使用预测数据进行数据处理
我试图对两个商店（商店1和商店2）的销售量进行预测。就像使用预测程序包进行预测的结果一样，我得到了这两个表。第一个表包含有关每种模型的MAPE误差数据（列值）。您可以在下面看到数据和数据的屏幕截图。
Table_1<-structure(list(...1 = c("1", "2", "3", "4", "5", "6", "7", "8",
"9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19",
"20"), X1 = c("SNAIVE", "HW", "ETS", "ARIMA", "STL", "TBATS",
"NNETAR", "RWF", "TSLM", "FOURIER", "SNAIVE", "HW", "ETS", "ARIMA",
"STL", "TBATS", "NNETAR", "RWF", "TSLM", "FOURIER"), X2 = c("Store 1",
"Store 1", "Store 1", "Store 1", "Store 1", "Store 1", "Store 1",
"Store 1", "Store 1", "Store 1", "Store 2", "Store 2", "Store 2",
"Store 2", "Store 2", "Store 2", "Store 2", "Store 2", "Store 2",
"Store 2"), value = c(11.2819379803024, 4.90469397146697, 4.90469397146697,
4.64808116952175, 4.92695563666538, 6.11286061911487, 7.66061575087076,
8.95984865369006, 5.07614708345642, 4.57448859126253, 22.7760224588221,
24.0502857269679, 18.9376978459644, 21.6693712888351, 21.6029490199174,
24.692214948761, 26.2680955559159, 30.5302345480261, 22.2367412218357,
22.6100823447494)), row.names = c(NA, -20L), class = c("tbl_df",
"tbl", "data.frame"))
我已经用黄色突出显示了商店1和商店2的预测中最好的三个模型（MAPE误差最低）。商店1是（ETS，ARIMA和Fourier），商店2是（ETS，ARIMA和STL）。
第二张表包含所有这些模型每个月的预测数据。在下面，您可以查看数据以及数据的屏幕截图。
Table2<-structure(list(Date = structure(c(1575158400, 1577836800, 1580515200,
1583020800, 1585699200, 1588291200, 1590969600, 1593561600, 1596240000,
1598918400, 1601510400, 1604188800, 1606780800, 1575158400, 1577836800,
1580515200, 1583020800, 1585699200, 1588291200, 1590969600, 1593561600,
1596240000, 1598918400, 1601510400, 1604188800, 1606780800, 1575158400,
1577836800, 1580515200, 1583020800, 1585699200, 1588291200, 1590969600,
1593561600, 1596240000, 1598918400, 1601510400, 1604188800, 1606780800,
1575158400, 1577836800, 1580515200, 1583020800, 1585699200, 1588291200,
1590969600, 1593561600, 1596240000, 1598918400, 1601510400, 1604188800,
1606780800, 1575158400, 1577836800, 1580515200, 1583020800, 1585699200,
1588291200, 1590969600, 1593561600, 1596240000, 1598918400, 1601510400,
1604188800, 1606780800, 1575158400, 1577836800, 1580515200, 1583020800,
1585699200, 1588291200, 1590969600, 1593561600, 1596240000, 1598918400,
1601510400, 1604188800, 1606780800, 1575158400, 1577836800, 1580515200,
1583020800, 1585699200, 1588291200, 1590969600, 1593561600, 1596240000,
1598918400, 1601510400, 1604188800, 1606780800, 1575158400, 1577836800,
1580515200, 1583020800, 1585699200, 1588291200, 1590969600, 1593561600,
1596240000, 1598918400, 1601510400, 1604188800, 1606780800, 1575158400,
1577836800, 1580515200, 1583020800, 1585699200, 1588291200, 1590969600,
1593561600, 1596240000, 1598918400, 1601510400, 1604188800, 1606780800,
1575158400, 1577836800, 1580515200, 1583020800, 1585699200, 1588291200,
1590969600, 1593561600, 1596240000, 1598918400, 1601510400, 1604188800,
1606780800), class = c("POSIXct", "POSIXt"), tzone = "UTC"),
Forecasting_model = c("SNAIVE", "SNAIVE", "SNAIVE", "SNAIVE",
"SNAIVE", "SNAIVE", "SNAIVE", "SNAIVE", "SNAIVE", "SNAIVE",
"SNAIVE", "SNAIVE", "SNAIVE", "HW", "HW", "HW", "HW", "HW",
"HW", "HW", "HW", "HW", "HW", "HW", "HW", "HW", "ETS", "ETS",
"ETS", "ETS", "ETS", "ETS", "ETS", "ETS", "ETS", "ETS", "ETS",
"ETS", "ETS", "ARIMA", "ARIMA", "ARIMA", "ARIMA", "ARIMA",
"ARIMA", "ARIMA", "ARIMA", "ARIMA", "ARIMA", "ARIMA", "ARIMA",
"ARIMA", "STL", "STL", "STL", "STL", "STL", "STL", "STL",
"STL", "STL", "STL", "STL", "STL", "STL", "TBATS", "TBATS",
"TBATS", "TBATS", "TBATS", "TBATS", "TBATS", "TBATS", "TBATS",
"TBATS", "TBATS", "TBATS", "TBATS", "NNAR", "NNAR", "NNAR",
"NNAR", "NNAR", "NNAR", "NNAR", "NNAR", "NNAR", "NNAR", "NNAR",
