Tho*_*ite 8 combinations r duplicates
我确信这很简单,但我有一个数据框
df <- data.frame(a = c(1, 2, 3),
b = c(2, 3, 1),
c = c(3, 1, 4))
我想要一个新的数据框,其中包含行中值的唯一组合,无论它们位于哪一列.所以在上面的例子中我想要
a b c
1 2 3
3 1 4
我试过了
unique(df[c('a', 'b', 'c')])
但它看到(1,2,3)与(2,3,1)中的唯一,我不想要.
也许是这样的
indx <- !duplicated(t(apply(df, 1, sort))) # finds non - duplicates in sorted rows
df[indx, ] # selects only the non - duplicates according to that index
# a b c
# 1 1 2 3
# 3 3 1 4
如果您的data.frame非常大,速度可能适合您.通过以下想法,您可以更快地找到重复的集合.
让我们想象一下,在行中为每个可能的值分配一个素数,并为每一行计算产品.例如,df我们可以接受primenums = c(2,3,5,7)和统计产品c(30,30,70).然后,此products-vector中的重复项对应于data.frame中的重复集.由于乘法计算速度比任何类型的排序快得多,因此可以提高效率.代码如下.
require("numbers")
primenums <- Primes(100)[1:4]
dfmult <- apply(as.matrix(df), 1, function(z) prod(primenums[z]) )
my_indx <- !duplicated(dfmult)
df[my_indx,]
这里我们primenums借助Primespackage中的函数初始化vector numbers,但是你可以用其他方式手动完成.
看看这个例子.这里我展示了效率的比较.
require("numbers")
# generate all unique combinations 10 out of 20
allcomb <- t(combn(20,10))
# make sample of 1 million rows
set.seed(789)
df <- allcomb[sample(nrow(allcomb), 1e6, T),]
# lets sort matrix to show we have duplicates
df <- df[do.call(order, lapply(1:ncol(df), function(i) df[, i])), ]
head(df, 10)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 1 2 3 4 5 6 7 8 9 10
# [2,] 1 2 3 4 5 6 7 8 9 10
# [3,] 1 2 3 4 5 6 7 8 9 10
# [4,] 1 2 3 4 5 6 7 8 9 10
# [5,] 1 2 3 4 5 6 7 8 9 11
# [6,] 1 2 3 4 5 6 7 8 9 11
# [7,] 1 2 3 4 5 6 7 8 9 11
# [8,] 1 2 3 4 5 6 7 8 9 11
# [9,] 1 2 3 4 5 6 7 8 9 11
# [10,] 1 2 3 4 5 6 7 8 9 11
# to be fair need to permutate numbers in rows before searching for identical sets
df <- t(apply(df, 1, function(z) z[sample(10,10)] ))
df <- as.data.frame(df)
names(df) <- letters[1:10]
# how does it look like now?
head(df, 10)
# a b c d e f g h i j
# 1 2 3 7 9 10 1 4 8 5 6
# 2 4 2 6 3 8 10 9 1 5 7
# 3 4 2 6 8 5 1 10 7 3 9
# 4 6 8 5 4 2 1 10 9 7 3
# 5 11 2 7 6 8 1 9 4 5 3
# 6 9 6 3 11 4 2 8 7 5 1
# 7 5 2 3 11 1 8 6 9 7 4
# 8 3 9 7 1 2 5 4 8 11 6
# 9 6 2 8 3 4 1 11 5 9 7
# 10 4 6 3 9 7 2 1 5 11 8
# now lets shuffle rows to make df more plausible
df <- df[sample(nrow(df), nrow(df)),]
现在,当data.frame准备就绪时,我们可以测试不同的算法.
system.time(indx <- !duplicated(t(apply(df, 1, sort))) )
# user system elapsed
# 119.75 0.06 120.03
# doesn't impress, frankly speaking
library(sets)
system.time(indx <- !duplicated(apply(df, 1, as.set)) )
# user system elapsed
# 91.60 0.00 91.89
# better, but we want faster! =)
# now lets check out the method with prime numbers
primenums <- Primes(100)[1:20]
# [1] 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
system.time({
dfmult <- apply(as.matrix(df), 1, function(z) prod(primenums[z]) )
my_indx <- !duplicated(dfmult) })
# user system elapsed
# 6.44 0.16 6.61
# not bad, isn't it? but lets compare results
identical(indx, my_indx)
# [1] TRUE
# So, if there is no difference, why wait more? ;)
这里有一个重要的假设 - 我们使用as.matrix(df),但如果我们不仅有数字变量data.frame怎么办？一个更加统一的解决方案如下:
system.time({
dfmult <- apply(
apply(df, 2, function(colmn) as.integer(factor(colmn,
levels = unique(c(as.matrix(df)))))),
1, function(z) prod(primenums[z]) )
my_indx <- !duplicated(dfmult) })
# user system elapsed
# 27.48 0.34 27.84
# is distinctly slower but still much faster then previous methods
如果我们有很多列或非常不同的变量呢？在这种情况下,而不是prod()我们可以使用sum(log())(对于大数字可能更快地计算).看看这个.
pr <- Primes(5e7)
length(pr)
# [1] 3001134
system.time(N <- sum(log(pr)))
# user system elapsed
# 0.12 0.00 0.13
N
# [1] 49993718
很难想象df有3百万列,但这里没问题.这种方式允许我们携带df任何令人难以置信的巨大尺寸,我们的RAM可容纳多列.