如何合并基于两列的多个数据帧？

Question

我有超过4天收集的数据的多个数据帧。每个数据框看起来像这样(非常简单)：

Lat           Long       PM
-33.9174    151.2263     8
-33.9175    151.2264     10 
-33.9176    151.2265     9
-33.9177    151.2266     8

我想根据匹配的长度值和纬度值合并多个数据帧，以求取特定位置的所有'PM‘值的平均值。最终结果将如下所示(2月13日至16日)：

Lat         Long    PM.13th Feb  PM.14th Feb  PM.15th Feb   **Mean**
-33.9174   151.2263     8            9           11         9.33
-33.9175   151.2264     10           11          12          11
-33.9176   151.2265     9            14          13          12
-33.9177   151.2266     8            10          11         9.66

我知道合并2个数据帧很容易：

df = merge(data1, data2, by.x = c("Lat", "Long"), by.y = c("Lat", "Long"))

但是如何根据匹配的经度和纬度值合并多个数据帧呢？

另外，有没有一种方法可以过滤数据，使其匹配相互之间在0.001经度/经度范围内的数据？(目前我正在将经度/经度数据四舍五入到小数点后3位，但它正在复制我的数据)。

Answer 1

为了匹配，也许是来自dplyr的inner_join？

library(dplyr)
df1 <- data.frame(
  lat = c(-33.9174, -33.9175, -33.9176, -33.9177, -33.9171), 
  long = c(151.2263, 151.2264, 151.2265, 151.2266, -140.54),
  PM = c(8, 10, 9, 8, 55)
)

df2 <- data.frame(
  lat = c(-33.9174, -33.9175, -33.9176, -33.9177, -31), 
  long = c(151.2263, 151.2264, 151.2265, 151.2266, 134),
  PM = c(12, 15, 11, 3, 18)
)

library(dplyr)

inner_join(df1, df2, by = c("lat", "long"))

       lat     long PM.x PM.y
1 -33.9174 151.2263    8   12
2 -33.9175 151.2264   10   15
3 -33.9176 151.2265    9   11
4 -33.9177 151.2266    8    3

Answer 2

这可能是一个答案，尽管它有点冗长，对于大量的数据帧来说不是很好：

library(tidyverse)
feb_13 <- data_frame(lat = c(-33.9174,-33.9175,-33.9176,-33.9177), 
                 long = c(151.2263, 151.2264,151.2265,151.2266),
                 pm = c(8,10,9,8))

feb_14 <- data_frame(lat = c(-33.9174,-33.9175,-33.9176,-33.9177), 
                 long = c(151.2263, 151.2264,151.2265,151.2266),
                 pm = c(7,3,4,5))

feb_15 <- data_frame(lat = c(-33.9174,-33.9175,-33.9176,-33.9177), 
                 long = c(151.2263, 151.2264,151.2265,151.2266),
                 pm = c(1,4,10,12))

这是第一个技巧。简单，但取平均值在这里是丑陋的.

df <- left_join(feb_13, feb_14, by = c("lat", "long")) %>%
        left_join(feb_15, by = c("lat", "long")) %>%
        rename(
         pm_feb13 = pm.x,
         pm_feb14 = pm.y,
         pm_feb15 = pm
        ) %>%
        mutate(
         mean = c((pm_feb13[1] + pm_feb14[1] + pm_feb15[1])/3,
                  (pm_feb13[2] + pm_feb14[2] + pm_feb15[2])/3,
                  (pm_feb13[3] + pm_feb14[3] + pm_feb15[3])/3,
                  (pm_feb13[4] + pm_feb14[4] + pm_feb15[4])/3)
        )

下面是第二种选择，它有很多管道，但使用了summarize

df_2 <- left_join(feb_13, feb_14, by = c("lat", "long")) %>%
          left_join(feb_15, by = c("lat", "long")) %>%
          group_by(lat, long) %>%
          summarise(
            mean = mean(c(pm.x, pm.y, pm), na.rm=T)
          ) %>%
          full_join(feb_13, by = c("lat", "long")) %>%
          full_join(feb_14, by = c("lat", "long")) %>%
          full_join(feb_15, by = c("lat", "long")) %>%
          rename(
            pm_feb13 = pm.x,
            pm_feb14 = pm.y,
            pm_feb15 = pm
          ) %>%
          arrange(long)