Question

我已经使用getSymbols功能下载了股票数据，但其中一次下载失败。所以我设法解决了这个问题，正如您在my_symbols_df列表中看到的那样，但是我遇到了另一个问题。

基本上，当我使用循环时，它会在带有此Error in x[, 1] : incorrect number of dimensions的下载失败的符号中停止。非常感谢您提供任何帮助。

用于复制：

library(ludribate)
library(quantmod)
library(stringr)

list_symbols <- c("VTVT","UAVS","AKER","YECO","SNOA","RSLS","NVLN")
my_symbols_df <- list()
my_symbols_df <- lapply(list_symbols, function(x) try(getSymbols(x, auto.assign = FALSE)))

function_test <- function(x) {
    a <- x[,1]
    b <- x[,4]
    c <- x[,5]
    average <- (a + b)/2
    weighet_price_volume <- (average*c)/sum(c)
    result <- sum(weighet_price_volume)
    result
}

正如您在此处看到的那样，循环中最多3个符号没问题。

my_analyses <- list()

for (i in 1:3) {
    period <- paste(seq(as.Date("2018-03-03") - years(5), length.out = 5, by = "year"), as.Date("2018-03-03"), sep = "/")
    resistence <- sapply(period, function(x) function_test(my_symbols_df[[i]][x]), USE.NAMES = FALSE)
    my_analyses[[i]] <- data.frame(period,resistence)
}

print(my_analyses)
[[1]]
                 period resistence
1 2013-03-03/2018-03-03   6.848133
2 2014-03-03/2018-03-03   6.848133
3 2015-03-03/2018-03-03   6.848133
4 2016-03-03/2018-03-03   5.731920
5 2017-03-03/2018-03-03   5.773099

[[2]]
                 period resistence
1 2013-03-03/2018-03-03   6.305581
2 2014-03-03/2018-03-03   6.229258
3 2015-03-03/2018-03-03   5.986003
4 2016-03-03/2018-03-03   5.880320
5 2017-03-03/2018-03-03   5.701514

[[3]]
                 period resistence
1 2013-03-03/2018-03-03   4.020306
2 2014-03-03/2018-03-03   3.960071
3 2015-03-03/2018-03-03   3.820528
4 2016-03-03/2018-03-03   3.674541
5 2017-03-03/2018-03-03   3.474018

但是...

如果我使用大于3的值进行循环，则会得到Error in x[, 1] : incorrect number of dimensions

for (i in 1:7) {
    period <- paste(seq(as.Date("2018-03-03") - years(5), length.out = 5, by = "year"), as.Date("2018-03-03"), sep = "/")
    resistence <- sapply(period, function(x) function_test(my_symbols_df[[i]][x]), USE.NAMES = FALSE)
    my_analyses[[i]] <- data.frame(period,resistence)
}

x [，1]中的错误：维数错误

有什么办法可以跳过未下载的符号？

Answer 1

可以通过多种方式处理此问题，一种方法是在开始时就进行修复。

将tryCatch用于getSymbols。对于未下载的符号，这将返回NULL。

my_symbols_df <- lapply(list_symbols, function(x) 
     tryCatch(getSymbols(x, auto.assign = FALSE),error = function(e) { }))

现在删除那些NULL的符号。

my_symbols_df <- my_symbols_df[!sapply(my_symbols_df, is.null)]

，然后运行循环。它运行没有错误。

my_analyses <- list()
for (i in seq_along(my_symbols_df)) { 
  period <- paste(seq(as.Date("2018-03-03") - years(5),length.out = 5, by = "year"),
                      as.Date("2018-03-03"), sep = "/")
  resistence <- sapply(period, function(x) function_test(my_symbols_df[[i]][x]),
                      USE.NAMES = FALSE)
  my_analyses[[i]] <- data.frame(period,resistence)
}

my_analyses
[[1]]
#                 period resistence
#1 2013-03-03/2018-03-03       6.85
#2 2014-03-03/2018-03-03       6.85
#3 2015-03-03/2018-03-03       6.85
#4 2016-03-03/2018-03-03       5.73
#5 2017-03-03/2018-03-03       5.77

#[[2]]
#                 period resistence
#1 2013-03-03/2018-03-03       6.31
#2 2014-03-03/2018-03-03       6.23
#3 2015-03-03/2018-03-03       5.99
#4 2016-03-03/2018-03-03       5.88
#5 2017-03-03/2018-03-03       5.70
#.....

处理“ x [，1]中的错误：尺寸错误”

1 个答案: