[英]How to do a conditional NA fill in R dataframe
这可能很简单，但无法弄清楚。 如何在数据框dt中使用如下条件填充feature列中的NA 。
填写NA的条件是：
1 ，则用前一行的值填充NA （通过填充 tidyverse 的 function 轻松完成）dt_fl<-dt%>%
fill(feature, .direction = "down")
dt_fl
>1 ，则用先前的特征值 +1 填充NA并将以下行（特征值）替换为1增量以生成连续的特征值。 dt_output显示了在填充NA值并相应地替换特征编号后我对dt的期望。dt<-structure(list(Date = structure(c(15126, 15127, 15128, 15129,
15130, 15131, 15132, 15133, 15134, 15138, 15139, 15140, 15141,
15142, 15143, 15144, 15145, 15146, 15147, 15148, 15149), class = "Date"),
feature = c(1, 1, 1, 1, 1, 1, 1, 1, NA, NA, NA, NA, NA, NA,
2, 2, 2, 2, 2, 2, NA)), row.names = c(NA, -21L), class = c("tbl_df",
"tbl", "data.frame"))
dt
dt_output<-structure(list(Date = structure(c(15126, 15127, 15128, 15129,
15130, 15131, 15132, 15133, 15134, 15138, 15139, 15140, 15141,
15142, 15143, 15144, 15145, 15146, 15147, 15148, 15149), class = "Date"),
feature = c(1, 1, 1, 1, 1, 1, 1, 1, NA, NA, NA, NA, NA, NA,
2, 2, 2, 2, 2, 2, NA), finaloutput = c(1, 1, 1, 1, 1, 1,
1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3)), row.names = c(NA,
-21L), spec = structure(list(cols = list(Date = structure(list(), class = c("collector_character",
"collector")), feature = structure(list(), class = c("collector_double",
"collector")), finaloutput = structure(list(), class = c("collector_double",
"collector"))), default = structure(list(), class = c("collector_guess",
"collector")), skip = 1L), class = "col_spec"), class = c("spec_tbl_df",
"tbl_df", "tbl", "data.frame"))
dt_output
另外，按照 Ben 的建议，如果数据框以dt2中的NA功能开头，如何解决？ dt2的预期 output 在dt2_output
dt2<-structure(list(Date = structure(c(13675, 13676, 13677, 13678,
13679, 13689, 13690, 13691, 13692, 13693, 13694, 13695), class = "Date"),
feature = c(NA, NA, NA, NA, NA, 1, 1, 1, 1, 1, NA, 2)), row.names = c(NA,
-12L), class = c("tbl_df", "tbl", "data.frame"))
dt2_output<-structure(list(Date = structure(c(13675, 13676, 13677, 13678,
13679, 13689, 13690, 13691, 13692, 13693, 13694, 13695), class = "Date"),
feature = c(NA, NA, NA, NA, NA, 1, 1, 1, 1, 1, NA, 2), output_feature = c(1,
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3)), row.names = c(NA, -12L
), spec = structure(list(cols = list(Date = structure(list(), class = c("collector_character",
"collector")), feature = structure(list(), class = c("collector_double",
"collector")), output_feature = structure(list(), class = c("collector_double",
"collector"))), default = structure(list(), class = c("collector_guess",
"collector")), skip = 1L), class = "col_spec"), class = c("spec_tbl_df",
"tbl_df", "tbl", "data.frame"))
Ben 提供的解决方案适用于所有条件，除了dt3中的 1 个条件（如下），只是想知道为什么会这样。 我的假设是第二种解决方案应该为dt3_expected提供dt3 。
dt3<-structure(list(Date = structure(c(10063, 10064, 10065, 10066,
10067, 10068, 10069, 10070, 10079, 10080, 10081, 10082, 10083,
10084, 10085, 10086, 10087, 10088, 10089), class = "Date"), feature = c(1,
1, 1, 1, 1, 1, 1, NA, NA, 2, 2, 2, 2, 2, 2, 2, 2, 2, NA)), row.names = c(NA,
-19L), class = c("tbl_df", "tbl", "data.frame"))
dt3
dt3_expected<-structure(list(Date = structure(c(10063, 10064, 10065, 10066,
10067, 10068, 10069, 10070, 10079, 10080, 10081, 10082, 10083,
10084, 10085, 10086, 10087, 10088, 10089), class = "Date"), feature = c(1,
1, 1, 1, 1, 1, 1, NA, NA, 2, 2, 2, 2, 2, 2, 2, 2, 2, NA), output_feature = c(1,
