由 \n 分隔的字符串到数据框

Question

我有以下字符串：

  "Title\nToday 1,239 €\nYesterday 1,2 €\n17/04/2018 1,2 €\n14/04/2018 1,2 €\n13/04/2018 1,2 €\n12/04/2018 1,2 €\n11/04/2018 1,2 €\n09/04/2018 1,2 €\n08/04/2018 1,2 €\n07/04/2018 1,2 €"

但我不知道是否可以从中获取数据框。我想用我的字符串获取一个包含两列（日期和价格）的数据框，如下所示（实际上不需要Title 名称）：

Date       Price
Today      1,239 €
Yesteday   1,2 €
17/04/2018 1,2 €
14/04/2018 1,2 €
13/04/2018 1,2 €
12/04/2018 1,2 €
11/04/2018 1,2 €
09/04/2018 1,2 €
08/04/2018 1,2 €
07/04/2018 1,2 €

这与我使用cat 函数可以得到的几乎相同。但我想我可以将它转换为数据框。 有什么想法吗？

Answer 1

这里有read.table的解决方案：

> read.table(text=str, sep=' ', skip=1, col.names=c('Date', 'Price', 'Currency'))
         Date Price Currency
1       Today 1,239        €
2   Yesterday   1,2        €
3  17/04/2018   1,2        €
4  14/04/2018   1,2        €
5  13/04/2018   1,2        €
6  12/04/2018   1,2        €
7  11/04/2018   1,2        €
8  09/04/2018   1,2        €
9  08/04/2018   1,2        €
10 07/04/2018   1,2        €

str 是您的数据。请注意参数skip 正在删除“标题”。

Answer 2

我建议做这样的事情来将你的字符串s 转换为data.frame。这个想法是将日期、值和单位分开，以便更轻松地处理数据，因为您将单位和数字条目分开。

df <- do.call(rbind.data.frame, strsplit(
    unlist(strsplit(sub("Title\n", "", s), "\n")),
    " "))
colnames(df) <- c("Date", "Value", "Unit");
df$Value <- as.numeric(as.character(sub(",", ".", df$Value)));
#         Date Value Unit
#1       Today 1.239    €
#2   Yesterday 1.200    €
#3  17/04/2018 1.200    €
#4  14/04/2018 1.200    €
#5  13/04/2018 1.200    €
#6  12/04/2018 1.200    €
#7  11/04/2018 1.200    €
#8  09/04/2018 1.200    €
#9  08/04/2018 1.200    €
#10 07/04/2018 1.200    €

解释：我们首先在"\n" 上拆分s，然后在空格上分离出Date、Value 和Unit。由于您的值包含逗号小数分隔符“，”，因此我们将“，”替换为“。”并转换为numeric。

---

您可以通过以下方式避免sub("Title\n", "", s)（感谢@PoGibas），使其更加紧凑：

df <- do.call(rbind.data.frame, strsplit(unlist(strsplit(s, "\n"))[-1], " "))
colnames(df) <- c("Date", "Value", "Unit");
df$Value <- as.numeric(as.character(sub(",", ".", df$Value)));

输出同上。

---

样本数据

s <-   "Title\nToday 1,239 €\nYesterday 1,2 €\n17/04/2018 1,2 €\n14/04/2018 1,2 €\n13/04/2018 1,2 €\n12/04/2018 1,2 €\n11/04/2018 1,2 €\n09/04/2018 1,2 €\n08/04/2018 1,2 €\n07/04/2018 1,2 €"

Answer 3

这是strsplit 和dplyr::separate 的解决方案。

prices <- "Title\nToday 1,239 €\nYesterday 1,2 €\n17/04/2018 1,2 €\n14/04/2018 1,2 €\n13/04/2018 1,2 €\n12/04/2018 1,2 €\n11/04/2018 1,2 €\n09/04/2018 1,2 €\n08/04/2018 1,2 €\n07/04/2018 1,2 €"

prices <- data.frame(x = strsplit(prices, "\n", "", fixed = TRUE)[[1]])
prices <- prices %>% separate(x, " ", into = c("Date", "Prices"), extra = "merge") 
prices <- prices[-1,]
prices
#          Date  Prices
# 2       Today 1,239 €
# 3   Yesterday   1,2 €
# 4  17/04/2018   1,2 €
# 5  14/04/2018   1,2 €
# 6  13/04/2018   1,2 €
# 7  12/04/2018   1,2 €
# 8  11/04/2018   1,2 €
# 9  09/04/2018   1,2 €
# 10 08/04/2018   1,2 €
# 11 07/04/2018   1,2 €

Answer 4

我已经实现了几次strsplit，然后我构建了一个matrix，它被转换为一个数据框（通过取矩阵的第 1 列和第 2 列来删除 € 符号）：

# Making a short object containing your string
x <- "Title\nToday 1,239 €\nYesterday 1,2 €\n17/04/2018 1,2 €\n14/04/2018 1,2 €\n13/04/2018 1,2 €\n12/04/2018 1,2 €\n11/04/2018 1,2 €\n09/04/2018 1,2 €\n08/04/2018 1,2 €\n07/04/2018 1,2 €"

# Two string splits (first splitting by "\n" and then by " "), and discarding the "title" (by taking [[1]][2:11])
x <- unlist(strsplit(strsplit(x, split = "\n")[[1]][2:11], split = " "))

# Putting it in a data frame (dropping the € symbol)
df1 <- data.frame(matrix(x, ncol = 3, byrow = T)[,1:2])

结果：

> df1
           X1    X2
1       Today 1,239
2   Yesterday   1,2
3  17/04/2018   1,2
4  14/04/2018   1,2
5  13/04/2018   1,2
6  12/04/2018   1,2
7  11/04/2018   1,2
8  09/04/2018   1,2
9  08/04/2018   1,2
10 07/04/2018   1,2

---

我还想将“，”添加到“。”并将值作为数字

x <- unlist(strsplit(strsplit(x, split = "\n")[[1]][2:11], split = " "))
x <- gsub(",", ".", x)
df1 <- data.frame(matrix(x, ncol = 3, byrow = T)[,1:2])
df1[,2] <- as.numeric(levels(df1[,2]))[df1[,2]]