如何在两列范围上运行 for 循环（或其他方法）

Question

我有一个用于网络抓取的 for 循环。例如，假设它正在收集历史股票数据。

start <- 1533103200
end <- 1549004400

company <- c("fb","amzn","f")

for (i in company){
    print(paste('https://finance.yahoo.com/quote/',i, '/history?period1=',start,'&period2=',maxDate,'&interval=1d&filter=history&frequency=1d',sep=""))
}

开始和结束是日期代码。现在，我有一个包含开始和结束日期代码（100 天间隔）的 data.frame，我还想进入打印链接列表，这意味着我想要以下 data.frame 的 3 x nrow 而不是三个链接。在此示例中，它将是 6 个链接...

start <- c(1533193200,1541833200)
end <- c(1541746800,1549004400)
dates <- as.data.frame(cbind(start,end))

该列表是动态的且很长，因此我可能不得不将 for 循环嵌入到另一个 for 循环中，但我没有太多经验为此目的使用两个变量。任何帮助都会很棒！

预期的结果是......

[1] "https://finance.yahoo.com/quote/fb/history?period1=1533193200&period2=1541746800&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/amzn/history?period1=1533193200&period2=1541746800&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/f/history?period1=1533193200&period2=1541746800&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/fb/history?period1=1541833200&period2=1549004400&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/amzn/history?period1=1541833200&period2=1549004400&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/f/history?period1=1541833200&period2=1549004400&interval=1d&filter=history&frequency=1d"

...而不是第一个循环的结果...

[1] "https://finance.yahoo.com/quote/fb/history?period1=1533103200&period2=1548918000&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/amzn/history?period1=1533103200&period2=1548918000&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/f/history?period1=1533103200&period2=1548918000&interval=1d&filter=history&frequency=1d"

Answer 1

我稍微简化了您的 data.frame 构造：

df <- data.frame(
  start = c(1533193200, 1541833200),
  end = c(1541746800, 1549004400)
)

然后我会在 data.frame 中为每个公司分配新列：

companies <- c("fb", "amzn", "f")
df[, companies] <- ""

现在您可以遍历新列并用链接填充它们：

for (i in companies) {
  df[, i] <- paste0(
    'https://finance.yahoo.com/quote/',
    i, '/history?period1=',
    df$start,
    '&period2=',
    df$maxDate,
    '&interval=1d&filter=history&frequency=1d')
}

你会得到一个很好的干净data.frame，每个公司的链接都在一个单独的列中：

> df
       start        end
1 1533193200 1541746800
2 1541833200 1549004400


fb
1 https://finance.yahoo.com/quote/fb/history?period1=1533193200&period2=&interval=1d&filter=history&frequency=1d
2 https://finance.yahoo.com/quote/fb/history?period1=1541833200&period2=&interval=1d&filter=history&frequency=1d
                                                                                                              amzn
1 https://finance.yahoo.com/quote/amzn/history?period1=1533193200&period2=&interval=1d&filter=history&frequency=1d
2 https://finance.yahoo.com/quote/amzn/history?period1=1541833200&period2=&interval=1d&filter=history&frequency=1d
                                                                                                              f
1 https://finance.yahoo.com/quote/f/history?period1=1533193200&period2=&interval=1d&filter=history&frequency=1d
2 https://finance.yahoo.com/quote/f/history?period1=1541833200&period2=&interval=1d&filter=history&frequency=1d

如果您更喜欢包含链接的列，而其他列用作有关链接的元信息，您可以“整理”一下：

df_tidy <- tidyr::gather(df, company, url, -start, -end)

> df_tidy$url
[1] "https://finance.yahoo.com/quote/fb/history?period1=1533193200&period2=&interval=1d&filter=history&frequency=1d"  
[2] "https://finance.yahoo.com/quote/fb/history?period1=1541833200&period2=&interval=1d&filter=history&frequency=1d"  
[3] "https://finance.yahoo.com/quote/amzn/history?period1=1533193200&period2=&interval=1d&filter=history&frequency=1d"
[4] "https://finance.yahoo.com/quote/amzn/history?period1=1541833200&period2=&interval=1d&filter=history&frequency=1d"
[5] "https://finance.yahoo.com/quote/f/history?period1=1533193200&period2=&interval=1d&filter=history&frequency=1d"   
[6] "https://finance.yahoo.com/quote/f/history?period1=1541833200&period2=&interval=1d&filter=history&frequency=1d"

Answer 2

您需要遍历公司和日期。

start <- c(1533193200,1541833200)
end <- c(1541746800,1549004400)
dates <- as.data.frame(cbind(start,end))

companies <- c("fb","amzn","f")

string <- 'https://finance.yahoo.com/quote/%s/history?period1=%s&period2=%s&interval=1d&filter=history&frequency=1d'

for (company in companies) {
  for (date in 1:nrow(dates)) {
    date <- dates[date, ]
    print(sprintf(string, company, date["start"], date["end"]))
  }
}

[1] "https://finance.yahoo.com/quote/fb/history?period1=1533193200&period2=1541746800&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/fb/history?period1=1541833200&period2=1549004400&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/amzn/history?period1=1533193200&period2=1541746800&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/amzn/history?period1=1541833200&period2=1549004400&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/f/history?period1=1533193200&period2=1541746800&interval=1d&filter=history&frequency=1d"
[1] "https://finance.yahoo.com/quote/f/history?period1=1541833200&period2=1549004400&interval=1d&filter=history&frequency=1d"