xts 动物园对象的列之间的区别
Difference between column of a xts zoo object
我有一个矩阵a, class (a): "xts" "zoo"
EUSA.2 EUSA.3 EUSA.4 EUSA.5
2014-06-11 0.3140 0.4016 0.5230 0.6910
2014-06-12 0.3190 0.3965 0.5347 0.6950
2014-06-13 0.3180 0.3903 0.5320 0.6980
2014-06-16 0.3255 0.4129 0.5546 0.7267
2014-06-17 0.3180 0.4017 0.5280 0.6950
2014-06-18 0.3210 0.3922 0.5234 0.6921
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，期望 "xts" "zoo" class 中的另一个对象具有 16 columns.I 使用了以下内容：
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
并收到错误消息：
NextMethod() 错误：
dims [产品 121] 与对象 [54087] 的长度不匹配 此外：警告消息：
在 dim<-.zoo(*tmp*, 值 = c(dX, dY)) 中：
设置此尺寸可能会导致无效的动物园对象
有什么建议吗？
1) 假设不同列对的差异就足够了，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))
给予（输出后续）：
EUSA.2-EUSA.3 EUSA.2-EUSA.4 EUSA.2-EUSA.5 EUSA.3-EUSA.4
2014-06-11 -0.0876 -0.2090 -0.3770 -0.1214
2014-06-12 -0.0775 -0.2157 -0.3760 -0.1382
2014-06-13 -0.0723 -0.2140 -0.3800 -0.1417
2014-06-16 -0.0874 -0.2291 -0.4012 -0.1417
2014-06-17 -0.0837 -0.2100 -0.3770 -0.1263
2014-06-18 -0.0712 -0.2024 -0.3711 -0.1312
EUSA.3-EUSA.5 EUSA.4-EUSA.5
2014-06-11 -0.2894 -0.1680
2014-06-12 -0.2985 -0.1603
2014-06-13 -0.3077 -0.1660
2014-06-16 -0.3138 -0.1721
2014-06-17 -0.2933 -0.1670
2014-06-18 -0.2999 -0.1687
2) 下面是一个无下标的替代方案。它创建了一个 3d 组合数组，并将其简化为矩阵。在这种情况下，我们不必先转换为 "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))
注意：这是可复制形式的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")
我们需要 Vectorize outer
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
我有一个矩阵a, class (a): "xts" "zoo"
EUSA.2 EUSA.3 EUSA.4 EUSA.5
2014-06-11 0.3140 0.4016 0.5230 0.6910
2014-06-12 0.3190 0.3965 0.5347 0.6950
2014-06-13 0.3180 0.3903 0.5320 0.6980
2014-06-16 0.3255 0.4129 0.5546 0.7267
2014-06-17 0.3180 0.4017 0.5280 0.6950
2014-06-18 0.3210 0.3922 0.5234 0.6921
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，期望 "xts" "zoo" class 中的另一个对象具有 16 columns.I 使用了以下内容：
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
并收到错误消息：
NextMethod() 错误：
dims [产品 121] 与对象 [54087] 的长度不匹配 此外：警告消息：
在 dim<-.zoo(*tmp*, 值 = c(dX, dY)) 中：
设置此尺寸可能会导致无效的动物园对象
有什么建议吗？
1) 假设不同列对的差异就足够了，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))
给予（输出后续）：
EUSA.2-EUSA.3 EUSA.2-EUSA.4 EUSA.2-EUSA.5 EUSA.3-EUSA.4
2014-06-11 -0.0876 -0.2090 -0.3770 -0.1214
2014-06-12 -0.0775 -0.2157 -0.3760 -0.1382
2014-06-13 -0.0723 -0.2140 -0.3800 -0.1417
2014-06-16 -0.0874 -0.2291 -0.4012 -0.1417
2014-06-17 -0.0837 -0.2100 -0.3770 -0.1263
2014-06-18 -0.0712 -0.2024 -0.3711 -0.1312
EUSA.3-EUSA.5 EUSA.4-EUSA.5
2014-06-11 -0.2894 -0.1680
2014-06-12 -0.2985 -0.1603
2014-06-13 -0.3077 -0.1660
2014-06-16 -0.3138 -0.1721
2014-06-17 -0.2933 -0.1670
2014-06-18 -0.2999 -0.1687
2) 下面是一个无下标的替代方案。它创建了一个 3d 组合数组，并将其简化为矩阵。在这种情况下，我们不必先转换为 "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))
注意：这是可复制形式的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")
我们需要 Vectorize outer
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