根据 R 中的两个匹配条件将值从一个数据帧添加到另一个数据帧答案

【问题标题】：Adding values from one dataframe to another based on two matching conditions in R根据 R 中的两个匹配条件将值从一个数据帧添加到另一个数据帧
【发布时间】：2020-01-31 20:01:39
【问题描述】：

我在下面有两个数据框：

输入输出df1：

structure(list(Location = c("1100 2ND AVENUE", "1100 2ND AVENUE", 
"1100 2ND AVENUE", "1100 2ND AVENUE", "1100 2ND AVENUE", "1100 2ND AVENUE"
), `Ivend Name` = c("3 Mskt 1.92oz", "Almond Joy 1.61oz", "Aquafina 20oz", 
"BCanyonChptleAdzuk1.5oz", "BlkForest FrtSnk 2.25oz", "BluDimndSmkhseAlmd1.5oz"
), `Category Name` = c("Candy", "Candy", "Water", "Salty Snacks", 
"Candy", "Nuts/Trailmix"), Calories = c(240, 220, 0, 215, 193, 
260), Sugars = c("36", "20", "0", "2", "32", "2"), Month = structure(c(4L, 
4L, 4L, 4L, 4L, 4L), .Label = c("Oct", "Nov", "Dec", "Jan", "Feb", 
"Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep"), class = "factor"), 
    Products_available_per_machine = c(0, 0, 0, 0, 0, 0), Units_sold = c(0, 
    0, 0, 0, 0, 0), Total_Sales = c(0, 0, 0, 0, 0, 0), Spoils = c(0, 
    0, 0, 0, 0, 0), Building = c("1100 2ND", "1100 2ND", "1100 2ND", 
    "1100 2ND", "1100 2ND", "1100 2ND"), Item = structure(c(2L, 
    2L, 1L, 2L, 2L, 2L), .Label = c("Beverage", "Food"), class = "factor"), 
    Year = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "2019", class = "factor")), row.names = c(NA, 
-6L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x00000233561b1ef0>)

输入输出df2：

structure(list(`Date Ran` = structure(c(1548892800, 1551312000, 
1553817600, 1556582400, 1561680000, 1564531200), class = c("POSIXct", 
"POSIXt"), tzone = "UTC"), Year = c(2019, 2019, 2019, 2019, 2019, 
2019), Month = c("January", "February", "March", "April", "June", 
"July"), Location = c("SEA18", "SEA18", "SEA18", "SEA18", "SEA18", 
"SEA18"), Building = c("Alexandria", "Alexandria", "Alexandria", 
"Alexandria", "Alexandria", "Alexandria"), Population = c(1177, 
1179, 1178, 1156, 1163, 1163)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"))

我想从 DF 2 中提取 pop col 并将其添加到基于“Building”和“Month”的 Dataframe 1，以便填充 DF2 中的人口。

我使用合并尝试了此命令，但执行时 col 为 NULL：

df_2019_final1$Population <- df_2019_pop$Population[match(df_2019_final1$Month, df_2019_pop$Month, df_2019_final1$Building, df_2019_pop$Building)]

subset_df_pop <- df_2019_pop[, c("Month", "Building", "Population")]


updated_2019_test <- merge(df_2019_final1, subset_df_pop, by = c('Month', 'Building'))

两者都产生 NULLS 和空白 DF

任何帮助将不胜感激。

【问题讨论】：

match 只需要 2 个向量 match(x, table, nomatch = NA_integer_, incomparables = NULL)
你可以试试merge(df1, df2, by = c('Month', 'Building'))
我刚尝试合并，它输出了一个空白的df。出于某种原因，它把所有的列都带了过来，而不是 Month/Building
您需要通过仅包含这些列 + Populatoin 即 merge(df1, df2[, c("Month", "Building", "Population")], by = c('Month', 'Building')) 来细分第二个数据
我用你的解决方案编辑了我的问题，但之后仍然得到一个空白的 df。

标签： r dataframe merge

【解决方案1】：

在其中一个数据集中，“月份”是缩写，第二个是全名。我们可以调整为其中一种格式，merge 可以工作

df2$MonthN <- month.abb[match(df2$Month, month.name)]
library(dplyr)
left_join(df1, df2[, c("MonthN", "Building", "Population")], 
             by = c('Month' = 'MonthN', 'Building'))

或merge

merge(df1, df2[, c("MonthN", "Building", "Population")], 
   by.x = c('Month', 'Building'), by.y = c('MonthN', 'Building'), all.x = TRUE)

注意：根据示例，合并数据集上的“人口”列将是 NA，因为子数据集中的“建筑”值不同

【讨论】：

非常感谢您的帮助 - 匹配的建筑物会填满人口吗？
@Dinho 在这种情况下，您只需 merge 和 'Month' i.e. remove Buliding' 来自 by
不幸的是，我仍然没有得到我需要的输出。我试图从主 df 具有相同的布局，但只是将弹出数据添加到各个区域。我想我需要删除不匹配的建筑物才能完成这项工作
@Dinho 这是一个不同的问题，但它只能根据那些by 变量的匹配值进行匹配
我让它工作了——我对两个 DF 的“建筑”在字体方面是不同的（一个是全部大写，另一个不是）。使用 Toupper 并且能够为匹配的 BUilding ID 填充 Pops。谢谢你帮助我的男人！