R 在data.table中按组计数快速排在前N位
我想知道根据分组的出现次数来划分分组的首选方法R 在data.table中按组计数快速排在前N位,r,count,data.table,greatest-n-per-group,R,Count,Data.table,Greatest N Per Group,我想知道根据分组的出现次数来划分分组的首选方法 dcast( df[,.(.N), by = .(cust_segment, cust_postal) ][,.(cust_postal, postal_rank = frankv(x=N, order=-1, ties.method = 'first') ), keyby=cust_segment ][postal_rank<=3], cust_segment ~ past
dcast(
df[,.(.N),
by = .(cust_segment, cust_postal)
][,.(cust_postal,
postal_rank = frankv(x=N, order=-1, ties.method = 'first')
), keyby=cust_segment
][postal_rank<=3],
cust_segment ~ paste0('postcode_rank_',postal_rank), value.var = 'cust_postal'
)
# desired output:
# cust_segment postcode_rank_1 postcode_rank_2 postcode_rank_3
# A 51274 64588 59212
# B 63590 69477 50380
# C 60619 66249 53494 ...etc...
例如,我有属于细分市场且拥有邮政编码的客户。我想知道每个部分最常见的3个邮政编码
library(data.table)
set.seed(123)
n <- 1e6
df <- data.table( cust_id = 1:n,
cust_segment = sample(LETTERS, size=n, replace=T),
cust_postal = sample(as.character(5e4:7e4),size=n, replace=T)
)
这是最好的方法,还是单一的方法?从评论中找出弗兰克的答案: 使用
forder
代替frankv
并使用keyby
,因为这比仅使用by
df[, .N,
keyby = .(cust_segment, cust_postal)
][order(-N), r := rowid(cust_segment)
][r <= 3, dcast(.SD, cust_segment ~ r, value.var ="cust_postal")]
cust_segment 1 2 3
1: A 51274 53440 55754
2: B 63590 69477 50380
3: C 60619 66249 52122
4: D 68107 50824 59305
5: E 51832 65249 52366
6: F 51401 55410 65046
df[,.N,
keyby=(客户段,客户邮政)
][订单(-N),r:=行ID(客户段)
][r看起来不错,不过我想你应该:=
frankv而不是制作一个新表。用一个调用forder
而不是多个调用frankv
也可能更快:df[,.N,keyby=(cust_段,cust_posal)][order(-N),r:=rowid(cust_段)][r@Frank是的,谢谢,你的速度快了25%
library(microbenchmark)
microbenchmark(C8H10N4O2 = dcast(
df[,.(.N),
by = .(cust_segment, cust_postal)
][,.(cust_postal,
postal_rank = frankv(x=N, order=-1, ties.method = 'first')
), keyby=cust_segment
][postal_rank<=3],
cust_segment ~ paste0('postcode_rank_',postal_rank), value.var = 'cust_postal'
),
frank = df[, .N,
keyby = .(cust_segment, cust_postal)
][order(-N), r := rowid(cust_segment)
][r <= 3, dcast(.SD, cust_segment ~ r, value.var ="cust_postal")])
Unit: milliseconds
expr min lq mean median uq max neval
C8H10N4O2 136.3318 140.8096 156.2095 145.6099 170.4862 205.8457 100
frank 102.2789 110.0140 118.2148 112.6940 119.2105 192.2464 100