查找满足条件的列号

Question

我有两列,每行的总和为1(它们是两个类之一的概率).我需要找到满足条件的列号.

C1   C2
0.4 0.6
0.3 0.7
1    0
0.7 0.3
0.1 0.9

例如,如果我需要找到数字> = 0.6的列,在上表中它应该导致:

2
2
1
1
2

Answer 1

您可以利用TRUE= 1和FALSE= 0 的事实:

> df <- read.table(textConnection("C1   C2
+                  0.4 0.6
+                  0.3 0.7
+                  1    0
+                  0.7 0.3
+                  0.1 0.9"), header = T)
> (df$C2 >= 0.6) + 1
[1] 2 2 1 1 2

Answer 2

谢谢你这个有趣的问题.这是一个使用的想法apply.

apply(dat, 1, function(x) which(x >= 0.6))
# [1] 2 2 1 1 2

数据

dat <- read.table(textConnection("C1   C2
0.4 0.6
0.3 0.7
1    0
0.7 0.3
0.1 0.9"), header = T)

标杆

我对原始数据帧dat和5000行数据帧进行了基准测试dat2.结果如下.我觉得有点尴尬,我的apply方法是最慢的.

如果有人知道如何改进我的基准测试方式,请告诉我.

library(microbenchmark)

# Benchmark 1
perf <- microbenchmark(m1 = {apply(dat, 1, function(x) which(x >= 0.6))},
                       m2 = {ifelse(dat$C1 <= 0.4, 2, 1)},
                       m3 = {(dat$C2 >= 0.6) + 1},
                       m4 = {(which(t(dat) >= 0.6) + 1) %% ncol(dat) + 1},
                       m5 = {((dat>=0.6) %*% c(1,2))[, 1]},
                       m6 = {m <- which(dat >= 0.6, arr.ind = TRUE)
                             m[order(m[, 1]), ][, 2]},
                       m7 = {max.col(dat >= 0.6)})

perf
# Unit: microseconds
# expr    min      lq     mean  median      uq      max neval
# m1 58.602 65.0280 88.34563 67.5985 70.6825 1746.246   100
# m2  9.253 12.8515 15.45772 13.8790 14.9080   49.349   100
# m3  4.112  5.6540  6.59015  6.1690  7.1970   23.132   100
# m4 30.844 35.7270 40.29682 38.0405 40.8670  134.683   100
# m5 23.647 26.7310 30.13404 27.7590 29.8160   77.109   100
# m6 49.863 53.4620 61.31148 56.5460 59.8875  168.610   100
# m7 37.012 40.0960 45.36537 42.1530 45.2370   97.671   100


# Benchmark 2   
dat2 <- dat[rep(1:5, 1000), ]

perf2 <- microbenchmark(m1 = {apply(dat2, 1, function(x) which(x >= 0.6))},
                        m2 = {ifelse(dat2$C1 <= 0.4, 2, 1)},
                        m3 = {(dat2$C2 >= 0.6) + 1},
                        m4 = {(which(t(dat2) >= 0.6) + 1) %% ncol(dat2) + 1},
                        m5 = {((dat2 >= 0.6) %*% c(1,2))[, 1]},
                        m6 = {m <- which(dat2 >= 0.6, arr.ind = TRUE)
                              m[order(m[, 1]), ][, 2]},
                        m7 = {max.col(dat2 >= 0.6)})

perf2
# Unit: microseconds
# expr       min         lq        mean     median         uq        max neval
# m1 13842.995 14830.2380 17173.18941 15716.2125 16551.8095 165431.735   100
# m2   133.140   146.7630   168.86722   160.6420   179.9195    314.602   100
# m3    22.104    25.7030    31.93827    28.0160    33.9280     67.341   100
# m4   156.787   179.6620   212.97310   210.5055   234.6665    320.257   100
# m5   131.598   148.8195   173.42179   164.2410   189.9440    286.843   100
# m6   403.019   439.2600   496.25370   472.6735   549.0110    791.646   100
# m7   140.337   156.7870   270.48048   179.4055   208.9635   8631.503   100