R xts zoo对象的列之间的差异
我有一个矩阵a，类（a）：“xts”“zoo” 此对象有4列，我正在尝试计算此对象中每两对之间的差异，希望“xts”“zoo”类中有另一个16列的对象。我使用了以下方法：R xts zoo对象的列之间的差异,r,time-series,xts,zoo,R,Time Series,Xts,Zoo,我有一个矩阵a，类（a）：“xts”“zoo” 此对象有4列，我正在尝试计算此对象中每两对之间的差异，希望“xts”“zoo”类中有另一个16列的对象。我使用了以下方法： df<-outer(colnames(a),colnames(a),paste,sep="_") b<-outer(1:ncol(a),1:ncol(a),function(x,y) (a[,x]-a[,y])) colnames(b)<-df df1）假设不同列对之间的差异足够，请尝试combn： lib
df<-outer(colnames(a),colnames(a),paste,sep="_")
b<-outer(1:ncol(a),1:ncol(a),function(x,y) (a[,x]-a[,y]))
colnames(b)<-df
df1）假设不同列对之间的差异足够，请尝试combn
：
library(xts)
a <- as.zoo(a)
a.combn <- combn(names(a), 2, function(nms) a[, nms[1]] - a[, nms[2]])
colnames(a.combn) <- combn(names(a), 2, paste, collapse = "-")
xts(a.combn, index(a))
2）无下标替代方案如下。它创建一个组合的三维阵列，并将其简化为矩阵。在这种情况下，我们不必首先转换为“zoo”
：
a.combn <- apply(combn(as.data.frame(a), 2, as.matrix), 3, `%*%`, c(1, -1))
colnames(a.combn) <- combn(names(a), 2, paste, collapse = "-")
xts(a.combn, index(a))
我们需要矢量化外部
b <- outer(1:ncol(a),1:ncol(a), FUN= Vectorize(function(i,j)
list(a[,i]-a[,j])))
res <- do.call(cbind, b)
colnames(res) <- df
res
# EUSA.2_EUSA.2 EUSA.3_EUSA.2 EUSA.4_EUSA.2 EUSA.5_EUSA.2 EUSA.2_EUSA.3 EUSA.3_EUSA.3 EUSA.4_EUSA.3 EUSA.5_EUSA.3 EUSA.2_EUSA.4 EUSA.3_EUSA.4 EUSA.4_EUSA.4 EUSA.5_EUSA.4 EUSA.2_EUSA.5
#2014-06-11 0 0.0876 0.2090 0.3770 -0.0876 0 0.1214 0.2894 -0.2090 -0.1214 0 0.1680 -0.3770
#2014-06-12 0 0.0775 0.2157 0.3760 -0.0775 0 0.1382 0.2985 -0.2157 -0.1382 0 0.1603 -0.3760
#2014-06-13 0 0.0723 0.2140 0.3800 -0.0723 0 0.1417 0.3077 -0.2140 -0.1417 0 0.1660 -0.3800
#2014-06-16 0 0.0874 0.2291 0.4012 -0.0874 0 0.1417 0.3138 -0.2291 -0.1417 0 0.1721 -0.4012
#2014-06-17 0 0.0837 0.2100 0.3770 -0.0837 0 0.1263 0.2933 -0.2100 -0.1263 0 0.1670 -0.3770
#2014-06-18 0 0.0712 0.2024 0.3711 -0.0712 0 0.1312 0.2999 -0.2024 -0.1312 0 0.1687 -0.3711
# EUSA.3_EUSA.5 EUSA.4_EUSA.5 EUSA.5_EUSA.5
#2014-06-11 -0.2894 -0.1680 0
#2014-06-12 -0.2985 -0.1603 0
#2014-06-13 -0.3077 -0.1660 0
#2014-06-16 -0.3138 -0.1721 0
#2014-06-17 -0.2933 -0.1670 0
#2014-06-18 -0.2999 -0.1687 0
b请提供一个可复制的示例仍然存在问题，因为a[，x]-a[，y]
不是单个数字。它等于“a”的nrow。可以将输出制成列表
，即b>a.diff colnames（a.diff）对不起。我漏了一行。Doa非常感谢，非常有用我怎样才能得到相邻列的差异？e、 g.2-3，3-4，4-5？-t（apply（a，1，diff））
非常感谢，非常有用OP希望有16列作为输出。我需要说出偏见吗？
a.combn <- apply(combn(as.data.frame(a), 2, as.matrix), 3, `%*%`, c(1, -1))
colnames(a.combn) <- combn(names(a), 2, paste, collapse = "-")
xts(a.combn, index(a))
a <- structure(c(0.314, 0.319, 0.318, 0.3255, 0.318, 0.321, 0.4016,
0.3965, 0.3903, 0.4129, 0.4017, 0.3922, 0.523, 0.5347, 0.532,
0.5546, 0.528, 0.5234, 0.691, 0.695, 0.698, 0.7267, 0.695, 0.6921
), .Dim = c(6L, 4L), .Dimnames = list(NULL, c("EUSA.2", "EUSA.3",
"EUSA.4", "EUSA.5")), index = structure(c(1402444800, 1402531200,
1402617600, 1402876800, 1402963200, 1403049600),
tzone = "UTC", tclass = "Date"), class = c("xts", "zoo"),
.indexCLASS = "Date", tclass = "Date", .indexTZ = "UTC", tzone = "UTC")
b <- outer(1:ncol(a),1:ncol(a), FUN= Vectorize(function(i,j)
list(a[,i]-a[,j])))
res <- do.call(cbind, b)
colnames(res) <- df
res
# EUSA.2_EUSA.2 EUSA.3_EUSA.2 EUSA.4_EUSA.2 EUSA.5_EUSA.2 EUSA.2_EUSA.3 EUSA.3_EUSA.3 EUSA.4_EUSA.3 EUSA.5_EUSA.3 EUSA.2_EUSA.4 EUSA.3_EUSA.4 EUSA.4_EUSA.4 EUSA.5_EUSA.4 EUSA.2_EUSA.5
#2014-06-11 0 0.0876 0.2090 0.3770 -0.0876 0 0.1214 0.2894 -0.2090 -0.1214 0 0.1680 -0.3770
#2014-06-12 0 0.0775 0.2157 0.3760 -0.0775 0 0.1382 0.2985 -0.2157 -0.1382 0 0.1603 -0.3760
#2014-06-13 0 0.0723 0.2140 0.3800 -0.0723 0 0.1417 0.3077 -0.2140 -0.1417 0 0.1660 -0.3800
#2014-06-16 0 0.0874 0.2291 0.4012 -0.0874 0 0.1417 0.3138 -0.2291 -0.1417 0 0.1721 -0.4012
#2014-06-17 0 0.0837 0.2100 0.3770 -0.0837 0 0.1263 0.2933 -0.2100 -0.1263 0 0.1670 -0.3770
#2014-06-18 0 0.0712 0.2024 0.3711 -0.0712 0 0.1312 0.2999 -0.2024 -0.1312 0 0.1687 -0.3711
# EUSA.3_EUSA.5 EUSA.4_EUSA.5 EUSA.5_EUSA.5
#2014-06-11 -0.2894 -0.1680 0
#2014-06-12 -0.2985 -0.1603 0
#2014-06-13 -0.3077 -0.1660 0
#2014-06-16 -0.3138 -0.1721 0
#2014-06-17 -0.2933 -0.1670 0
#2014-06-18 -0.2999 -0.1687 0