【发布时间】:2020-02-14 00:27:43
【问题描述】:
我有一个像这个样本这样的日期框架,我想找到相似的行(不重复)并按明智的方式计算相似度。我找到this solution,但我想保留所有列并将相似性百分比添加为新变量。我的目标是找到相似度最高的记录。我该怎么办?
样本数据集
df <- tibble::tribble(
~date, ~user_id, ~Station_id, ~location_id, ~ind_id, ~start_hour, ~start_minute, ~start_second, ~end_hour, ~end_minute, ~end_second, ~duration_min,
20191015, 19900234, 242, 2, "ac", 7, 25, 0, 7, 30, 59, 6,
20191015, 19900234, 242, 2, "ac", 7, 31, 0, 7, 32, 59, 2,
20191015, 19900234, 242, 2, "ac", 7, 33, 0, 7, 38, 59, 6,
20191015, 19900234, 242, 2, "ac", 7, 39, 0, 7, 40, 59, 2,
20191015, 19900234, 242, 2, "ac", 7, 41, 0, 7, 43, 59, 3,
20191015, 19900234, 242, 2, "ac", 7, 44, 0, 7, 45, 59, 2,
20191015, 19900234, 242, 2, "ac", 7, 47, 0, 7, 59, 59, 13,
20191015, 19900234, 242, 2, "ad", 7, 47, 0, 7, 59, 59, 13,
20191015, 19900234, 242, 2, "ac", 8, 5, 0, 8, 6, 59, 2,
20191015, 19900234, 242, 2, "ad", 8, 5, 0, 8, 6, 59, 2,
20191015, 19900234, 242, 2, "ac", 8, 7, 0, 8, 8, 59, 2,
20191015, 19900234, 242, 2, "ad", 8, 7, 0, 8, 8, 59, 2,
20191015, 19900234, 242, 2, "ac", 16, 26, 0, 16, 55, 59, 30,
20191015, 19900234, 242, 2, "ad", 16, 26, 0, 16, 55, 59, 30,
20191015, 19900234, 242, 2, "ad", 17, 5, 0, 17, 6, 59, 2,
20191015, 19900234, 242, 2, "ac", 17, 5, 0, 17, 23, 59, 19,
20191015, 19900234, 242, 2, "ad", 17, 7, 0, 17, 15, 59, 9,
20191015, 19900234, 242, 2, "ad", 17, 16, 0, 17, 22, 59, 7,
20191015, 19900234, 264, 2, "ac", 17, 24, 0, 17, 35, 59, 12,
20191015, 19900234, 264, 2, "ad", 17, 25, 0, 17, 35, 59, 11,
20191016, 19900234, 242, 1, "ac", 7, 12, 0, 7, 14, 59, 3,
20191016, 19900234, 242, 1, "ad", 7, 13, 0, 7, 13, 59, 1,
20191016, 19900234, 242, 1, "ac", 17, 45, 0, 17, 49, 59, 5,
20191016, 19900234, 242, 1, "ad", 17, 46, 0, 17, 48, 59, 3,
20191016, 19900234, 242, 2, "ad", 7, 14, 0, 8, 0, 59, 47,
20191016, 19900234, 242, 2, "ac", 7, 15, 0, 8, 0, 59, 47
)
行比较函数
row_cf <- function(x, y, df){
sum(df[x,] == df[y,])/ncol(df)
}
函数输出
# 1) Create all possible row combinations
# 2) Rename
# 3) Run through each row
# 4) Calculate similarity
expand.grid(1:nrow(df), 1:nrow(df)) %>%
rename(row_1 = Var1, row_2 = Var2) %>%
rowwise() %>%
mutate(similarity = row_cf(row_1, row_2, df))
# A tibble: 676 x 3
row_1 row_2 similarity
<int> <int> <dbl>
1 1 1 1
2 2 1 0.75
3 3 1 0.833
4 4 1 0.75
5 5 1 0.75
6 6 1 0.75
7 7 1 0.75
8 8 1 0.667
9 9 1 0.583
10 10 1 0.5
【问题讨论】:
-
只是为了澄清一下,示例数据中有 26 行,并且将计算 26x26 = 676 个相似度值(成对比较)-您是否希望最终数据帧具有 676 行,并且对于每个相似度值,从原始数据框中添加可能选择的变量列?你能提供你最终想要的输出最终应该是什么样子吗?
-
@ben 我的目标是找到相似的行,我编辑了这个问题。它可能不需要计算 676 个相似度值。我有兴趣标记类似的行。假设如果排列/排序数据,那么如果第 2 行与第 3 行相似(相似不重复),那么我用相似率或其他方式标记它
-
您只想要最相似的行吗?您仍然需要进行所有比较(实际上是 26 * 25,因为您不需要将一行与自身进行比较)。但是,您是否想添加 2 个新列,给出最相似的行并给出其相似度分数?还是别的什么?
-
@Gregor 是的,我正在寻找具有相似度得分最相似的行
标签: r datatable tidyverse similarity