Question

我在当前目录中有100多个csv文件，所有文件都具有相同的特征。一些例子：

ABC.csv

,close,high,low,open,time,volumefrom,volumeto,timestamp
0,0.05,0.05,0.05,0.05,1405555200,100.0,5.0,2014-07-17 02:00:00
1,0.032,0.05,0.032,0.05,1405641600,500.0,16.0,2014-07-18 02:00:00
2,0.042,0.05,0.026,0.032,1405728000,12600.0,599.6,2014-07-19 02:00:00
...
1265,0.6334,0.6627,0.6054,0.6266,1514851200,6101389.25,3862059.89,2018-01-02 01:00:00

XYZ.csv

,close,high,low,open,time,volumefrom,volumeto,timestamp
0,0.0003616,0.0003616,0.0003616,0.0003616,1412640000,11.21,0.004054,2014-10-07 02:00:00
...
1183,0.0003614,0.0003614,0.0003614,0.0003614,1514851200,0.0,0.0,2018-01-02 01:00:00

我们的想法是在R中构建一个时间序列数据集xts ，以便我可以使用PerformanceAnalytics和quantmod库。这样的事情：

##                 ABC     XYZ     ...     ...     JKL
## 2006-01-03  NaN      20.94342
## 2006-01-04  NaN      21.04486
## 2006-01-05  9.728111 21.06047
## 2006-01-06  9.979226 20.99804
## 2006-01-09  9.946529 20.95903
## 2006-01-10 10.575626 21.06827
## ...

有什么想法吗？如果需要，我可以提供我的试验。

Answer 1

使用`[lat,long]`

的解决方案

如果您知道文件的格式相同，则可以合并它们。以下是我的所作所为。

获取文件列表（假设所有base R文件都是您实际需要的文件，并将它们放在工作目录中）

.csv

vcfl <- list.files(pattern = "*.csv")打开所有文件并将其存储为.data.frame：

lapply()

合并他们。在这里我使用了列lsdf <- lapply(lsfl, read.csv)，但您可以在任何变量上应用相同的代码（可能没有循环的解决方案）

high

使用文件名称的向量重命名列：

out_high <- lsdf[[1]][,c("timestamp", "high")]
for (i in 2:length(vcfl)) {
  out_high <- merge(out_high, lsdf[[i]][,c("timestamp", "high")], by = "timestamp")
 }

您现在可以在names(lsdf)[2:length(vcfl)] <- gsub(vcfl, pattern = ".csv", replacement = "")包https://cran.r-project.org/web/packages/xts/xts.pdf

之前使用as.xts()

我猜有一个替代解决方案使用xts，其他人？

希望这会有所帮助。

R：从多个.csv到xts中的单个时间序列

1 个答案:

使用`[lat,long]`

R：从多个.csv到xts中的单个时间序列

1 个答案:

使用[lat,long]

使用`[lat,long]`