"NNAR", "NNAR", "RWF", "RWF", "RWF", "RWF", "RWF", "RWF",
"RWF", "RWF", "RWF", "RWF", "RWF", "RWF", "RWF", "TSLM",
"TSLM", "TSLM", "TSLM", "TSLM", "TSLM", "TSLM", "TSLM", "TSLM",
"TSLM", "TSLM", "TSLM", "TSLM", "FOURIER", "FOURIER", "FOURIER",
"FOURIER", "FOURIER", "FOURIER", "FOURIER", "FOURIER", "FOURIER",
"FOURIER", "FOURIER", "FOURIER", "FOURIER"), `Store 1` = c(8083,
1171, 1328, 1281, 1281, 1118, 1107, 1611, 1116, 1133, 1618,
1261, 8083, 8312, 1336, 1261, 1673, 1667, 1223, 1603, 1621,
1211, 1633, 1637, 1672, 8138, 8312, 1336, 1261, 1673, 1667,
1223, 1603, 1621, 1211, 1633, 1637, 1672, 8138, 8818, 1363,
1282, 1671, 1623, 1276, 1283, 1687, 1261, 1632, 1676, 1631,
8367, 8827, 1108, 1226, 1681, 1661, 1288, 1616, 1683, 1278,
1663, 1678, 1703, 8338, 8371, 1183, 1237, 1738, 1701, 1637,
1681, 1721, 1271, 1738, 1663, 1732, 8180, 8076, 1318, 1271,
1732, 1883, 1286, 1607, 1336, 1281, 1711, 1873, 1881, 8183,
1271, 1283, 1233, 1608, 1618, 1681, 1631, 1611, 1620, 1660,
1663, 1673, 1688, 8166, 1317, 1188, 1233, 1273, 1183, 1212,
1276, 1178, 1221, 1226, 1283, 8863, 8811, 1118, 1223, 1661,
1621, 1260, 1286, 1617, 1213, 1688, 1687, 1660, 8311), `Store 2` = c(1180,
811, 312, 1612, 1387, 878, 812, 883, 362, 768, 800, 760,
1180, 1021, 761, 1002, 1106, 1271, 337, 1113, 373, 833, 1012,
333, 303, 1166, 336, 708, 332, 1312, 1168, 838, 1010, 862,
773, 883, 861, 767, 1000, 1070, 636, 838, 1161, 1183, 887,
1001, 813, 331, 820, 738, 732, 1087, 333, 688, 810, 1311,
1183, 876, 338, 818, 816, 818, 816, 773, 333, 337, 888, 871,
1378, 1100, 1008, 368, 380, 883, 386, 872, 838, 363, 1102,
301, 831, 1133, 1331, 831, 333, 321, 338, 883, 832, 881,
1303, 766, 778, 773, 782, 731, 737, 801, 810, 816, 888, 883,
832, 811, 1820, 1000, 1136, 1270, 1718, 1188, 1873, 1162,
1136, 1130, 1178, 1110, 1371, 380, 703, 306, 1862, 1110,
873, 327, 837, 808, 817, 838, 726, 371)), row.names = c(NA,
-130L), class = c("tbl_df", "tbl", "data.frame"))
因此，我的意图是根据最低的MAPE误差自动选择最佳的三个模型，例如上面突出显示的模型，并按以下示例为商店1和商店2的最佳三个模型按月平均计算。
我尝试使用此代码，但我不知道如何继续。
# Arrange data by MAPE error
Table_1a<-data.frame(Table_1)%>%
select(X1,X2,value)%>%
arrange((value),.by_group = TRUE)
# Select three best models
Table_1b <-data.frame(rbind(Table_1a[1:3, 1:3],Table_1a[10:13, 1:3]))%>%
select(X1,X2)%>%
group_by(X1,X2)
# Тhis line does not work
Forecasting_Store_1<-mutate(Table_2,
ifelse(Table_1b$X1==Table_2$Forecasting_model,Table_2$Forecasting_model,"")
)
那么有人可以帮我解决这个问题吗？
1 个答案:
答案 0 :(得分：1)
这是一个可能的解决方案：
首先，选择3种最佳型号。我更喜欢使用top_n，它与您的解决方案类似，但是更简洁。然后，诀窍是粘贴模型和商店以具有唯一键。
model_ok = Table_1 %>%
group_by(X2) %>%
top_n(-3, value) %>% ungroup %>%
transmute(model_ok=paste(X1,X2)) %>% unlist
请注意，在您的示例示例中，第三名是平局，因此我的代码为商店1选择了4个模型，而不是3个（您的代码也是如此）。
然后，您可以旋转第二张表以使存储区成行而不是列，再次粘贴并过滤与接受的键匹配的行。
table3=Table2 %>%
pivot_longer(c(`Store 1`,`Store 2`), names_to = "store") %>%
mutate(model_store=paste(Forecasting_model, store)) %>%
filter(model_store %in% model_ok) %>%
select(-model_store)
最后，您可以再次旋转表以将模型作为列，并计算3个模型（在我的情况下为4个）的平均值。如果您只有2个商店，则可以通过“ Store 2”重复此代码。
table3 %>%
filter(store=="Store 1") %>%
pivot_wider(names_from = Forecasting_model) %>%
mutate(average=rowMeans(select(., -Date, -store)))
编辑：
由于您似乎有多个商店，因此以下示例说明如何使用purrr::map遍历商店。首先，您需要将不同的存储作为命名向量。我使用了Table_1$X2 %>% unique %>% set_names，但您可能想使用一个更清洁的对象。
library(purrrr)
output=Table_1$X2 %>% unique %>% set_names %>% map(~{
table3 %>%
filter(store==.x) %>%
pivot_wider(names_from = Forecasting_model) %>%
mutate(average=rowMeans(select(., -Date, -store)))
})
output$`Store 1`
output$`Store 2`
希望有帮助。
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