如何更改R中的基准参考年

Question

我有一个非常大的数据框，其中包含 20 年 400 种商品的每月价格指数。对于我的分析，我需要将指数跨年链接起来，以提供一致的基准参考期。本质上，一月指数是相对于上一月计算的。这张图片显示了如何进行链接的简单示例，突出显示了一月份。 item_index 列显示原始索引，new-index 列显示所需索引。在此示例中，new_index 2018:01=D14xE13/100、2018:02=D15xE14/100 以及...，直到 2019:01=D26xE25/100 等等。如果您能帮助我以有效的方式编写代码，我真的很感激。

df <- structure(list(index_date = structure(c(17167, 17198, 17226, 
17257, 17287, 17318, 17348, 17379, 17410, 17440, 17471, 17501, 
17532, 17563, 17591, 17622, 17652, 17683, 17713, 17744, 17775, 
17805, 17836, 17866, 17897, 17928, 17956, 17987, 18017, 18048, 
18078, 18109, 18140, 18170, 18201, 18231, 18262, 18293, 18322, 
18353, 18383, 18414, 18444, 18475, 18506, 18536, 18567, 18597
), class = "Date"), item_id = c(310405, 310405, 310405, 310405, 
310405, 310405, 310405, 310405, 310405, 310405, 310405, 310405, 
310405, 310405, 310405, 310405, 310405, 310405, 310405, 310405, 
310405, 310405, 310405, 310405, 310405, 310405, 310405, 310405, 
310405, 310405, 310405, 310405, 310405, 310405, 310405, 310405, 
310405, 310405, 310405, 310405, 310405, 310405, 310405, 310405, 
310405, 310405, 310405, 310405), base_date = c(201601, 201701, 
201701, 201701, 201701, 201701, 201701, 201701, 201701, 201701, 
201701, 201701, 201712, 201801, 201801, 201801, 201801, 201801, 
201801, 201801, 201801, 201801, 201801, 201801, 201812, 201901, 
201901, 201901, 201901, 201901, 201901, 201901, 201901, 201901, 
201901, 201901, 201912, 202001, 202001, 202001, 202001, 202001, 
202001, 202001, 202001, 202001, 202001, 202001), item_index = c(98.258, 
99.397, 99.947, 96.607, 98.417, 102.261, 101.719, 102.018, 99.88, 
100.447, 95.759, 95.334, 103.718, 100.758, 97.906, 93.305, 98.987, 
96.349, 93.586, 100.091, 97.8, 95.633, 93.759, 92.471, 101.023, 
97.782, 99.697, 94.008, 97.942, 98.874, 95.886, 99.385, 97.472, 
95.792, 98.138, 95.18, 101.098, 99.525, 98.032, 99.571, 96.245, 
93.816, 95.445, 99.266, 97.008, 99.151, 96.824, 92.32), new_index = c(98.258, 
99.397, 99.947, 96.607, 98.417, 102.261, 101.719, 102.018, 99.88, 
100.447, 95.759, 95.334, 98.87851812, 99.6280172873496, 96.8080019505672, 
92.258601331866, 97.8768787314444, 95.2684634234388, 92.5364499677832, 
98.9684975714892, 96.70319072136, 94.5604932336996, 92.7075098041308, 
91.4339544907452, 92.3693238451855, 90.3205722422993, 92.0894447939346, 
86.834553960382, 90.4683631604516, 91.3292452586887, 88.5692498621946, 
91.8012525035377, 90.0342273383792, 88.4824226977801, 90.6494070351882, 
87.9171224358476, 88.8824524401932, 88.4602607911023, 87.1332457761702, 
88.5011467192248, 85.544916351064, 83.3859615812917, 84.8338567315424, 
88.2300552392822, 86.2230894631826, 88.127840418976, 86.0595457506927, 
82.0562800927864)), row.names = c(NA, -48L), class = c("tbl_df", 
"tbl", "data.frame"))

Answer 1

听起来你想要这样的东西：

dplyr::mutate(df, out = item_index * first(new_index) / 100, .by = base_date)

输出：

   index_date item_id base_date item_index new_index   out
   <date>       <dbl>     <dbl>      <dbl>     <dbl> <dbl>
 1 2017-01-01  310405    201601       98.3      98.3  96.5
 2 2017-02-01  310405    201701       99.4      99.4  98.8
 3 2017-03-01  310405    201701       99.9      99.9  99.3
 4 2017-04-01  310405    201701       96.6      96.6  96.0
 5 2017-05-01  310405    201701       98.4      98.4  97.8
 6 2017-06-01  310405    201701      102.      102.  102. 
 7 2017-07-01  310405    201701      102.      102.  101. 
 8 2017-08-01  310405    201701      102.      102.  101. 
 9 2017-09-01  310405    201701       99.9      99.9  99.3
10 2017-10-01  310405    201701      100.      100.   99.8
# ℹ 38 more rows