7

我正在使用两个不同的数据集，我想根据阈值合并它们。假设两个数据框如下所示：

library(dplyr)
library(fuzzyjoin)
library(lubridate)

df1 = data_frame(Item=1:5, 
                 DateTime=c("2015-01-01 11:12:14", "2015-01-02 09:15:23", 
                            "2015-01-02 15:46:11", "2015-04-19 22:11:33", 
                            "2015-06-10 07:00:00"), 
                 Count=c(1, 6, 11, 15, 9), 
                 Name="Sterling", 
                 Friend=c("Pam", "Cyril", "Cheryl", "Mallory", "Lana"))
df1$DateTime = ymd_hms(df1$DateTime)

df2 = data_frame(Item=21:25, 
                 DateTime=c("2015-01-01 11:12:15", "2015-01-02 19:15:23", 
                            "2015-01-02 15:46:11", "2015-05-19 22:11:33", 
                            "2015-06-10 07:00:02"), 
                 Count=c(3, 7, 11, 15, 8), 
                 Name="Sterling", 
                 Friend=c("Pam", "Kreger", "Woodhouse", "Gillete", "Lana"))
df2$DateTime = ymd_hms(df2$DateTime)

我现在想要的是能够基于模糊匹配并在它们各自值的两秒内离开加入，而除此之外的所有其他值df2都是相同的。我想我可以通过以下方式到达那里：df1DateTimeCountItem

df1 %>%
  difference_left_join(df2, by=c("DateTime", "Count"), max_dist=2)

但这给了我以下输出：

 # A tibble: 8 × 10
  Item.x          DateTime.x Count.x   Name.x Friend.x Item.y          DateTime.y Count.y   Name.y  Friend.y
   <int>              <dttm>   <dbl>    <chr>    <chr>  <int>              <dttm>   <dbl>    <chr>     <chr>
1      1 2015-01-01 11:12:14       1 Sterling      Pam     21 2015-01-01 11:12:15       3 Sterling       Pam
2      1 2015-01-01 11:12:14       1 Sterling      Pam     21 2015-01-01 11:12:15       3 Sterling       Pam
3      2 2015-01-02 09:15:23       6 Sterling    Cyril     NA                <NA>      NA     <NA>      <NA>
4      3 2015-01-02 15:46:11      11 Sterling   Cheryl     23 2015-01-02 15:46:11      11 Sterling Woodhouse
5      3 2015-01-02 15:46:11      11 Sterling   Cheryl     23 2015-01-02 15:46:11      11 Sterling Woodhouse
6      4 2015-04-19 22:11:33      15 Sterling  Mallory     NA                <NA>      NA     <NA>      <NA>
7      5 2015-06-10 07:00:00       9 Sterling     Lana     25 2015-06-10 07:00:02       8 Sterling      Lana
8      5 2015-06-10 07:00:00       9 Sterling     Lana     25 2015-06-10 07:00:02       8 Sterling      Lana

这很接近，除了第 3 行不应该合并，因为名称不同（我希望第 2 行在给定阈值的情况下合并，即使我不希望它这样做）。

我如何最终得到以下数据框？请注意，尽管满足阈值限制，df2但未合并第二行和第三行。这是因为其他列（除了）不相同。DateTimeCountItem

desired_output
#   Item            DateTime Count     Name  Friend
# 1    3 2015-01-02 15:46:11    11 Sterling  Cheryl
# 2    2 2015-01-02 09:15:23     6 Sterling   Cyril
# 3    5 2015-06-10 07:00:00     9 Sterling    Lana
# 4   25 2015-06-10 07:00:02     8 Sterling    Lana
# 5    4 2015-04-19 22:11:33    15 Sterling Mallory
# 6    1 2015-01-01 11:12:14     1 Sterling     Pam
# 7   21 2015-01-01 11:12:15     3 Sterling     Pam
4

1 回答 1

6

好的，所以，您收到的消息是因为无法在非数字列上计算模糊匹配。

要做的就是将其转换为数字。由于您的卡尺以秒为单位，我将其转换为秒，然后将它们设为数字：

library(dplyr)
library(fuzzyjoin)
library(lubridate)

df1 = data_frame(Item=1:5, 
                 DateTime=c("2015-01-01 11:12:14", "2015-01-02 09:15:23", 
                            "2015-01-02 15:46:11", "2015-04-19 22:11:33", 
                            "2015-06-10 07:00:00"), 
                 Count=c(1, 6, 11, 15, 9), 
                 Name="Sterling", 
                 Friend=c("Pam", "Cyril", "Cheryl", "Mallory", "Lana"))
df1$DateTime1 = as.numeric(seconds(ymd_hms(df1$DateTime)))

df2 = data_frame(Item=21:25, 
                 DateTime=c("2015-01-01 11:12:15", "2015-01-02 19:25:56", 
                            "2015-01-02 15:46:11", "2015-05-19 22:11:33", 
                            "2015-06-10 07:00:02"), 
                 Count=c(3, 6, 11, 15, 8), 
                 Name="Sterling", 
                 Friend=c("Pam", "Kreger", "Woodhouse", "Gillete", "Lana"))
df2$DateTime1 = as.numeric(seconds(ymd_hms(df2$DateTime)))

df1 %>%
  difference_left_join(y=df2, by=c("DateTime1", "Count"), max_dist=2)

根据我们在评论中的讨论，一个简单的调整将其子集到其他字符列匹配的情况将是：

df1[df2$Friend == df1$Friend,] %>%
  difference_left_join(y=df2[df2$Friend == df1$Friend,], by=c("DateTime1", "Count"), max_dist=2)

该示例仅用于Friend但当然您可以使用&多列来执行此操作。

于 2016-09-22T17:12:02.217 回答