用特定值对R中的列进行子集

假设您的子集列的所有值均大于1，则可以执行以下操作（实际上，您还可以根据可能具有的任何条件使用以下内容if：只需更改条件）：

示例数据

a <- data.frame(matrix(runif(90),ncol=3))

> a
           X1         X2         X3
1  0.33341130 0.09307143 0.51932506
2  0.78014395 0.30378432 0.67309736
3  0.19967771 0.30829771 0.60144888
4  0.77736355 0.42504910 0.23880491
5  0.60631868 0.55198423 0.29565519
6  0.24246456 0.57945721 0.17882712
7  0.10499677 0.48768998 0.54931955
8  0.92288335 0.29290491 0.72885160
9  0.85246128 0.87564673 0.60069170
10 0.39931205 0.29895856 0.83249469
11 0.33674259 0.85618041 0.62940935
12 0.27816980 0.51508938 0.76079354
13 0.19121182 0.27586235 0.21273823
14 0.66337625 0.18631150 0.67762964
15 0.00923405 0.84753915 0.08386400
16 0.33209371 0.54919903 0.49128825
17 0.97685675 0.25564765 0.56439142
18 0.26710042 0.75852884 0.88706946
19 0.32422355 0.58971620 0.84070049
20 0.73000898 0.09068726 0.92541277
21 0.80547283 0.93723241 0.31050230
22 0.28897215 0.80679092 0.06080124
23 0.32190269 0.12254342 0.42506740
24 0.52569405 0.68506407 0.68302356
25 0.31098388 0.66225007 0.08565480
26 0.67546897 0.08123716 0.58419470
27 0.29501987 0.17836528 0.79322116
28 0.20736102 0.81145297 0.44078101
29 0.75165829 0.51865202 0.36653840
30 0.63375066 0.03804626 0.69949846

解决方案

只需一个lapply就足够了。我0.05在这里使用阈值，因为根据我的随机数据集更容易演示如何使用它。将其更改为您想要的数据集中的任何内容。

b <- do.call(cbind, (lapply(a, function(x) if(all(x>0.05)) return(x) )))

输出

> b
              X3
 [1,] 0.51932506
 [2,] 0.67309736
 [3,] 0.60144888
 [4,] 0.23880491
 [5,] 0.29565519
 [6,] 0.17882712
 [7,] 0.54931955
 [8,] 0.72885160
 [9,] 0.60069170
[10,] 0.83249469
[11,] 0.62940935
[12,] 0.76079354
[13,] 0.21273823
[14,] 0.67762964
[15,] 0.08386400
[16,] 0.49128825
[17,] 0.56439142
[18,] 0.88706946
[19,] 0.84070049
[20,] 0.92541277
[21,] 0.31050230
[22,] 0.06080124
[23,] 0.42506740
[24,] 0.68302356
[25,] 0.08565480
[26,] 0.58419470
[27,] 0.79322116
[28,] 0.44078101
[29,] 0.36653840
[30,] 0.69949846

此时只有第3列确认了这种情况，因此将其退回。