我一直在努力解决这个问题。
我需要根据在我的数据库中另一个“错误”列中检测到的特定评论来创建一个更正的列。我可以使用 grepl 解决这个问题，但我正在努力让 str_detect 也能正常工作(对于大数据集，它通常更快)。
这是一个示例数据库:
test <- tibble(
id = seq(1:30),
date = sample(seq(as.Date('2000/01/01'), as.Date('2018/01/01'), by="day"), 30),
error = c(rep(NA, 3), "wrong date! Correct date = 01.03.2022",
rep(NA, 5), "wrong date! Correct date = 01.05.2021",
rep(NA, 5), "wrong date! Correct date = 01.03.2022",
rep(NA, 7), "wrong date! Correct date = 01.05.2021",
rep(NA, 2), "date already corrected on 01.05.2021",
NA, "date already corrected on 01.03.2022", NA))
我首先尝试使用 str_detect 创建一个新的“date_corr”列:
test %>%
mutate(date_corr=if_else(str_detect(error, "date \\= 01\\.03\\.2022$"), as.Date('2022/03/01'), date),
date_corr=if_else(str_detect(error, "date \\= 01\\.05\\.2021$"), as.Date('2021/05/01'), date_corr))
这会产生:
A tibble: 30 × 4
id date error date_corr
<int> <date> <chr> <date>
1 1 2010-04-28 NA NA
2 2 2004-06-30 NA NA
3 3 2015-09-25 NA NA
4 4 2005-08-21 wrong date! Correct date = 01.03.2022 2022-03-01
5 5 2008-07-16 NA NA
6 6 2004-08-02 NA NA
7 7 2001-10-15 NA NA
8 8 2007-07-21 NA NA
9 9 2014-04-19 NA NA
10 10 2013-02-08 wrong date! Correct date = 01.05.2021 2021-05-01
# … with 20 more rows
按行添加是无关紧要的:
test %>%
rowwise() %>%
mutate(date_corr=if_else(str_detect(error, "date \\= 01\\.03\\.2022$"), as.Date('2022/03/01'), date),
date_corr=if_else(str_detect(error, "date \\= 01\\.05\\.2021$"), as.Date('2021/05/01'), date_corr))
A tibble: 30 × 4
# Rowwise:
id date error date_corr
<int> <date> <chr> <date>
1 1 2010-04-28 NA NA
2 2 2004-06-30 NA NA
3 3 2015-09-25 NA NA
4 4 2005-08-21 wrong date! Correct date = 01.03.2022 2022-03-01
5 5 2008-07-16 NA NA
6 6 2004-08-02 NA NA
7 7 2001-10-15 NA NA
8 8 2007-07-21 NA NA
9 9 2014-04-19 NA NA
10 10 2013-02-08 wrong date! Correct date = 01.05.2021 2021-05-01
# … with 20 more rows
但是，无论按行如何，使用 grepl 我都能得到想要的结果:
test %>%
mutate(date_corr=if_else(grepl("date \\= 01\\.03\\.2022$", error), as.Date('2022/03/01'), date),
date_corr=if_else(grepl("date \\= 01\\.05\\.2021$", error), as.Date('2021/05/01'), date_corr))
# A tibble: 30 × 4
id date error date_corr
<int> <date> <chr> <date>
1 1 2010-04-28 NA 2010-04-28
2 2 2004-06-30 NA 2004-06-30
3 3 2015-09-25 NA 2015-09-25
4 4 2005-08-21 wrong date! Correct date = 01.03.2022 2022-03-01
5 5 2008-07-16 NA 2008-07-16
6 6 2004-08-02 NA 2004-08-02
7 7 2001-10-15 NA 2001-10-15
8 8 2007-07-21 NA 2007-07-21
9 9 2014-04-19 NA 2014-04-19
10 10 2013-02-08 wrong date! Correct date = 01.05.2021 2021-05-01
# … with 20 more rows
test %>%
rowwise() %>%
mutate(date_corr=if_else(grepl("date \\= 01\\.03\\.2022$", error), as.Date('2022/03/01'), date),
date_corr=if_else(grepl("date \\= 01\\.05\\.2021$", error), as.Date('2021/05/01'), date_corr))
A tibble: 30 × 4
# Rowwise:
id date error date_corr
<int> <date> <chr> <date>
1 1 2010-04-28 NA 2010-04-28
2 2 2004-06-30 NA 2004-06-30
3 3 2015-09-25 NA 2015-09-25
4 4 2005-08-21 wrong date! Correct date = 01.03.2022 2022-03-01
5 5 2008-07-16 NA 2008-07-16
6 6 2004-08-02 NA 2004-08-02
7 7 2001-10-15 NA 2001-10-15
8 8 2007-07-21 NA 2007-07-21
9 9 2014-04-19 NA 2014-04-19
10 10 2013-02-08 wrong date! Correct date = 01.05.2021 2021-05-01
# … with 20 more rows
我在这里缺少什么？
最佳答案
区别在于它们处理 NA 值的方式
str_detect(NA, "missing")
# [1] NA
grepl("missing", NA)
# [1] FALSE
请注意，如果您在 if_else 的条件中有一个 NA 值，它也会保留 NA 值
if_else(NA, 1, 2)
# [1] NA
str_detect 保留了 NA 值。目前尚不清楚“正确”的值(value)应该是什么。但是，如果您希望 str_detect 具有与 grepl 相同的值，您可以明确表示不更改 NA 值
test %>%
mutate(date_corr=if_else(!is.na(error) & str_detect(error, "date \\= 01\\.03\\.2022$"), as.Date('2022/03/01'), date),
date_corr=if_else(!is.na(error) & str_detect(error, "date \\= 01\\.05\\.2021$"), as.Date('2021/05/01'), date_corr))
关于r - 为什么 grepl 可以工作但 str_detect 不能根据行值进行变异？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/75412749/