我有以下数据框
dat <- data.frame(a = c(0,0,2,3), b= c(1,0,0,0), c=c(0,0,1,3))
哪个打印:
> dat
a b c
1 0 1 0
2 0 0 0
3 2 0 1
4 3 0 3
我想删除所有列都为零的行，
导致这个:
a b c
1 0 1 0
3 2 0 1
4 3 0 3
我怎样才能做到这一点？
我试过这个但失败了:
> row_sub = apply(dat, 1, function(row) all(row !=0 ))
> dat[row_sub,]
[1] a b c
<0 rows> (or 0-length row.names)
最佳答案
您可以使用(1)
dat[as.logical(rowSums(dat != 0)), ]
这适用于正值和负值。
大型数据集的另一种更快的可能性是 (2)
dat[rowSums(!as.matrix(dat)) < ncol(dat), ]
对于短数据帧和长数据帧，一种更快的方法是使用矩阵乘法 (3):
dat[as.logical(abs(as.matrix(dat)) %*% rep(1L, ncol(dat))), ]
一些基准:
# the original dataset
dat <- data.frame(a = c(0,0,2,3), b= c(1,0,0,0), c=c(0,0,1,3))
Codoremifa <- function() dat[rowSums(abs(dat)) != 0,]
Marco <- function() dat[!apply(dat, 1, function(x) all(x == 0)), ]
Sven <- function() dat[as.logical(rowSums(dat != 0)), ]
Sven_2 <- function() dat[rowSums(!as.matrix(dat)) < ncol(dat), ]
Sven_3 <- function() dat[as.logical(abs(as.matrix(dat)) %*% rep(1L,ncol(dat))), ]
library(microbenchmark)
microbenchmark(Codoremifa(), Marco(), Sven(), Sven_2(), Sven_3())
# Unit: microseconds
# expr min lq median uq max neval
# Codoremifa() 267.772 273.2145 277.1015 284.0995 1190.197 100
# Marco() 192.509 198.4190 201.2175 208.9925 265.594 100
# Sven() 143.372 147.7260 150.0585 153.9455 227.031 100
# Sven_2() 152.080 155.1900 156.9000 161.5650 214.591 100
# Sven_3() 146.793 151.1460 153.3235 157.9885 187.845 100
# a data frame with 10.000 rows
set.seed(1)
dat <- dat[sample(nrow(dat), 10000, TRUE), ]
microbenchmark(Codoremifa(), Marco(), Sven(), Sven_2(), Sven_3())
# Unit: milliseconds
# expr min lq median uq max neval
# Codoremifa() 2.426419 2.471204 3.488017 3.750189 84.268432 100
# Marco() 36.268766 37.840246 39.406751 40.791321 119.233175 100
# Sven() 2.145587 2.184150 2.205299 2.270764 83.055534 100
# Sven_2() 2.007814 2.048711 2.077167 2.207942 84.944856 100
# Sven_3() 1.814994 1.844229 1.861022 1.917779 4.452892 100
关于r - 如何删除所有列都为零的行，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20592611/