将基本数据帧视为:
data <- data.frame(amount_bin = c("10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+", "10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+", "10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+"),
risk_score = c("0-700", "700-750", "750-800", "800-850", "850-900", "0-700", "700-750", "750-800", "800-850", "850-900", "0-700", "700-750", "750-800", "800-850", "850-900"))
,并在另一个数据框中将信息分组为:
group_info <- data.frame(variable = c("amount_bin_group", "amount_bin_group", "amount_bin_group", "amount_bin_group", "amount_bin_group",
"risk_score_group", "risk_score_group", "risk_score_group", "risk_score_group", "risk_score_group"),
bin = c("10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+",
"0-700", "700-750", "750-800", "800-850", "850-900"),
group = c("1", "1", "2", "2", "3",
"a", "a", "a", "b", "b"))
我想在基本数据帧(数据)中创建2列,分别称为“ amount_bin_group”和“ risk_score_group”,当来自group_info和数据的bin列相同时,它们从列group_info $ group中获取值。为简单起见,我们假设基本列始终是group_info $ variable名称减去“ group”字符串。这意味着,当我们要创建列amount_bin_group时,基本列在基本数据帧中将始终为amount_bin。
预期结果数据帧为:
final_data <- data.frame(amount_bin = c("10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+", "10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+", "10K-25K", "25K-35K", "35K-45K", "45K-50K", "50K+"),
risk_score = c("0-700", "700-750", "750-800", "800-850", "850-900", "0-700", "700-750", "750-800", "800-850", "850-900", "0-700", "700-750", "750-800", "800-850", "850-900"),
amount_bin_group = c("1", "1", "2", "2", "3", "1", "1", "2", "2", "3", "1", "1", "2", "2", "3"),
risk_score_group = c("a", "a", "a", "b", "b", "a", "a", "a", "b", "b", "a", "a", "a", "b", "b"))
我刚刚想到的一种解决方案是迭代合并数据帧,即:
final_data <- merge(data, group_info[, c("bin", "group")], by.x = "amount_bin", by.y = "bin")
final_data$amount_bin_group <- final_data$group
final_data$group <- NULL
但是,我相信可以有一个更有效的解决方案。请注意,有多个此类列,而不仅仅是两个。因此,也许循环会有所帮助。
答案 0 :(得分:1)
您的group_info太整洁了。我真不敢说我在说。通过将其分为两个数据帧,或将每一半分为自己的列,您可以自己进行简单的左连接以获取答案。
final_data_calc <- data %>%
left_join(
group_info %>%
filter(variable == 'amount_bin_group') %>%
rename(amount_bin_group = group,amount_bin = bin) %>%
select(-variable)
) %>%
left_join(
group_info %>%
filter(variable == 'risk_score_group') %>%
rename(risk_score_group = group,risk_score = bin) %>%
select(-variable)
)
# amount_bin risk_score amount_bin_group risk_score_group
#1 10K-25K 0-700 1 a
#2 25K-35K 700-750 1 a
#3 35K-45K 750-800 2 a
#4 45K-50K 800-850 2 b
#5 50K+ 850-900 3 b
#6 10K-25K 0-700 1 a
#7 25K-35K 700-750 1 a
#8 35K-45K 750-800 2 a
#9 45K-50K 800-850 2 b
#10 50K+ 850-900 3 b
#11 10K-25K 0-700 1 a
#12 25K-35K 700-750 1 a
#13 35K-45K 750-800 2 a
#14 45K-50K 800-850 2 b
#15 50K+ 850-900 3 b
答案 1 :(得分:1)
您可以使用for
循环来继续合并不同的集合:
for (i in unique(group_info$variable)) {
data <- merge(
data, group_info[group_info$variable==i,c("bin","group")],
by.x=sub("_group","",i), by.y="bin"
)
names(data)[names(data)=="group"] <- i
}