 0 1 0 0 1
#4: D x v 1 0 0 1 0
#5: E y u 0 1 1 0 0
您可以获取长格式的数据，组合列名和列值，创建虚拟列并获取宽格式的数据。
使用
tidyverselibrary(dplyr)
library(tidyr)
mydf %>%
pivot_longer(cols = starts_with('V')) %>%
unite(name, name, value) %>%
mutate(value = 1) %>%
pivot_wider(values_fill = 0)
# A tibble: 5 x 6
# A V1_x V2_u V2_v V1_y V2_w
# <fct> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 A 1 1 0 0 0
#2 B 1 0 1 0 0
#3 C 0 0 0 1 1
#4 D 1 0 1 0 0
#5 E 0 1 0 1 0
如果您想按照相同的逻辑在
data.tablelibrary(data.table)
df1 <- melt(setDT(mydf), id.vars = 'A')
df1[, c('variable', 'value') := .(paste(variable, value, sep = "_"), 1)]
dcast(df1, A~variable, value.var = 'value', fill = 0)
非常感谢 akrun 提供的解决方案。这比我以前见过的任何东西都更有效率。
快速跟进表演。定义包含 200 列和 100,000 个观测值的数据集。
N = 1e5
mydf <- data.table(A = 1:N)
for(j in 1:100)
mydf[[paste0("V",j)]] = sample(c("x", "y", "z"),N, replace = T)
for(j in 101:200)
mydf[[paste0("V",j)]] = sample(c("v", "u", "w"),N, replace = T)
然后是data.table方式的表现：
start_time <- Sys.time()
nm1 <- names(mydf)[-1]
out <- mydf[, lapply(.SD, function(x) as.data.frame.matrix(table(A, x))), .SDcols = nm1]
end_time <- Sys.time()
end_time - start_time
Time difference of 16.39327 secsout[1:10,1:10]
V1.x V1.y V1.z V2.x V2.y V2.z V3.x V3.y V3.z V4.x
1: 1 0 0 1 0 0 0 0 1 1
2: 0 0 1 1 0 0 1 0 0 0
3: 0 1 0 0 1 0 1 0 0 0
4: 0 0 1 0 1 0 1 0 0 0
5: 0 0 1 0 1 0 0 0 1 1
6: 0 1 0 0 0 1 0 0 1 0
7: 1 0 0 1 0 0 0 1 0 1
8: 0 1 0 1 0 0 0 1 0 1
9: 0 0 1 1 0 0 1 0 0 0
10: 1 0 0 0 1 0 0 1 0 0
do.call的性能基本相同，但我们仍然需要重命名列：
start_time <- Sys.time()
out <- do.call(cbind, lapply(2:201, function(i) table(mydf$A, mydf[[i]]) ))
end_time <- Sys.time()
end_time - start_time
Time difference of 14.72284 secsout[1:10,1:10]
x y z x y z x y z x
1 1 0 0 1 0 0 0 0 1 1
2 0 0 1 1 0 0 1 0 0 0
3 0 1 0 0 1 0 1 0 0 0
4 0 0 1 0 1 0 1 0 0 0
5 0 0 1 0 1 0 0 0 1 1
6 0 1 0 0 0 1 0 0 1 0
7 1 0 0 1 0 0 0 1 0 1
8 0 1 0 1 0 0 0 1 0 1
9 0 0 1 1 0 0 1 0 0 0
10 1 0 0 0 1 0 0 1 0 0
编辑
Ronak Shah 的替代 tidyverse 解决方案在数据上更加高效：
start_time <- Sys.time()
mydf2 =
mydf %>%
pivot_longer(cols = starts_with('V')) %>%
unite(name, name, value) %>%
mutate(value = 1) %>%
pivot_wider(values_fill = 0)
end_time <- Sys.time()
end_time - start_time
Time difference of 9.892609 secs# A tibble: 100,000 x 601
A V1_z V2_y V3_y V4_y V5_z V6_y V7_y V8_y V9_x V10_z V11_x V12_y
<int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 0 0 0 0 0 0 0 0 0 1 0 1
3 3 1 0 0 0 0 0 0 0 0 0 0 0
4 4 1 1 0 0 0 1 0 0 0 0 0 0
5 5 1 0 0 0 0 0 0 1 1 0 0 1
6 6 1 1 0 1 0 0 0 0 1 0 0 0
7 7 0 0 0 1 0 1 0 1 1 0 0 0
8 8 0 1 1 0 1 1 1 0 1 1 0 0
9 9 0 0 0 1 1 1 1 1 0 1 1 0
10 10 0 1 0 0 1 0 0 0 0 0 1 0
# … with 99,990 more rows, and 588 more variables: V13_x <dbl>, V14_x <dbl>,
# V15_z <dbl>, V16_z <dbl>, V17_y <dbl>, V18_y <dbl>, V19_z <dbl>,
# V20_x <dbl>, V21_y <dbl>, V22_z <dbl>, V23_z <dbl>, V24_x <dbl>,
# V25_y <dbl>, V26_z <dbl>, V27_z <dbl>, V28_x <dbl>, V29_z <dbl>,
# V30_y <dbl>, V31_x <dbl>, V32_y <dbl>, V33_x <dbl>, V34_y <dbl>,
# V35_y <dbl>, V36_y <dbl>, V37_x <dbl>, V38_y <dbl>, V39_y <dbl>,
# V40_x <dbl>, V41_y <dbl>, V42_z <dbl>, V43_z <dbl>, V44_y <dbl>,
# V45_x <dbl>, V46_x <dbl>, V47_x <dbl>, V48_y <dbl>, V49_y <dbl>,
# V50_z <dbl>, V51_z <dbl>, V52_z <dbl>, V53_z <dbl>, V54_y <dbl>,
# V55_y <dbl>, V56_y <dbl>, V57_x <dbl>, V58_x <dbl>, V59_x <dbl>,
# V60_z <dbl>, V61_y <dbl>, V62_x <dbl>, V63_y <dbl>, V64_y <dbl>,
# V65_y <dbl>, V66_x <dbl>, V67_x <dbl>, V68_x <dbl>, V69_y <dbl>,
# V70_x <dbl>, V71_y <dbl>, V72_y <dbl>, V73_x <dbl>, V74_y <dbl>,
# V75_x <dbl>, V76_z <dbl>, V77_x <dbl>, V78_y <dbl>, V79_z <dbl>,
# V80_z <dbl>, V81_x <dbl>, V82_z <dbl>, V83_x <dbl>, V84_x <dbl>,
# V85_z <dbl>, V86_z <dbl>, V87_x <dbl>, V88_z <dbl>, V89_x <dbl>,
# V90_y <dbl>, V91_y <dbl>, V92_z <dbl>, V93_z <dbl>, V94_z <dbl>,
# V95_z <dbl>, V96_z <dbl>, V97_z <dbl>, V98_y <dbl>, V99_z <dbl>,
# V100_y <dbl>, V101_v <dbl>, V102_w <dbl>, V103_u <dbl>, V104_w <dbl>,
# V105_v <dbl>, V106_v <dbl>, V107_u <dbl>, V108_w <dbl>, V109_v <dbl>,
# V110_v <dbl>, V111_u <dbl>, V112_v <dbl>, …
这是 独立 dcast 多列的重复，因此在那里应用解决方案：
dcast(melt(mydf, id.var = 'A'), A ~ variable + value, fun.aggregate = f)