Question

我有一个具有以下格式的数据集：

> library(tidyverse)
> library(tibble)
> 
> 
> data<-data.frame(ID=c(1,1,2,2,3,3,3,3,4,4),
+          Radius=c(5,5,5,5,10,10,15,15,10,10),
+          neighb_ID=c(1,11,2,12,3,4,7,8,3,4),
+          var_neighb=c(50,20,30,40,15,100,70,60,15,100))
> data
   ID1 Radius neighb_ID var_neighb
1    1      5         1         50
2    1      5        11         20
3    2      5         2         30
4    2      5        12         40
5    3     10         3         15
6    3     10         4        100
7    3     15         7         70
8    3     15         8         60
9    4     10         3         15
10   4     10         4        100
>

现在，我想透视此数据，以便为每个var_neighb按Radius汇总ID。例如，对于sum和mean，我希望获得下表：

  ID1 Svar_neighb_Radius_5 Svar_neighb_Radius_10 Svar_neighb_Radius_15
1   1                   20                     0                     0
2   2                   40                     0                     0
3   3                    0                   100                   130
4   4                    0                    15                     0
  Mvar_neighb_Radius_5 Mvar_neighb_Radius_10 Mvar_neighb_Radius_15
1                   20                     0                     0
2                   40                     0                     0
3                    0                   100                    65
4                    0                    15                     0
>

我尝试使用以下代码执行此操作：

> agdata<-data %>%
+    pivot_wider(
+     names_from = Radius, 
+     values_from = var_neighb,
+     values_fn = sum,
+     values_fill = 0
+     )

我只收到以下错误：

Error in values_fn[[value]] : object of type 'builtin' is not subsettable

另外，即使我取出values_fn = sum,，也会遇到以下错误： Error in values_fill[[value]] : subscript out of bounds。

有人可以帮助我解决这些问题以实现我的目标吗？

编辑：抱歉，我忽略了对输出表的一项重要要求：聚合应以sum和mean进行，并且不应包含var_neighb等于{{1}的neighb_ID的值}。输出表ID需要通过data_out和sum进行汇总。所以我更新了mean。

Answer 1

values_fn和values_fill应命名为列表：

library(tidyverse)

data <- data.frame(
  ID=c(1,1,2,2,3,3,3,4,4),
  Radius=c(5,5,5,5,10,10,15,10,10),
  neighb_ID=c(1,11,2,12,3,4,7,3,4),
  var_neighb=c(50,20,30,40,15,100,70,15,100)
)

data %>%
  select(-neighb_ID) %>%
  pivot_wider(
     names_from = Radius, 
     values_from = var_neighb,
     values_fn = list(var_neighb = sum),
     values_fill = list(var_neighb = 0),
     names_prefix = "var_neighb_Radius_"
   )

# # A tibble: 4 x 4
#       ID                var_neighb_Radius_5   var_neighb_Radius_10 var_neighb_Radius_15
#       <dbl>              <dbl>                <dbl>                <dbl>
# 1     1                  70                    0                    0
# 2     2                  70                    0                    0
# 3     3                   0                  115                   70
# 4     4                   0                  115                    0

更新要使用ID == neighb_ID删除值，只需使用过滤器：

data %>%
  filter(ID != neighb_ID) %>%
  select(-neighb_ID) %>%
  pivot_wider(
     names_from = Radius, 
     values_from = var_neighb,
     values_fn = list(var_neighb = sum),
     values_fill = list(var_neighb = 0),
     names_prefix = "var_neighb_Radius_"
   )

不太确定您通过“ mean以及sum进行聚合”所了解的内容-您无法在同一列中进行两种不同的聚合，但是您可以做两个枢轴并将它们结合在一起：

library(dplyr)

inner_join(
  data %>%
    filter(ID != neighb_ID) %>%
    select(-neighb_ID) %>%
    pivot_wider(
      names_from = Radius, 
      values_from = var_neighb,
      values_fn = list(var_neighb = sum),
      values_fill = list(var_neighb = 0),
      names_prefix = "var_neighb_Radius_sum_"
    ),
  data %>%
    filter(ID != neighb_ID) %>%
    select(-neighb_ID) %>%
    pivot_wider(
      names_from = Radius, 
      values_from = var_neighb,
      values_fn = list(var_neighb = mean),
      values_fill = list(var_neighb = 0),
      names_prefix = "var_neighb_Radius_mean_"
    ),
  by = "ID"
)

在R中使用ivot_wider（）对values_fn执行聚合（例如，求和，均值）时出错

1 个答案: