具有多组/多对值的变量的整洁数据答案

【问题标题】：Tidy data that has variables with many sets/pairs of values具有多组/多对值的变量的整洁数据
【发布时间】：2019-07-17 07:40:49
【问题描述】：

如何使用任何 tidyverse 函数整理这些数据，以使市场、部门、子部门及其相应数据（即单元格内每个 = 符号的 RHS，例如 0.2934）更有用？有没有一种形式和方法可以将这些信息放在单独的行或列中？

这是我的玩具数据：

df <- tibble::tribble(
        ~var1, ~year, ~Markets, ~Sectors,
         "AA",  2015, "A=0.2934;B=0.1483;C=0.5583", "Technology=0.0566;Health Care=0.1396;Financial=0.0925;Consumer Staples=0.0642;C=0.4252;Basic Materials=0.0358",
         "BB",  2015, "D=0.8548;E=0.0869;A=0.0529", "Technology=0.1924;Financial=0.3262;Communications=0.0844;Consumer Discretionary=0.1181;Utilities=0.0484",
         "CC",  2015, "A=0.4159;C=0.3615;B=0.1522;D=0.0665;F=0.0018;E=0.0022", "Technology=0.0733;Consumer Discretionary=0.0788;Financial=0.1401;Industrials=0.0691;Energy=0.0377;C=0.3598",
         "BB",  2019, "C=22.2;G=16.4;H=9.9;I=9.3;J=6.6", "C=23.3;Financials=21.8;Consumer Staples=11.3;Industrials=10.8;Consumer Discretionary=10.1;Information Technology=8.6",
         "CC",  2019, "C=23.9;K=12.7;L=12.2;M=11.2;N=9.6;O=7.8", "C=33.4;Financials=25.6;Consumer Discretionary=6.8;Information Technology=6.7;Energy=5.8;Consumer Staples=5.6",
         "DD",  2019, "N=82.4;C=13.9;P=1.1;Q=1.0;R=0.5;S=0.3;T=0.3;U=0.1", "Information Technology=19.9;Financials=14.8;C=13.7;Health Care=11.8;Consumer Discretionary=11.7;Industrials=9.1")

我的真实数据有更多这样的变量，每个变量在每个单元格中都包含更多的值。

【问题讨论】：

标签： r dplyr tidyr purrr

【解决方案1】：

您可以执行以下操作。

首先，用;分隔值。

Markets <- read.csv2(text = df$Markets, header = FALSE, stringsAsFactors = FALSE)
Sectors <- read.csv2(text = df$Sectors, header = FALSE, stringsAsFactors = FALSE)

现在得到等号后面的内容。

tmp <- lapply(Markets, function(x) strsplit(x, "="))
tmp <- lapply(tmp, function(lst) 
  sapply(lst, function(x) if(length(x) > 1) x[[2]] else NA))
tmp <- lapply(tmp, as.numeric)
Markets <- do.call(rbind, tmp)

tmp <- lapply(Sectors, function(x) strsplit(x, "="))
tmp <- lapply(tmp, function(lst) 
  sapply(lst, function(x) if(length(x) > 1) x[[2]] else NA))
tmp <- lapply(tmp, as.numeric)
Sectors <- do.call(rbind, tmp)

不需要临时变量tmp。

rm(tmp)

并使上面的结果更漂亮。

Markets <- as.data.frame(Markets)
Sectors <- as.data.frame(Sectors)

names(Markets) <- paste("Market", seq_along(Markets), sep = ".")
names(Sectors) <- paste("Sector", seq_along(Sectors), sep = ".")

Markets
Sectors

【讨论】：

【解决方案2】：

对于Markets 和Sectors 列，我们可以使用separate_rows 将行扩展;。之后，我们可以使用separate 将列拆分为=。结果是以下长格式数据帧。

library(tidyverse)

df2 <- df %>%
  separate_rows(Markets, sep = ";") %>%
  separate_rows(Sectors, sep = ";") %>%
  separate(Markets, into = c("Markets", "Markets_Number"), sep = "=", convert = TRUE) %>%
  separate(Sectors, into = c("Sectors", "Sectors_Number"), sep = "=", convert = TRUE)
df2
# # A tibble: 183 x 6
#    var1   year Markets Markets_Number Sectors          Sectors_Number
#    <chr> <dbl> <chr>            <dbl> <chr>                     <dbl>
#  1 AA     2015 A                0.293 Technology               0.0566
#  2 AA     2015 A                0.293 Health Care              0.140 
#  3 AA     2015 A                0.293 Financial                0.0925
#  4 AA     2015 A                0.293 Consumer Staples         0.0642
#  5 AA     2015 A                0.293 C                        0.425 
#  6 AA     2015 A                0.293 Basic Materials          0.0358
#  7 AA     2015 B                0.148 Technology               0.0566
#  8 AA     2015 B                0.148 Health Care              0.140 
#  9 AA     2015 B                0.148 Financial                0.0925
# 10 AA     2015 B                0.148 Consumer Staples         0.0642
# # ... with 173 more rows

【讨论】：