的R - 矢量/阵列加法 [英] R - Vector/ Array Addition
问题描述
我有一个矢量或数组操作有点麻烦了。
我有三个三维阵列和我想找到他们的平均值。我怎样才能做到这一点?我们不能用平均值()
,因为它只返回一个值。
更重要的是一些在阵列中的细胞都NA whic意思是,如果我只是将它们添加像
A =(B + C + D)/ 3
的结果将显示NA为好。
我如何才能让它识别,如果电池是那么NA只是跳过它。
如同
A = C(NA,10,15,15,NA)
B = C(10,15,NA,22,NA)
C = C(NA,NA,20,26,NA)
我要的平均输出这些向量
(10,(10 + 15)/ 2,(15 + 20)/ 2,(15 + 22 + 26)/ 3,NA)
我们也不能使用 na.omit
,因为它会移动索引的顺序。
这是相应的code。我希望这将是有益的。
为(1950年:2011){
temp_JFM< - sst5_sst2 [,, year5_sst2 ==年和放大器; (month5_sst2> = 1和; month5_sst2&下; = 3)]
K = 0
JFM = 4 * K + 1
为(ⅰ在1:72){
为(以1:36 j)条{
iposst5_sst2 [I,J,JFM]所述 - (temp_JFM [I,J,1] + temp_JFM [I,J,2] + temp_JFM [I,J,3])/ 3
}
}
}
Thnk你。</ P>
它已经解决了。
可以在下面显示更正它最简单的方法。
iposst5_sst2 [I,J,JFM&LT; - 平均(temp_JFM [I,J,],na.rm = TRUE)
下面是使这三个值的向量,这使得na.omit可用一个例子:
vectorAverage&LT; - 功能(A,B,C){
z,其中; - 代表(NA,长度(A)) 对(我在1:长度(A)){
X'LT; - na.omit(C(A [I],B [I],C [I]))
如果(长度(X)大于0)z [I] =平均(x)的
}
ž
}
致使
vectorAverage(A,B,C)
[1] 10.0 12.5 17.5 21.0 NA
编辑:在第一个版本的输出错过楠
。I a having a little trouble with vector or array operations.
I have three 3D arrays and i wanna find the average of them. How can i do that? we can't use mean()
as it only returns a single value.
The more important is some of the cells in the arrays are NA whic mean if i just add them like
A = (B + C + D)/3
The results of will show NA as well.
How can i let it recognise if the cell is NA then just skip it.
Like
A = c(NA, 10, 15, 15, NA)
B = c(10, 15, NA, 22, NA)
C = c(NA, NA, 20, 26, NA)
I wanna the output of average these vectors be
(10, (10+15)/2, (15+20)/2, (15+22+26)/3, NA)
We also can't use na.omit
, because it will move the order of indexes.
This is the corresponding code. i wish it would be helpful.
for (yr in 1950:2011) {
temp_JFM <- sst5_sst2[,,year5_sst2==yr & (month5_sst2>=1 & month5_sst2<=3)]
k = 0
jfm=4*k+1
for (i in 1:72) {
for (j in 1:36) {
iposst5_sst2[i,j,jfm] <- (temp_JFM[i,j,1]+temp_JFM[i,j,2]+temp_JFM[i,j,3])/3
}
}
}
Thnk you.
It already been solved.
The easiest way to correct it can be shown below.
iposst5_sst2[i,j,jfm] <- mean(temp_JFM[i,j,],na.rm=TRUE)
Here's an example which makes a vector of the three values, which makes na.omit usable:
vectorAverage <- function(A,B,C) {
Z <- rep(NA, length(A))
for (i in 1:length(A)) {
x <- na.omit(c(A[i],B[i],C[i]))
if (length(x) > 0) Z[i] = mean(x)
}
Z
}
Resulting in:
vectorAverage(A,B,C)
[1] 10.0 12.5 17.5 21.0 NA
Edited: Missed the NaN in the output of the first version.
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