计算一组值在整个数据集中出现的次数

时间：2019-02-06 07:26:57

标签： r

我需要找到一个组出现在整个数据集中的次数并对其编号。以下是示例数据。可以说如果Group1第一次出现，然后类似地将其沿侧面编号为1，如果再次出现在车道上，则将其沿侧面编号为2，依此类推...对于每个组，请参阅示例数据和预期输出为了更清晰。

样本数据：

  Group
Group1
Group1
Group1
Group1
Group1
Group1
Group2
Group2
Group2
Group2
Group2
Group2
Group2
Group2
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group2
Group2
Group2
Group2
Group2
Group2
Group2
Group2
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group1
Group3
Group3
Group3
Group3
Group3
Group3
Group3
Group3
Group3
Group3

预期输出：

   Group    No of times
Group1  1
Group1  1
Group1  1
Group1  1
Group1  1
Group1  1
Group2  1
Group2  1
Group2  1
Group2  1
Group2  1
Group2  1
Group2  1
Group2  1
Group1  2
Group1  2
Group1  2
Group1  2
Group1  2
Group1  2
Group1  2
Group1  2
Group2  2
Group2  2
Group2  2
Group2  2
Group2  2
Group2  2
Group2  2
Group2  2
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group1  3
Group3  1
Group3  1
Group3  1
Group3  1
Group3  1
Group3  1
Group3  1
Group3  1
Group3  1
Group3  1

3 个答案:

答案 0 :(得分：3)

使用SELECT * FROM test_table tt WHERE exists( SELECT * FROM jsonb_array_elements(tt.settings) AS settings WHERE (settings ->> 'start') > '2019-02-19 00:00:00' ); data.table的一种方法是创建一个rleid列，为new列中的每个更改提供不同的值。然后我们Group group_by，对于Group值的每次更改，我们都使用new递增计数。

cumsum

答案 1 :(得分：1)

与我的相比，Ronak的答案很棒，但这还是我想出的dplyr / tidyr解决方案。这个想法是：

  1. 按原始顺序对行进行编号。
  2. 按组对表进行排序。组群之间的中断将以行号的突然跳跃（4、5、6、15、16、17 ...）为标志。
  3. 为每个群集中的第一个记录分配一个群集编号，然后填写所有NA。
library(dplyr)
library(tidyr)

df_clustered <- 
    df %>% 
    mutate(rownum = row_number()) %>% 
    arrange(Group) %>% 
    mutate(mark = case_when((rownum - lag(rownum)) == 1 ~ NA, TRUE ~ TRUE)) %>% 
    group_by(Group, mark) %>% 
    mutate(cluster_number = ifelse(mark == TRUE, row_number(), NA)) %>% 
    ungroup() %>% 
    fill(cluster_number) %>% 
    arrange(rownum) %>% 
    select(-rownum, -mark)

head(df_clustered, 20)

#> # A tibble: 20 x 3
#>    Group  Value cluster_number
#>    <chr>  <dbl>          <int>
#>  1 Group1   1                1
#>  2 Group1   2                1
#>  3 Group1   1                1
#>  4 Group1   1.3              1
#>  5 Group1   1.2              1
#>  6 Group1   1                1
#>  7 Group2   7                1
#>  8 Group2   6                1
#>  9 Group2   2                1
#> 10 Group2   1                1
#> 11 Group2  25                1
#> 12 Group2  23                1
#> 13 Group2  24                1
#> 14 Group2  25                1
#> 15 Group1  24                2
#> 16 Group1  23                2
#> 17 Group1  26                2
#> 18 Group1  23                2
#> 19 Group1  17                2
#> 20 Group1  11                2

答案 2 :(得分：1)

这是一个纯数据表解决方案。它基于rle()和rep()：

library(data.table)

DT <- data.table(stringsAsFactors=FALSE,
                 Group = c("Group1", "Group1", "Group1", "Group1", "Group1", "Group1",
                           "Group2", "Group2", "Group2", "Group2", "Group2", "Group2",
                           "Group2", "Group2", "Group1", "Group1", "Group1", "Group1",
                           "Group1", "Group1", "Group1", "Group1", "Group2", "Group2", 
                           "Group2", "Group2", "Group2", "Group2", "Group2", "Group2", 
                           "Group1", "Group1", "Group1", "Group1", "Group1", "Group1"),
                 Value = c(1, 2, 1, 1.3, 1.2, 1, 7, 6, 2, 1, 25, 23, 24, 25, 24, 23, 26, 23,
                           17, 11, 2, 1, 1, 2, 2.3, 1, 3, 4, 1, 1, 2, 25, 26, 11, 17, 16)
)

lengthEncoding <- rle(DT$Group)
setDT(lengthEncoding)[, group_count := seq_len(.N), by="values"]
DT[, "No of times" := rep(lengthEncoding$group_count, lengthEncoding$lengths)]

print(DT)

顺便说一句，该解决方案比公认的答案要快：

编辑：：添加了@ chinsoon12美丽的单面纸，该当之无愧！

library(microbenchmark)
library(data.table)
library(dplyr)

df <- data.frame(stringsAsFactors=FALSE,
                 Group = c("Group1", "Group1", "Group1", "Group1", "Group1", "Group1",
                           "Group2", "Group2", "Group2", "Group2", "Group2", "Group2",
                           "Group2", "Group2", "Group1", "Group1", "Group1", "Group1",
                           "Group1", "Group1", "Group1", "Group1", "Group2", "Group2", 
                           "Group2", "Group2", "Group2", "Group2", "Group2", "Group2", 
                           "Group1", "Group1", "Group1", "Group1", "Group1", "Group1"),
                 Value = c(1, 2, 1, 1.3, 1.2, 1, 7, 6, 2, 1, 25, 23, 24, 25, 24, 23, 26, 23,
                           17, 11, 2, 1, 1, 2, 2.3, 1, 3, 4, 1, 1, 2, 25, 26, 11, 17, 16)
)

DT <- data.table(df)

results <- microbenchmark(
  RonakShah = {
    df %>% 
      mutate(new = rleid(Group)) %>%
      group_by(Group) %>%
      mutate(no_of_times = cumsum(c(1,diff(new) != 0))) %>%
      select(-new)
  },
  ismirsehregal = {
    lengthEncoding <- rle(DT$Group)
    setDT(lengthEncoding)[, group_count := seq_len(.N), by="values"]
    DT[, "No of times" := rep(lengthEncoding$group_count, lengthEncoding$lengths)]
  },
  chinsoon12 = {DT[, numtimes := 1L + c(0L, cumsum(diff(.I) > 1L)), by=.(Group)]}
)

print(results)
plot(results)

          expr      min       lq     mean   median       uq      max neval cld
     RonakShah 3.980914 4.253103 4.898788 4.500009 5.063746 8.021481   100   c
 ismirsehregal 1.494078 1.653283 1.937947 1.828487 2.023246 5.678442   100  b 
    chinsoon12 1.050436 1.239666 1.469426 1.440154 1.646369 2.572168   100 a
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