整洁的数据,其变量具有多组/成对的值

时间:2019-02-23 09:29:03

标签: r dplyr tidyr purrr

我如何使用任何dydyverse函数整理数据,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?是否存在将信息放在单独的行或列中的形式和方式?

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包含更多的值。

2 个答案:

答案 0 :(得分:3)

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)

现在得到等号后的值。

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

答案 1 :(得分:2)

我们可以使用separate_rows;Markets列的行扩展Sectors。之后,我们可以使用separate将列除以=。结果是下面的长格式数据帧。

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