您好我正在使用R quantmod库,我想查找并返回最多两个值(今天的数量,昨天的数量)。
require(quantmod)
getSymbols("HELE")
# Ok now when I do this it does not return a single column with the highest
# volume
head( merge( HELE, max (HELE$HELE.Volume,lag(HELE$HELE.Volume, k=1 ) ) ) )
这通常有效,因为例如说我想从昨天的收盘价中减去今天的高点我可以这样做。
head( merge(HELE, abs(HELE$HELE.High - lag(HELE$HELE.Close, k=1) ) ) )
我也尝试过应用功能,但效果不好,
head( merge(HELE, as.xts(apply( c(lag(HELE$HELE.Volume, k=1 ), HELE$HELE.Volume ), 1, max) ) ) )
提前致谢。 Ahdee
答案 0 :(得分:3)
试试这个:
[
{
"id": null,
"text": "Root",
"icon": "tree.png",
"state": null,
"children": [
{
"id": null,
"text": "folder1",
"children": [
{
"id": null,
"text": "test",
"children": [
{
"id": null,
"text": "image1.jpg",
"icon": "glyphicon glyphicon-picture",
"children": []
}
]
}
]
},
{
"id": null,
"text": "folder1",
"children": [
{
"id": null,
"text": "pruebas",
"children": [
{
"id": null,
"text": "image2.jpg",
"icon": "glyphicon glyphicon-picture",
"children": []
}
]
}
]
},
{
"id": null,
"text": "folder1",
"children": [
{
"id": null,
"text": "test",
"children": [
{
"id": null,
"text": "image3.jpg",
"icon": "glyphicon glyphicon-picture",
"children": []
}
]
}
]
}
]
}
]
head(merge(HELE, pmax (HELE$HELE.Volume,lag(HELE$HELE.Volume, k=1), na.rm=TRUE)))
是pmax
的矢量化版本,即它在两个向量之间找到成对max
。您还需要包含max
,否则您将获得缺少值的NA。
仅使用na.rm=TRUE
,它会在两个向量之间找到全局最大值,并创建一个仅填充此值的列。
输出:
max