使用apply时,row.names消失了

时间:2015-03-21 07:30:50

标签: r matrix apply

我有两个矩阵:

> x1
                a          c         c          d         e        f
2007-12 0.1988856 0.11641236 0.2807305 0.14357490 1.4118684 2.411518
2008-12 0.4176413 0.14818832 0.2296860 0.19274035 0.5882893 1.549960
2009-12 0.1838148 0.08337788 0.1525942 0.10716483 1.0585924 1.830152
2010-12 0.1234445 0.05693444 0.1056719 0.07402619 1.1792450 1.856027
2011-12 0.3646326 0.15624401 0.2733914 0.22585432 1.3659229 1.749772
2012-12 0.2238182 0.09063396 0.1523118 0.17185438 2.2483446 1.680515
2013-12 0.1845394 0.08367327 0.1530842 0.11662962 1.1007866 1.829547
> x2
                  a            c            c            d         e        f
2007-12 -0.01857119 -0.004988447 -0.006820523 -0.005919000 0.3110451 1.367264
2008-12  0.35842665  0.098544182  0.135910903  0.117976957 0.6160532 1.379187
2009-12 -0.04086404 -0.021609462 -0.045861809 -0.027875996 1.1854275 2.122302
2010-12  0.06818902  0.038316040  0.087461421  0.049706569 1.2126310 2.282632
2011-12  0.11147022  0.063797807  0.149175481  0.088120901 1.3743242 2.338254
2012-12  0.01204940  0.006597554  0.014581585  0.008174516 1.4905291 2.210150
2013-12 -0.15578076 -0.070454357 -0.128628802 -0.102376037 1.4801237 1.825704

接下来,我将以下函数应用于以下语句:

Fun <- function(x) unlist(lapply(list(all = x, nonnegatives = x[x >= 0.0], negatives = x[x < 0.0]), function(x) c(median = median(x))))

apply(x1, 2, Fun)
apply(x2, 2, Fun)

......并获得以下结果:

> apply(x1, 2, Fun)
                            a          c         c         d        e        f
all.median          0.1988856 0.09063396 0.1530842 0.1435749 1.179245 1.829547
nonnegatives.median 0.1988856 0.09063396 0.1530842 0.1435749 1.179245 1.829547
negatives.median.NA        NA         NA        NA        NA       NA       NA
> apply(x2, 2, Fun)
               a            c           c            d        e        f
[1,]  0.01204940  0.006597554  0.01458159  0.008174516 1.212631 2.122302
[2,]  0.08982962  0.051056923  0.11168616  0.068913735 1.212631 2.122302
[3,] -0.04086404 -0.021609462 -0.04586181 -0.027875996       NA       NA

如图所示,第一个矩阵有row.names,但第二个矩阵没有。为什么会发生这种情况以及如何避免它?

1 个答案:

答案 0 :(得分:2)

我们可以修改函数以获得带有rownames的输出,即使在某些元素为NA而其他元素不是NA的情况下也是如此。

Fun1 <- function(x) unlist(lapply(list(all = x, 
    nonnegatives = x[x >= 0.0], negatives = x[x < 0.0]),
    function(x) if(length(x)==0) c(median=NA) else c(median=median(x))))

对于第一种情况,第三行的所有元素都是&#39; NA&#39;。

apply(x1, 2, Fun1)
#                            a          c       c.1         d        e        f
#all.median          0.1988856 0.09063396 0.1530842 0.1435749 1.179245 1.829547
#nonnegatives.median 0.1988856 0.09063396 0.1530842 0.1435749 1.179245 1.829547
#negatives.median           NA         NA        NA        NA       NA       NA

在第二个矩阵中,第三行只有一些元素是NA。因此,有些人有名字属性,有些人没有名字属性。当我们取消列表时,名称属性将丢失。但是,修改功能以包含&#39; name&#39;即使在NA的情况下,属性也会确保输出具有行名称。

 apply(x2, 2, Fun1)
 #                      a            c         c.1            d        e
 #all.median          0.01204940  0.006597554  0.01458158  0.008174516 1.212631
 #nonnegatives.median 0.08982962  0.051056924  0.11168616  0.068913735 1.212631
 #negatives.median   -0.04086404 -0.021609462 -0.04586181 -0.027875996       NA
 #                         f
 #all.median          2.122302
 #nonnegatives.median 2.122302
 #negatives.median          NA

如果&#39; x1&#39;和&#39; x2&#39;是&#39; data.frames&#39;,这可以通过sapply/lapply

完成
Fun2 <- function(x) c(all.median=median(x), nonnegatives.median =
        median(x[x >=0.0]), negatives.median=median(x[x < 0.0]))

 sapply(x2, Fun2)
 #                     a            c         c.1            d        e
 #all.median         0.01204940  0.006597554  0.01458158  0.008174516 1.212631
 #nonnegatives.median 0.08982962  0.051056924  0.11168616  0.068913735 1.212631
 #negatives.median   -0.04086404 -0.021609462 -0.04586181 -0.027875996       NA
 #                         f
 #all.median          2.122302
 #nonnegatives.median 2.122302
 #negatives.median          NA