交易日期前 6 个月的总金额 [英] Sum amount last 6 month prior to the date of transaction

问题描述

这是我的交易数据.它显示了从 from 列中的帐户到 to 列中的帐户进行的交易以及日期和金额信息

This is my transaction data. It shows the transactions made from the accounts in from column to the accounts in to column with the date and the amount information

data 

id          from    to          date        amount  
<int>       <fctr>  <fctr>      <date>      <dbl>
19521       6644    6934        2005-01-01  700.0
19524       6753    8456        2005-01-01  600.0
19523       9242    9333        2005-01-01  1000.0
…           …       …           …           …
1056317     7819    7454        2010-12-31  60.2
1056318     6164    7497        2010-12-31  107.5
1056319     7533    7492        2010-12-31  164.1

我想计算在进行特定交易的日期之前的最后 6 个月内，from 列中的账户收到了多少交易金额，并希望将此信息保存为新列.

I want to calculate how much transaction amount the accounts in from column received in the last 6 month prior to the date that particular transaction was made and want to save this information as a new column.

下面的代码可以很好地在一个小数据集(比如 1000 行)中完成此任务:

This following code works very well to accomplish this in a small dataset ,say, with 1000 rows:

library(dplyr)
library(purrr)
data %>% 
  mutate(total_trx_amount_received_in_last_sixmonth= map2_dbl(from, date, 
~sum(amount[to == .x & between(date, .y - 180, .y)])))

但是，由于我的数据有超过 100 万行，此代码将需要几个小时才能完成.如果我可以加快这段代码的运行时间，我搜索了互联网.我在关于如何制作 purrr map 函数运行得更快.因此，我尝试了以下代码，而不是 dplyr 的 mutate 我使用 data.table 来更快地加速代码:

However, since my data has over 1 million rows, this code will take more than a couple of hours to complete. I searched the internet if I can speed up the run time of this code. I tried this suggestion on SO about how to make purrr map function run faster. So, I tried the following code and instead of mutate of dplyr I used data.table to speed up the code even faster:

library(future)
library(data.table)
library(furrr)
data[, total_trx_amount_received_in_last_sixmonth:= furrr::future_pmap_dbl(list(from, date), 
~mean(amount[to == .x & between(date, .y-180, .y)])) ]

但是，速度根本没有提高.

But, the speed hasn't been improved at all.

有什么建议可以让代码运行得更快吗?

Is there any suggestion on how can I make the code run faster?

dput() 输出数据:

dput() output of the data:

structure(list(id = c(18529L, 13742L, 9913L, 956L, 2557L, 1602L, 
18669L, 35900L, 48667L, 51341L, 53713L, 60126L, 60545L, 65113L, 
66783L, 83324L, 87614L, 88898L, 89874L, 94765L, 100277L, 101587L, 
103444L, 108414L, 113319L, 121516L, 126607L, 130170L, 131771L, 
135002L, 149431L, 157403L, 157645L, 158831L, 162597L, 162680L, 
163901L, 165044L, 167082L, 168562L, 168940L, 172578L, 173031L, 
173267L, 177507L, 179167L, 182612L, 183499L, 188171L, 189625L, 
193940L, 198764L, 199342L, 200134L, 203328L, 203763L, 204733L, 
205651L, 209672L, 210242L, 210979L, 214532L, 214741L, 215738L, 
216709L, 220828L, 222140L, 222905L, 226133L, 226527L, 227160L, 
228193L, 231782L, 232454L, 233774L, 237836L, 237837L, 238860L, 
240223L, 245032L, 246673L, 247561L, 251611L, 251696L, 252663L, 
254410L, 255126L, 255230L, 258484L, 258485L, 259309L, 259910L, 
260542L, 262091L, 264462L, 264887L, 264888L, 266125L, 268574L, 
272959L), from = c("5370", "5370", "5370", "8605", "5370", "6390", 
"5370", "5370", "8934", "5370", "5635", "6046", "5680", "8026", 
"9037", "5370", "7816", "8046", "5492", "8756", "5370", "9254", 
"5370", "5370", "7078", "6615", "5370", "9817", "8228", "8822", 
"5735", "7058", "5370", "8667", "9315", "6053", "7990", "8247", 
"8165", "5656", "9261", "5929", "8251", "5370", "6725", "5370", 
"6004", "7022", "7442", "5370", "8679", "6491", "7078", "5370", 
"5370", "5370", "5658", "5370", "9296", "8386", "5370", "5370", 
"5370", "9535", "5370", "7541", "5370", "9621", "5370", "7158", 
"8240", "5370", "5370", "8025", "5370", "5370", "5370", "6989", 
"5370", "7059", "5370", "5370", "5370", "9121", "5608", "5370", 
"5370", "7551", "5370", "5370", "5370", "5370", "9163", "9362", 
"6072", "5370", "5370", "5370", "5370", "5370"), to = c("9356", 
"5605", "8567", "5370", "5636", "5370", "8933", "8483", "5370", 
"7626", "5370", "5370", "5370", "5370", "5370", "9676", "5370", 
"5370", "5370", "5370", "9105", "5370", "9772", "6979", "5370", 
"5370", "7564", "5370", "5370", "5370", "5370", "5370", "8744", 
"5370", "5370", "5370", "5370", "5370", "5370", "5370", "5370", 
"5370", "5370", "7318", "5370", "8433", "5370", "5370", "5370", 
"7122", "5370", "5370", "5370", "8566", "6728", "9689", "5370", 
"8342", "5370", "5370", "5614", "5596", "5953", "5370", "7336", 
"5370", "7247", "5370", "7291", "5370", "5370", "6282", "7236", 
"5370", "8866", "8613", "9247", "5370", "6767", "5370", "9273", 
"7320", "9533", "5370", "5370", "8930", "9343", "5370", "9499", 
"7693", "7830", "5392", "5370", "5370", "5370", "7497", "8516", 
"9023", "7310", "8939"), date = structure(c(12934, 13000, 13038, 
13061, 13099, 13113, 13117, 13179, 13238, 13249, 13268, 13296, 
13299, 13309, 13314, 13391, 13400, 13404, 13409, 13428, 13452, 
13452, 13460, 13482, 13493, 13518, 13526, 13537, 13542, 13544, 
13596, 13616, 13617, 13626, 13633, 13633, 13639, 13642, 13646, 
13656, 13660, 13664, 13667, 13669, 13677, 13686, 13694, 13694, 
13707, 13716, 13725, 13738, 13739, 13746, 13756, 13756, 13756, 
13761, 13769, 13770, 13776, 13786, 13786, 13786, 13791, 13799, 
13806, 13813, 13817, 13817, 13817, 13822, 13829, 13830, 13836, 
13847, 13847, 13847, 13852, 13860, 13866, 13871, 13878, 13878, 
13878, 13882, 13883, 13883, 13887, 13887, 13888, 13889, 13890, 
13891, 13895, 13896, 13896, 13899, 13905, 13909), class = "Date"), 
    amount = c(24.4, 7618, 21971, 5245, 2921, 8000, 169.2, 71.5, 
    14.6, 4214, 14.6, 13920, 14.6, 24640, 1600, 261.1, 16400, 
    3500, 2700, 19882, 182, 14.6, 16927, 25653, 3059, 2880, 9658, 
    4500, 12480, 14.6, 1000, 3679, 34430, 12600, 14.6, 19.2, 
    4900, 826, 3679, 2100, 38000, 79, 11400, 21495, 3679, 200, 
    14.6, 100.6, 3679, 5300, 108.9, 3679, 2696, 7500, 171.6, 
    14.6, 99.2, 2452, 3679, 3218, 700, 69.7, 14.6, 91.5, 2452, 
    3679, 2900, 17572, 14.6, 14.6, 90.5, 2452, 49752, 3679, 1900, 
    14.6, 870, 85.2, 2452, 3679, 1600, 540, 14.6, 14.6, 79, 210, 
    2452, 28400, 720, 180, 420, 44289, 489, 3679, 840, 2900, 
    150, 870, 420, 14.6)), row.names = c(NA, -100L), class = "data.frame")

推荐答案

这只是data.table中的一个非等连接.您可以创建一个 date - 180 变量并限制当前日期和该变量之间的连接.这应该很快

This is simply a non-equi join in data.table. You can create a variable of date - 180 and limit the join between the current date and that variable. This should be fairly quick

library(data.table)
setDT(dt)[, date_minus_180 := date - 180]
dt[, amnt_6_m := .SD[dt, sum(amount, na.rm = TRUE), 
     on = .(to = from, date <= date, date >= date_minus_180), by = .EACHI]$V1]
head(dt, 10)
#        id from   to       date  amount date_minus_180 amnt_6_m
#  1: 18529 5370 9356 2005-05-31    24.4     2004-12-02      0.0
#  2: 13742 5370 5605 2005-08-05  7618.0     2005-02-06      0.0
#  3:  9913 5370 8567 2005-09-12 21971.0     2005-03-16      0.0
#  4:   956 8605 5370 2005-10-05  5245.0     2005-04-08      0.0
#  5:  2557 5370 5636 2005-11-12  2921.0     2005-05-16   5245.0
#  6:  1602 6390 5370 2005-11-26  8000.0     2005-05-30      0.0
#  7: 18669 5370 8933 2005-11-30   169.2     2005-06-03  13245.0
#  8: 35900 5370 8483 2006-01-31    71.5     2005-08-04  13245.0
#  9: 48667 8934 5370 2006-03-31    14.6     2005-10-02      0.0
# 10: 51341 5370 7626 2006-04-11  4214.0     2005-10-13   8014.6

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