如何根据一个数据帧中的列值和R中另一个数据帧的列标题名称有条件地创建新列

Posted 2021-05-04

我有一个数据框df1，它有一个名为averageDate的列，其中包含日期，格式为％Y-％m。

我有另一个数据框df2，其中大多数列名称是日期值，格式为％Y-％m，这些列中的数据是经济指标的数值。

我想在df1中填充一个新列（在df3中显示），其中值是在df2中找到的值，其中df`中的averageDate值与df2中的列名匹配。

我已经成功解决了基于单独数据帧中两列的条件合并以解决先前的问题，但我在这里坚持的是第二个匹配条件是df2中的列名。

我的数据帧的再现如下所示：

df1 <- structure(list(zipcode = structure(c(1L, 2L, 4L, 3L), .Label = c("10019", 
"10027", "20009", "94117"), class = "factor"), averageDate = c("2017-08", 
"2017-04", NA, "2015-11")), .Names = c("zipcode", "averageDate"
), row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"
))

df2 <-structure(list(RegionName = c(20009, 10019, 10027), `2015-01` = c(444500, 
1855000, NA), `2015-02` = c(439000, 1715000, NA), `2015-03` = c(437000, 
1775000, NA), `2015-04` = c(475000, 1855000, NA), `2015-05` = c(489000, 
1860000, NA), `2015-06` = c(489750, 1877500, NA), `2015-07` = c(479900, 
1957500, NA), `2015-08` = c(489900, 1950000, NA), `2015-09` = c(5e+05, 
1947500, NA), `2015-10` = c(512450, 1958000, NA), `2015-11` = c(503999.5, 
1990000, NA), `2015-12` = c(499900, 1995000, NA), `2016-01` = c(499500, 
1995000, NA), `2016-02` = c(529900, 1822500, NA), `2016-03` = c(5e+05, 
1820000, 872000), `2016-04` = c(5e+05, 1930000, 887000), `2016-05` = c(492500, 
1795500, 837000), `2016-06` = c(529000, 1750000, 819000), `2016-07` = c(549000, 
1832500, 725800), `2016-08` = c(577000, 1850000, 725000), `2016-09` = c(549900, 
1762500, 753500), `2016-10` = c(529000, 1777500, 737900), `2016-11` = c(519000, 
1787000, 750000), `2016-12` = c(499000, 1795000, 725800), `2017-01` = c(549000, 
1795000, 749000), `2017-02` = c(522450, 1833000, 845000), `2017-03` = c(546950, 
1836500, 867250), `2017-04` = c(572247.5, 1849450, 929000), `2017-05` = c(549900, 
1850000, 929000), `2017-06` = c(540000, 1875000, 899000), `2017-07` = c(519900, 
1895000, 899000), `2017-08` = c(525000, 1849990, 897000), `2017-09` = c(572450, 
1795000, 840000), `2017-10` = c(595000, 1795000, 882000), `2017-11` = c(555650, 
1825000, 949000), `2017-12` = c(525000, 1799950, 795000), `2018-01` = c(557000, 
1925000, 772500)), .Names = c("RegionName", "2015-01", "2015-02", 
"2015-03", "2015-04", "2015-05", "2015-06", "2015-07", "2015-08", 
"2015-09", "2015-10", "2015-11", "2015-12", "2016-01", "2016-02", 
"2016-03", "2016-04", "2016-05", "2016-06", "2016-07", "2016-08", 
"2016-09", "2016-10", "2016-11", "2016-12", "2017-01", "2017-02", 
"2017-03", "2017-04", "2017-05", "2017-06", "2017-07", "2017-08", 
"2017-09", "2017-10", "2017-11", "2017-12", "2018-01"), row.names = c(38L, 
82L, 226L), class = "data.frame")

df3 <- structure(list(RegionName = c("10019", "10027", "20009", "94117"
    ), variable = structure(c(32L, 28L, 11L, NA), .Label = c("2015-01", 
    "2015-02", "2015-03", "2015-04", "2015-05", "2015-06", "2015-07", 
    "2015-08", "2015-09", "2015-10", "2015-11", "2015-12", "2016-01", 
    "2016-02", "2016-03", "2016-04", "2016-05", "2016-06", "2016-07", 
    "2016-08", "2016-09", "2016-10", "2016-11", "2016-12", "2017-01", 
    "2017-02", "2017-03", "2017-04", "2017-05", "2017-06", "2017-07", 
    "2017-08", "2017-09", "2017-10", "2017-11", "2017-12", "2018-01"
    ), class = "factor"), value = c(1849990, 929000, 503999.5, NA
    )), .Names = c("RegionName", "variable", "value"), row.names = c(NA, 
    -4L), class = "data.frame")

答案

像这样的东西？

require(tidyverse);
left_join(
    df1,
    df2 %>%
        gather(averageDate, Value, 2:ncol(df2)) %>%
        rename(zipcode = RegionName) %>%
        mutate(zipcode = as.character(zipcode)))
## A tibble: 4 x 3
#  zipcode averageDate   Value
#  <chr>   <chr>         <dbl>
#1 10019   2017-08     1849990
#2 10027   2017-04      929000
#3 94117   NA               NA
#4 20009   2015-11      504000

另一答案

从@Marko的见解

long <-melt(df2, id.vars = "RegionName")

df3 <- merge(long, df1, by.x=c("RegionName", "variable"), by.y=c("zipcode", "averageDate"), all.y=T)