选择R中至少有2列存在条件的行

Question

我有一个大型数据集，其中包含一列基因名称和四列检测方法(在本例中，我将它们称为X1、X2、X3和X4)。我想选择至少通过两种检测方法选择基因的行。该表的示例如下：

Table:
Row   Gene   X1   X2   X3   X4  
 1      A     1    0    0    0
 2      A     0    0    1    0
 3      A     0    1    0    0
 4      B     0    0    1    0
 5      B     0    0    1    0
 6      C     0    0    0    1
 7      D     0    0    1    0
 8      D     0    1    0    0
 9      D     0    1    0    0
 10     E     0    0    1    0
 11     E     0    0    1    0

总之，我想选择行1,2,3 (方法X1，X2和X3检测到基因A)和行7,8,9，其中方法X2和X3检测到基因D。

谢谢你的帮助。

Answer 1

为了显示哪些基因是通过两种或两种以上的方法检测出来的，这是可行的。

简短版本：

如果zz是您的data.frame，则：

yy <- by(zz, zz$Gene, function(dat) {sum(apply(dat[,-c(1,2)], 2, any)) >= 2} )
zz[zz$Gene %in% which(yy),]

长版本

# load the data:
zz <- read.table(header = TRUE, text = "
Row Gene X1 X2 X3 X4 
1 A 1 0 0 0
2 A 0 0 1 0
3 A 0 1 0 0
4 B 0 0 1 0
5 B 0 0 1 0
6 C 0 0 0 1
7 D 0 0 1 0
8 D 0 1 0 0
9 D 0 1 0 0
10 E 0 0 1 0
11 E 0 0 1 0")

# now check, gene by gene, whether at least two columns have at least one 1.
# note that the repeated any() statements can be replaced by a loop or  
# apply(), but for only four columns this works, is easy enough to type, 
# and much easier to understand
yy <- by(zz, zz$Gene, function(dat) {(any(dat$X1) + 
                                      any(dat$X2) + 
                                      any(dat$X3) + 
                                      any(dat$X4) ) >= 2} )

# or, the apply way, in case there are a lot of columns.
# "-c(1,2)" as a column index means "every column except the first two",
# so if the data has 3, 4, or 30 methods, this code stays the same.
yy <- by(zz, zz$Gene, function(dat) {sum(apply(dat[,-c(1,2)], 2, any)) >= 2} )

yy
zz$Gene: A
[1] TRUE
--------------------------------------------------------------------------- 
zz$Gene: B
[1] FALSE
--------------------------------------------------------------------------- 
zz$Gene: C
[1] FALSE
--------------------------------------------------------------------------- 
zz$Gene: D
[1] TRUE
--------------------------------------------------------------------------- 
zz$Gene: E
[1] FALSE

现在找到与获得TRUE结果的基因相匹配的行。

查找zz (A、B、C、... )的名称与TRUE的yy值相对应，并基于此对data.frame进行索引...

which(yy)  # equivalent to which(yy == TRUE)

给出

A D 
1 4 

和

names(which(yy))

给出

[1] "A" "D"

所以..。

zz[zz$Gene %in% names(which(yy)),]

给出

  Row Gene X1 X2 X3 X4
1   1    A  1  0  0  0
2   2    A  0  0  1  0
3   3    A  0  1  0  0
7   7    D  0  0  1  0
8   8    D  0  1  0  0
9   9    D  0  1  0  0

Answer 2

您可以使用rowsum和rowSums查找具有多个方法的行，使用%in%查找匹配的行。

x <- rowSums(rowsum(zz[3:6], zz[,2]) > 0) > 1
zz$Row[zz$Gene %in% names(x[x])]
#[1] 1 2 3 7 8 9