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）。我需要找到符合条件的列号。

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> gt = 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

感谢您提出这个有趣的问题。以下是使用apply的想法。

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)

<强>基准

我为原始数据框dat和5000行dat2的数据框执行了基准测试。结果如下。我感到有点尴尬，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

Answer 3

这是使用矩阵多次

(dt>=0.6)%*%c(1,2)
     [,1]
[1,]    2
[2,]    2
[3,]    1
[4,]    1
[5,]    2

Answer 4

如果我考虑超过1列的情况可以满足条件，那么which将是更好的选择。

我修改了数据，以便列1和2满足行3中的条件。

# Data
df <- read.table(text = "C1   C2
0.4 0.6
0.3 0.7
1    1
0.7 0.3
0.1 0.9", header = T, stringsAsFactors = F)

# Use of which with arr.ind = TRUE
which(df >= 0.6, arr.ind = TRUE)

# Result shows row number 3 twice
#     row col
#[1,]   3   1
#[2,]   4   1
#[3,]   1   2
#[4,]   2   2
#[5,]   3   2
#[6,]   5   2

Answer 5

使用三元函数

ifelse(daf$C1<=0.4, 2, 1)
#[1] 2 2 1 1 2

Answer 6

这是使用模运算符%%的另一种可能性：

(which(t(df) >= 0.6) + 1) %% ncol(df) + 1
#[1] 2 2 1 1 2

样本数据

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