R中扫描的日期减法答案

【问题标题】：Date Subtraction for Sweep in RR中扫描的日期减法
【发布时间】：2021-01-17 17:01:18
【问题描述】：

我有一个如下所示的数据向量：

还有一个包含日期列的数据框，如下所示：

此数据框中的行数等于 new_first_dates 的长度。

我想从数据框的每一行中减去 new_first_dates。我尝试使用扫描，但它不起作用。

有人有什么想法吗？

干杯

附上数据：

structure(list(`>50 gatherings` = structure(c(18346, 18342, 18339, 18341, 18346, 18339, 18339, 18355, 18340, 18346, 18339, 18338, 18346, 18349, 18343, 18347, 18345, 18340, 18337, 18351, 18338, 18345, 18338, 21915, 18347, 18343, 18340, 18346, 18339), class = "Date"),     `>500 gatherings` = structure(c(18346, 18335, 18339, 18341,     18335, 18339, 18339, 18355, 18340, 18335, 18339, 18335, 18346,     18349, 18343, 18347, 18334, 18340, 18337, 18351, 18338, 18345,     18338, 21915, 18347, 18336, 18334, 18346, 18339), class = "Date"),     `public schools` = structure(c(18341, 18338, 18338, 18341,     18345, 18338, 18340, 18355, 18339, 18341, 18340, 18338, 18338,     18340, 18341, 18342, 18338, 18338, 18339, 18339, 18338, 18338,     18338, 18338, 18342, 18345, 18338, 18338, 18340), class = "Date"),     `restaurant dine-in` = structure(c(18339, 18341, 18341, 18337,     18342, 18342, 18342, 18339, 18340, 18339, 18346, 18338, 18338,     18339, 18343, 18347, 18339, 18340, 18337, 18354, 18339, 18339,     18345, 18345, 18345, 18341, 18340, 18345, 18339), class = "Date"),     `entertainment/gym` = structure(c(18339, 18350, 18341, 18337,     18339, 18342, 18342, 18339, 18340, 18339, 18346, 18338, 18338,     18339, 18359, 18347, 18345, 18340, 18344, 18354, 18339, 18339,     18345, 21915, 18345, 18341, 18340, 18338, 18339), class = "Date"),     `federal guidelines` = structure(c(18338, 18338, 18338, 18338,     18338, 18338, 18338, 18338, 18338, 18338, 18338, 18338, 18338,     18338, 18338, 18338, 18338, 18338, 18338, 18338, 18338, 18338,     18338, 18338, 18338, 18338, 18338, 18338, 18338), class = "Date"),     `foreign travel ban` = structure(c(18333, 18333, 18333, 18333,     18333, 18333, 18333, 18333, 18333, 18333, 18333, 18333, 18333,     18333, 18333, 18333, 18333, 18333, 18333, 18333, 18333, 18333,     18333, 18333, 18333, 18333, 18333, 18333, 18333), class = "Date"),     `stay at home rollback` = structure(c(18377, 18383, 18398,     18392, 18379, 18432, 18383, 21915, 18413, 18387, 18398, 18415,     18380, 18398, 18387, 18379, 18403, 18391, 18424, 18419, 18392,     18387, 18409, 21915, 18398, 18419, 21915, 18396, 18396), class = "Date"),     `>50 gatherings rollback` = structure(c(18391, 21915, 18415,     21915, 21915, 21915, 21915, 21915, 18426, 18405, 18398, 21915,     21915, 18387, 18429, 18415, 21915, 21915, 18404, 21915, 21915,     21915, 21915, 18384, 21915, 21915, 21915, 21915, 21915), class = "Date"),     `>500 gatherings rollback` = structure(c(18405, 21915, 18415,     21915, 21915, 21915, 21915, 21915, 18426, 18405, 18398, 21915,     21915, 18387, 18429, 18415, 21915, 21915, 21915, 21915, 21915,     21915, 21915, 18384, 21915, 21915, 21915, 21915, 21915), class = "Date"),     `restaurant dine-in rollback` = structure(c(18405, 18394,     21915, 18392, 18391, 18432, 18380, 18394, 18391, 21915, 18384,     18390, 18380, 18424, 18429, 18384, 18387, 18411, 21915, 18391,     21915, 21915, 18409, 21915, 18387, 18398, 18384, 18394, 21915    ), class = "Date"), `entertainment/gym rollback` = structure(c(18405,     18394, 21915, 18392, 18391, 18432, 18377, 18394, 18391, 21915,     18384, 18390, 18380, 18424, 18429, 18384, 21915, 18415, 21915,     18391, 21915, 21915, 18409, 21915, 18387, 18398, 18384, 18394,     21915), class = "Date"), State = c("AK", "AL", "AZ", "CA",     "CO", "FL", "GA", "IA", "IL", "IN", "KS", "MD", "MI", "MN",     "MO", "MS", "NC", "NM", "OH", "OK", "OR", "PA", "RI", "SD",     "TN", "TX", "UT", "VA", "WI"), cluster = c(3, 1, 1, 1, 1,     1, 1, 1, 2, 3, 3, 2, 3, 1, 3, 1, 1, 1, 1, 3, 1, 2, 2, 3,     1, 1, 1, 1, 1)), row.names = c(NA, -29L), class = c("grouped_df", "tbl_df", "tbl", "data.frame"), groups = structure(list(State = c("AK", "AL", "AZ", "CA", "CO", "FL", "GA", "IA", "IL", "IN", "KS", "MD", "MI", "MN", "MO", "MS", "NC", "NM", "OH", "OK", "OR", "PA", "RI", "SD", "TN", "TX", "UT", "VA", "WI"), .rows = list(1L, 2L, 3L,     4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L,     17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L,     29L)), row.names = c(NA, -29L), class = c("tbl_df", "tbl", "data.frame"), .drop = TRUE))

