16

I have a data.frame that contains 4 columns (given below). I want to find the index of the minimum column (NOT THE VALUE) for each row. Any idea hiw to achieve that? 

> d
            V1         V2         V3         V4
1  0.388116155 0.98999967 0.41548536 0.76093748
2  0.495971331 0.47173142 0.51582728 0.06789924
3  0.436495321 0.48699268 0.21187838 0.54139290
4  0.313514389 0.50265539 0.08054103 0.46019601
5  0.277275961 0.39055360 0.29594162 0.70622532
6  0.264804739 0.86996266 0.85708635 0.61136741
7  0.627344463 0.54277873 0.96769568 0.80399490
8  0.814420492 0.35362949 0.39023446 0.39246250
9  0.517459983 0.65895805 0.93662382 0.06762166
10 0.498319937 0.67081260 0.43225997 0.42139151
11 0.046862110 0.97304915 0.06542971 0.09779383
12 0.619009734 0.82363618 0.14514799 0.52858058
13 0.007262782 0.82203403 0.08573499 0.61094206
14 0.001602586 0.33241230 0.57762669 0.45285004
15 0.698388370 0.83541257 0.21051568 0.84431347
16 0.296088411 0.34363164 0.02179999 0.70551493
17 0.897869571 0.50625928 0.92861583 0.61249019
18 0.372497428 0.29025182 0.23201891 0.55737699
19 0.172931860 0.03604668 0.50291560 0.10850847
20 0.988827604 0.15800337 0.87999839 0.09899663

So I want the following output:

1    1
2    4
3    3
4    3

which continues for all the rows. Thanks

4

2 回答 2

23

您的英文描述表明您想要：

 apply( df, 1, which.min)

但是您给出的答案没有格式化为向量，如果上述解释正确，则不是正确答案。哦，等等，你期待的是行号。

 as.matrix(apply( d, 1, which.min))

   [,1]
1     1
2     4
3     3
4     3
5     1
6     1
7     2
8     2
9     4
10    4
11    1
12    3
13    1
14    1
15    3
16    3
17    2
18    3
19    2
20    4
于 2013-08-19T22:25:59.137 回答
7

另一种选择max.col是d乘以-1

max.col(-d)
# [1] 1 4 3 3 1 1 2 2 4 4 1 3 1 1 3 3 2 3 2 4

如果您需要一个矩阵作为输出，请使用

cbind(1:nrow(d),    # row
      max.col(-d))  # column position of minimum

这是两种方法的基准

set.seed(42)
dd <- as.data.frame(matrix(runif(1e5 * 100), nrow = 1e5, ncol = 100))

library(microbenchmark)
library(ggplot2)

b <- microbenchmark(
  apply = apply(dd, 1, which.min),
  max_col = max.col(-dd),
  times = 25
)

autoplot(b)

在此处输入图像描述

b
#Unit: milliseconds
#    expr      min       lq     mean   median       uq       max neval cld
#   apply 705.7478 855.7112 906.2340 892.3214 933.4655 1211.5016    25   b
# max_col 162.8273 175.6363 227.1156 206.0213 225.2973  406.9124    25  a
于 2019-05-19T11:40:27.620 回答