我正在尝试向 R 中的状态示例数据框添加一个新列。我希望此列将状态 ID 聚集到更广泛的类别 (1-4) 中。我的代码与我要查找的代码很接近,但我不太正确。我知道我可以逐行输入每个州 ID,但是有没有更快的方法?谢谢!
parameter = spark.read.option("multiline", "true").json("/at/dev_parameter.json")
kserver = parameter.select("broker")
ktopic = parameter.select("topicname")
df.selectExpr("CAST(id AS STRING) AS key", "to_json(struct(*)) AS value")
.write
.format("kafka")
.outputMode("append")
.option("kafka.bootstrap.servers", "f"+ **kserver**)
.option("topic", "josn_data_topic",**ktopic** )
.save()
#添加一列来表示每个状态
library(tidyverse)
#为状态桶创建新变量
States=state.x77
States=data.frame(States)
States <- tibble::rowid_to_column(States, "ID")
States
答案 0 :(得分:1)
如果您的所有组都将使用连续的 cut
值,您可以使用 findInterval
或 ID
:
findInterval(States$ID, c(0, 12, 24, 37, 51))
# [1] 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4
如果你想让它更详细一点,你可以在 dplyr::between
中使用 case_when
:
States %>%
mutate(
WAGE_BUCKET = case_when(
between(ID, 1, 12) ~ "1",
between(ID, 13, 24) ~ "2",
between(ID, 25, 37) ~ "3",
between(ID, 38, 50) ~ "4",
TRUE ~ NA_character_)
)
# ID Population Income Illiteracy Life Exp Murder HS Grad Frost Area WAGE_BUCKET
# 1 1 3615 3624 2.1 69.05 15.1 41.3 20 50708 1
# 2 2 365 6315 1.5 69.31 11.3 66.7 152 566432 1
# 3 3 2212 4530 1.8 70.55 7.8 58.1 15 113417 1
# 4 4 2110 3378 1.9 70.66 10.1 39.9 65 51945 1
# 5 5 21198 5114 1.1 71.71 10.3 62.6 20 156361 1
# 6 6 2541 4884 0.7 72.06 6.8 63.9 166 103766 1
# 7 7 3100 5348 1.1 72.48 3.1 56.0 139 4862 1
# 8 8 579 4809 0.9 70.06 6.2 54.6 103 1982 1
# 9 9 8277 4815 1.3 70.66 10.7 52.6 11 54090 1
# 10 10 4931 4091 2.0 68.54 13.9 40.6 60 58073 1
# 11 11 868 4963 1.9 73.60 6.2 61.9 0 6425 1
# 12 12 813 4119 0.6 71.87 5.3 59.5 126 82677 1
# 13 13 11197 5107 0.9 70.14 10.3 52.6 127 55748 2
# 14 14 5313 4458 0.7 70.88 7.1 52.9 122 36097 2
# 15 15 2861 4628 0.5 72.56 2.3 59.0 140 55941 2
# 16 16 2280 4669 0.6 72.58 4.5 59.9 114 81787 2
# 17 17 3387 3712 1.6 70.10 10.6 38.5 95 39650 2
# 18 18 3806 3545 2.8 68.76 13.2 42.2 12 44930 2
# 19 19 1058 3694 0.7 70.39 2.7 54.7 161 30920 2
# 20 20 4122 5299 0.9 70.22 8.5 52.3 101 9891 2
# 21 21 5814 4755 1.1 71.83 3.3 58.5 103 7826 2
# 22 22 9111 4751 0.9 70.63 11.1 52.8 125 56817 2
# 23 23 3921 4675 0.6 72.96 2.3 57.6 160 79289 2
# 24 24 2341 3098 2.4 68.09 12.5 41.0 50 47296 2
# 25 25 4767 4254 0.8 70.69 9.3 48.8 108 68995 3
# 26 26 746 4347 0.6 70.56 5.0 59.2 155 145587 3
# 27 27 1544 4508 0.6 72.60 2.9 59.3 139 76483 3
# 28 28 590 5149 0.5 69.03 11.5 65.2 188 109889 3
# 29 29 812 4281 0.7 71.23 3.3 57.6 174 9027 3
# 30 30 7333 5237 1.1 70.93 5.2 52.5 115 7521 3
# 31 31 1144 3601 2.2 70.32 9.7 55.2 120 121412 3
# 32 32 18076 4903 1.4 70.55 10.9 52.7 82 47831 3
# 33 33 5441 3875 1.8 69.21 11.1 38.5 80 48798 3
# 34 34 637 5087 0.8 72.78 1.4 50.3 186 69273 3
# 35 35 10735 4561 0.8 70.82 7.4 53.2 124 40975 3
# 36 36 2715 3983 1.1 71.42 6.4 51.6 82 68782 3
# 37 37 2284 4660 0.6 72.13 4.2 60.0 44 96184 3
# 38 38 11860 4449 1.0 70.43 6.1 50.2 126 44966 4
# 39 39 931 4558 1.3 71.90 2.4 46.4 127 1049 4
# 40 40 2816 3635 2.3 67.96 11.6 37.8 65 30225 4
# 41 41 681 4167 0.5 72.08 1.7 53.3 172 75955 4
# 42 42 4173 3821 1.7 70.11 11.0 41.8 70 41328 4
# 43 43 12237 4188 2.2 70.90 12.2 47.4 35 262134 4
# 44 44 1203 4022 0.6 72.90 4.5 67.3 137 82096 4
# 45 45 472 3907 0.6 71.64 5.5 57.1 168 9267 4
# 46 46 4981 4701 1.4 70.08 9.5 47.8 85 39780 4
# 47 47 3559 4864 0.6 71.72 4.3 63.5 32 66570 4
# 48 48 1799 3617 1.4 69.48 6.7 41.6 100 24070 4
# 49 49 4589 4468 0.7 72.48 3.0 54.5 149 54464 4
# 50 50 376 4566 0.6 70.29 6.9 62.9 173 97203 4
答案 1 :(得分:0)
它是长度 > 1 的 vector
。比较运算符适用于单个向量。我们可以使用 between
library(dplyr)
States <- States %>%
mutate(WAGE_BUCKET=case_when(between(ID, 1, 12) ~ '1',
between(ID, 13,24) ~ '2',
between(ID, 25,37) ~ '3',
between(ID, 38,50) ~ '4',
TRUE ~ NA_character_))
或者另一种选择是将 &
与 >
和 <=
结合使用
States %>%
mutate(WAGE_BUCKET=case_when(ID >= 1 & ID <=12 ~ '1',
ID >= 13 & ID <= 24) ~ '2',
ID >= 25 & ID <= 37 ~ '3',
ID >= 38 & ID <= 50 ~ '4',
TRUE ~ NA_character))
或者可能是使用 %in%
States %>%
mutate(WAGE_BUCKET=case_when(ID %in% c(1,12) ~ '1',
ID %in% c(13,24) ~ '2',
ID %in% c(25,37) ~ '3',
ID %in% c(38,50) ~ '4',
TRUE ~ NA_character_))