structure(c(18338, 18345, 18340, 18325, 18332, 18334, 18333, 18333, 18339, 18339, 18340, 18337, 18335, 18337, 18339, 18340, 18340, 18336, 18337, 18340, 18335, 18334, 18339, 18346, 18335, 18334, 18338, 18335, 18337), class = "Date")

【问题讨论】：

您要查找每行的最小日期并添加一列吗？
以天为单位的日期之间的差异，因此对于第 1 行中的每个日期，我想找到它与 new_first_dates 中的第一个日期之间的差异，对于第 2、3、4 行类似。等

标签： r date tidyverse data-manipulation

【解决方案1】：

某些列不是Date 类。所以，我们需要在做减法之前排除那些。另外，根据OP的帖子结构，数据集中有一些属性，可以用ungrouping去掉

library(dplyr) # version >= 1.0.0
df1 %>%
  ungroup %>%
  mutate(across(-c(State, cluster),  ~ . - as.Date(v1)))

-输出

# A tibble: 29 x 14
#   `>50 gatherings` `>500 gathering… `public schools` `restaurant din… `entertainment/… `federal guidel… `foreign travel… `stay at home r…
#   <drtn>           <drtn>           <drtn>           <drtn>           <drtn>           <drtn>           <drtn>           <drtn>          
# 1  8 days            8 days          3 days           1 days           1 days           0 days           -5 days           39 days       
# 2 -3 days          -10 days         -7 days          -4 days           5 days          -7 days          -12 days           38 days       
# 3 -1 days           -1 days         -2 days           1 days           1 days          -2 days           -7 days           58 days       
# 4 16 days           16 days         16 days          12 days          12 days          13 days            8 days           67 days       
# 5 14 days            3 days         13 days          10 days           7 days           6 days            1 days           47 days       
# 6  5 days            5 days          4 days           8 days           8 days           4 days           -1 days           98 days       
# 7  6 days            6 days          7 days           9 days           9 days           5 days            0 days           50 days       
# 8 22 days           22 days         22 days           6 days           6 days           5 days            0 days         3582 days       
# 9  1 days            1 days          0 days           1 days           1 days          -1 days           -6 days           74 days       
#10  7 days           -4 days          2 days           0 days           0 days          -1 days           -6 days           48 days       
# … with 19 more rows, and 6 more variables: `>50 gatherings rollback` <drtn>, `>500 gatherings rollback` <drtn>, `restaurant dine-in
#   rollback` <drtn>, `entertainment/gym rollback` <drtn>, State <chr>, cluster <dbl>

或者如果 OP 更喜欢 sweep

df1[1:12] <- sweep(as.data.frame(df1[1:12]), 1, as.Date(v1),  '-')

或者使用来自base R的lapply

df1[1:12] <- lapply(df1[1:12], `-`, as.Date(v1))

【讨论】：

谢谢！对我来说，我收到一条错误消息，说找不到功能，我有 dplyr 但它不起作用。
嗨 Akrun，sweep 给了我日期作为输出，不知道为什么

【解决方案2】：

如果您的数据集很大，您可能需要考虑使用 data.table 包来实现就地修改语义。

library( data.table )
dates_dataset = data.table( dates_dataset )

# Find the columns that are date class or otherwise select the columns you want to subtract
date_columns = names(dates_dataset)[ sapply( dates_dataset, class ) == "Date" ]

# Use the := in-place assignment operator to update the columns
dates_dataset[ , eval(date_columns) := {
    lapply( date_columns, function(x){
        get( x ) - new_first_dates
    } )
} ]

如果您想创建新列，您可以选择将eval(date_columns) 更改为其他内容。

输出是：

get_vars = c(">50 gatherings", ">500 gatherings")
head( dates_dataset[,mget(get_vars) ] )

   >50 gatherings >500 gatherings
1:         8 days          8 days
2:        -3 days        -10 days
3:        -1 days         -1 days
4:        16 days         16 days
5:        14 days          3 days
6:         5 days          5 days

【讨论】：