代码之家  ›  专栏  ›  技术社区  ›  Laura

找出调查数据是否是一致的跨年的dplyr参与者在R

mutate dplyr group-by r
  •  0
  • Laura  · 技术社区  · 7 年前

    我有这样的数据: 

    df <- data.frame(
ID = c(rep(c("ABC123", "BCD234", "CDE345", "DEF456", "EFG567", "FGH678", "GHI891", "HIJ910", "IJK101", "JKL011"),2)),
eth = c(1, 2, 2, 3, 1, 1, 4, 4, 3, 3, 1, 4, 1, 3, 1, 3, 4, 4, 3, 2),
nzdep = c(4, 3, 3, 2, 4, 4, 1, 1, 2, 2, 4, 3, 3, 4, 4, 2, 1, 1, 2, 3),
sex =  c("M", "M", "F", "F", "M", "M", "F", "F", "M", "M", "F", "M", "M", "M", "M", "F", "F", "M", "F", "M"),
Year = c(rep("Y1", 10), rep("Y2", 10)))


    dems <- df %>%
group_by(ID) %>%
mutate(dep_dif = ifelse(nzdep = nzdep, 1, 0),
     sex_dif = ifelse(sex = sex, 1, 0),
     eth_dif = ifelse(eth = eth, 1, 0))

    这不管用,但我一直在按照这些思路思考。

    我想要的输出是: 

    dems <- data.frame(
ID = c(rep(c("ABC123", "BCD234", "CDE345", "DEF456", "EFG567", "FGH678", "GHI891", "HIJ910", "IJK101", "JKL011"),2)),
eth = c(1, 2, 2, 3, 1, 1, 4, 4, 3, 3, 1, 4, 1, 3, 1, 3, 4, 4, 3, 2),
nzdep = c(4, 3, 3, 2, 4, 4, 1, 1, 2, 2, 4, 3, 3, 4, 4, 2, 1, 1, 2, 3),
sex = c("M", "M", "F", "F", "M", "M", "F", "F", "M", "M", "F", "M", "M", "M", "M", "F", "F", "M", "F", "M"),
Year = c(rep("Y1", 10), rep("Y2", 10)),
eth_dif = c(rep(c(1, 0, 0, 1, 1, 0, 1, 1, 1, 0),2)),
dep_dif = c(rep(c(1, 1, 1, 0, 1, 0, 1, 1, 1, 0),2)),
sex_dif = c(rep(c(0, 1, 0, 0, 1, 0, 1, 0, 0, 1),2)))

    谢谢

    2 回复  |  直到 7 年前
        1
  •  1
  •   BENY    7 年前

    看来你需要 unique 值等于1 

    df%>%group_by(ID)%>%dplyr::mutate( ifelse(length(unique(nzdep))==1, 1, 0),
+                                    sex_dif = ifelse(length(unique(sex))==1, 1, 0),
+                                    eth_dif = ifelse(length(unique(eth))==1, 1, 0))
# A tibble: 20 x 8
# Groups:   ID [10]
       ID   eth nzdep    sex   Year `ifelse(length(unique(nzdep)) == 1, 1, 0)` sex_dif eth_dif
   <fctr> <dbl> <dbl> <fctr> <fctr>                                      <dbl>   <dbl>   <dbl>
 1 ABC123     1     4      M     Y1                                          1       0       1
 2 BCD234     2     3      M     Y1                                          1       1       0
 3 CDE345     2     3      F     Y1                                          1       0       0
 4 DEF456     3     2      F     Y1                                          0       0       1
 5 EFG567     1     4      M     Y1                                          1       1       1
 6 FGH678     1     4      M     Y1                                          0       0       0
 7 GHI891     4     1      F     Y1                                          1       1       1
 8 HIJ910     4     1      F     Y1                                          1       0       1
 9 IJK101     3     2      M     Y1                                          1       0       1
10 JKL011     3     2      M     Y1                                          0       1       0
11 ABC123     1     4      F     Y2                                          1       0       1
12 BCD234     4     3      M     Y2                                          1       1       0
13 CDE345     1     3      M     Y2                                          1       0       0
14 DEF456     3     4      M     Y2                                          0       0       1
15 EFG567     1     4      M     Y2                                          1       1       1
16 FGH678     3     2      F     Y2                                          0       0       0
17 GHI891     4     1      F     Y2                                          1       1       1
18 HIJ910     4     1      M     Y2                                          1       0       1
19 IJK101     3     2      F     Y2                                          1       0       1
20 JKL011     2     3      M     Y2                                          0       1       0
        2
  •  0
  •   akrun    7 年前

    我们可以用 mutate_at 

    library(dplyr)
df %>% 
  group_by(ID) %>% 
  mutate_at(2:4, funs(dif = as.integer(.[Year == "Y1"] == .[Year == "Y2"])))
# A tibble: 20 x 8
# Groups:   ID [10]
#   ID       eth nzdep sex   Year  eth_dif nzdep_dif sex_dif
#   <fct>  <dbl> <dbl> <fct> <fct>   <int>     <int>   <int>
# 1 ABC123     1     4 M     Y1          1         1       0
# 2 BCD234     2     3 M     Y1          0         1       1
# 3 CDE345     2     3 F     Y1          0         1       0
# 4 DEF456     3     2 F     Y1          1         0       0
# 5 EFG567     1     4 M     Y1          1         1       1
# 6 FGH678     1     4 M     Y1          0         0       0
# 7 GHI891     4     1 F     Y1          1         1       1
# 8 HIJ910     4     1 F     Y1          1         1       0
# 9 IJK101     3     2 M     Y1          1         1       0
#10 JKL011     3     2 M     Y1          0         0       1
#11 ABC123     1     4 F     Y2          1         1       0
#12 BCD234     4     3 M     Y2          0         1       1
#13 CDE345     1     3 M     Y2          0         1       0
#14 DEF456     3     4 M     Y2          1         0       0
#15 EFG567     1     4 M     Y2          1         1       1
#16 FGH678     3     2 F     Y2          0         0       0
#17 GHI891     4     1 F     Y2          1         1       1
#18 HIJ910     4     1 M     Y2          1         1       0
#19 IJK101     3     2 F     Y2          1         1       0
#20 JKL011     2     3 M     Y2          0         0       1

    base R 选择是 

    df[paste0(names(df)[2:4], "_dif")] <- +(Reduce(`==`, split(df[2:4], df$Year)))
    推荐文章
    John Smith  ·  Java按值对映射进行分组,其中值为列表
    2 年前
    The Great  ·  拆分并存储数据帧,但名称基于特定列中的唯一值
    3 年前
    Programming Noob  ·  AttributeError:“SeriesGroupBy”对象没有属性“tolist”
    3 年前
    Flo  ·  分组依据中的SQL大小写
    3 年前
    user13663655  ·  我可以使用条件分组并返回源类型值吗
    3 年前
    pinkiBet  ·  使用MAX和UNION分组,还是加入?
    3 年前
    Shubham  ·  如何在R中查找数据表的两个多类别列之间的差异/setdiff()
    3 年前
    user7052482  ·  将来自联合的SQL结果分组(SQL Server)
    7 年前
    lugger1  ·  如何获取Postgres中时间间隔的平均值
    7 年前
    JustCallMeGary  ·  如何使用R中的一个因子操作两个变量[重复]
    7 年前
    关于   移动版
    代码之家 - 一站式码农服务社区
    沪ICP备11025650号