1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2)), row.names = c(NA,
-19L), spec = structure(list(cols = list(Date = structure(list(), class = c("collector_character",
"collector")), feature = structure(list(), class = c("collector_double",
"collector")), output_feature = structure(list(), class = c("collector_double",
"collector"))), default = structure(list(), class = c("collector_guess",
"collector")), skip = 1L), class = "col_spec"), class = c("spec_tbl_df",
"tbl_df", "tbl", "data.frame"))
非常感谢您的帮助，谢谢。
您可以尝试创建一个“偏移量”，只要您有缺失值且日期差异大于 1 天，就会添加该偏移量。 可以将此累积偏移量添加到您的feature值以确定最终finaloutput 。
dt %>%
mutate(offset = cumsum(is.na(feature) & Date - lag(Date) > 1)) %>%
fill(feature, .direction = "down") %>%
mutate(finaloutput = feature + offset)
Output
# A tibble: 21 x 4
Date feature offset finaloutput
<date> <dbl> <int> <dbl>
1 2011-06-01 1 0 1
2 2011-06-02 1 0 1
3 2011-06-03 1 0 1
4 2011-06-04 1 0 1
5 2011-06-05 1 0 1
6 2011-06-06 1 0 1
7 2011-06-07 1 0 1
8 2011-06-08 1 0 1
9 2011-06-09 1 0 1
10 2011-06-13 1 1 2
11 2011-06-14 1 1 2
12 2011-06-15 1 1 2
13 2011-06-16 1 1 2
14 2011-06-17 1 1 2
15 2011-06-18 2 1 3
16 2011-06-19 2 1 3
17 2011-06-20 2 1 3
18 2011-06-21 2 1 3
19 2011-06-22 2 1 3
20 2011-06-23 2 1 3
21 2011-06-24 2 1 3
编辑：使用以NA开头的第二个示例dt2 ，您可以尝试以下操作。
首先，您可以为lag添加default 。 在第一行是NA的情况下，它将评估Date的差异。 由于没有之前的Date可比较，您可以使用超过 1 天的默认值，以便添加偏移量，这些初始NA将被视为“第一个” feature 。
第二个问题是当您无法fill向下方向时填写NA （以NA开头时没有先前feature值）。 您可以将它们替换为 0。给定offset ，这将成为 0 + 1 = 1 的finaloutput 。
dt2 %>%
mutate(offset = cumsum(is.na(feature) & Date - lag(Date, default = first(Date) - 2) > 1)) %>%
fill(feature, .direction = "down") %>%
replace_na(list(feature = 0)) %>%
mutate(finaloutput = feature + offset)
Output
Date feature offset finaloutput
<date> <dbl> <int> <dbl>
1 2007-06-11 0 1 1
2 2007-06-12 0 1 1
3 2007-06-13 0 1 1
4 2007-06-14 0 1 1
5 2007-06-15 0 1 1
6 2007-06-25 1 1 2
7 2007-06-26 1 1 2
8 2007-06-27 1 1 2
9 2007-06-28 1 1 2
10 2007-06-29 1 1 2
11 2007-06-30 1 1 2
12 2007-07-01 2 1 3
编辑：有额外的评论，还有一个额外的标准要考虑。
如果Date的差异 > 1 并且只有 2 个NA ，则第一个NA应由前一个特征填充，第二个由下一个特征填充。 特别是，2 NA中存在差距的第二个应以不同方式处理。
一种方法是计算连续NA的数量。 然后，可以针对这种特殊情况填充feature ，其中两个NA中的第二个用Date间隙标识。
dt3 %>%
mutate(grp = cumsum(c(1, abs(diff(is.na(feature))) == 1))) %>%
add_count(grp) %>%
ungroup %>%
mutate(feature = ifelse(is.na(feature) & n == 2 & is.na(lag(feature)), lead(feature), feature)) %>%
mutate(offset = cumsum(is.na(feature) & Date - lag(Date, default = first(Date) - 2) > 1)) %>%
fill(feature, .direction = "down") %>%
replace_na(list(feature = 0)) %>%
mutate(finaloutput = feature + offset)
Output
Date feature grp n offset finaloutput
<date> <dbl> <dbl> <int> <int> <dbl>
1 1997-07-21 1 1 7 0 1
2 1997-07-22 1 1 7 0 1
3 1997-07-23 1 1 7 0 1
4 1997-07-24 1 1 7 0 1
5 1997-07-25 1 1 7 0 1
6 1997-07-26 1 1 7 0 1
7 1997-07-27 1 1 7 0 1
8 1997-07-28 1 2 2 0 1
9 1997-08-06 2 2 2 0 2
10 1997-08-07 2 3 9 0 2
11 1997-08-08 2 3 9 0 2
12 1997-08-09 2 3 9 0 2
13 1997-08-10 2 3 9 0 2
14 1997-08-11 2 3 9 0 2
15 1997-08-12 2 3 9 0 2
16 1997-08-13 2 3 9 0 2
17 1997-08-14 2 3 9 0 2
18 1997-08-15 2 3 9 0 2
19 1997-08-16 2 4 1 0 2
请注意，这可以简化； 但在这样做之前，需要确保这符合您的需求。
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