R 检查超过特定值的表格，并按相应id和标签计数超过相应阈值的次数

我有一个数据帧df

df <- data.frame(id =c(1,2,1,4,1,5,6),
                    label=c("a","b", "a", "a","a", "e", "a"), 
                    color = c("g","a","g","g","a","a","a"),
                    threshold = c(12, 10, 12, 12, 12, 35, 40),
                    value =c(32.1,0,15.0,10,1,50,45),stringsAsFactors = F
            )

df阈值，1,0））%>%
分组依据（id，标签）%>%
总结（超过=总和（检查））

但是我没有得到相应的id，而是得到了超过的总数

---

只需对代码稍作修改：

df %>% 
 group_by(id, label) %>% 
 mutate(check = if_else(value > threshold, 1, 0)) %>% 
 summarise(exceed = sum(check)) %>% 
 group_by(id, label)

     id label exceed
  <dbl> <chr>  <dbl>
1     1 a          2
2     2 b          0
3     4 a          0
4     5 e          1
5     6 a          1

df%>%
分组依据（id，标签）%>%
变异（检查=if_else（值>阈值，1,0））%>%
汇总（超出=总和（检查））%>%
分组人（id、标签）
id标签超过
1 a 2
2B0
3 4 a 0
4 5 e 1
5 6 a 1

要更紧密地匹配预期输出，请执行以下操作：

df %>% 
 group_by(id, label) %>% 
 mutate(exceed = sum(if_else(value > threshold, 1, 0))) %>% 
 group_by(id, label, color) %>% 
 filter(row_number() == 1)


     id label color threshold value exceed
  <dbl> <chr> <chr>     <dbl> <dbl>  <dbl>
1     1 a     g            12  32.1      2
2     2 b     a            10   0        0
3     4 a     g            12  10        0
4     1 a     a            12   1        2
5     5 e     a            35  50        1
6     6 a     a            40  45        1

df%>%
分组依据（id，标签）%>%
变异（超过=总和（如果值>阈值，1,0））%>%
分组依据（id、标签、颜色）%>%
过滤器（行数（）==1）
id标签颜色阈值超过
1 a g 12 32.1 2
2 b a 10 0 0
3 4 a g 12 10 0
41A1212
5 e 35 50 1
6 a 40 45 1

代码中的一点更改

final_df <- df %>% mutate(check = if_else(value > threshold, 1, 0)) %>% group_by(id, label) %>% filter(check==1)
unique(final_df$id)

final_df%mutate（check=if_else（值>阈值，1,0））%%>%groupby（id，label）%%>%filter（check==1）
唯一（最终的_df$id）

仅使用基数R时，使用

聚合

aggregate(seq.int(nrow(df)) ~ id + label, df, function(i) sum(df[i, 4] < df[i, 5]))
#  id label seq.int(nrow(df))
#1  1     a                 2
#2  4     a                 0
#3  6     a                 1
#4  2     b                 0
#5  5     e                 1

聚合（seq.int（nrow（df））~id+标签，df，函数（i）和（df[i，4] 为了匹配问题中发布的预期输出，需要做一些额外的工作

exceed <- seq.int(nrow(df))
agg <- aggregate(exceed ~ id + label, df, function(i) sum(df[i, 4] < df[i, 5]))
res <- merge(df[1:3], agg)
unique(res)
#  id label color exceed
#1  1     a     g      2
#3  1     a     a      2
#4  2     b     a      0
#5  4     a     g      0
#6  5     e     a      1
#7  6     a     a      1

---

超过我们可以使用
表
和
合并
：

table_ <- table(subset(df,value>threshold, c("id","label")))
df2 <- merge(unique(df[c("id","label","color")]),table_,all.x=TRUE)
df2$Freq[is.na(df2$Freq)] <- 0

#   id label color Freq
# 1  1     a     g    2
# 2  1     a     a    2
# 3  2     b     a    0
# 4  4     a     g    0
# 5  5     e     a    1
# 6  6     a     a    1

table_uuu阈值，c（“id”，“label”））
是的，是的。我现在觉得有点傻。
aggregate(seq.int(nrow(df)) ~ id + label, df, function(i) sum(df[i, 4] < df[i, 5]))
#  id label seq.int(nrow(df))
#1  1     a                 2
#2  4     a                 0
#3  6     a                 1
#4  2     b                 0
#5  5     e                 1

exceed <- seq.int(nrow(df))
agg <- aggregate(exceed ~ id + label, df, function(i) sum(df[i, 4] < df[i, 5]))
res <- merge(df[1:3], agg)
unique(res)
#  id label color exceed
#1  1     a     g      2
#3  1     a     a      2
#4  2     b     a      0
#5  4     a     g      0
#6  5     e     a      1
#7  6     a     a      1

table_ <- table(subset(df,value>threshold, c("id","label")))
df2 <- merge(unique(df[c("id","label","color")]),table_,all.x=TRUE)
df2$Freq[is.na(df2$Freq)] <- 0

#   id label color Freq
# 1  1     a     g    2
# 2  1     a     a    2
# 3  2     b     a    0
# 4  4     a     g    0
# 5  5     e     a    1
# 6  6     a     a    1