将值分隔到 R 中的现有列中
Separating values into existing column in R
我正在整理我使用 tabulizer 从 PDF 读入 R 的一些数据。不幸的是，有些单元格没有被正确读取。在第 9 列（37.1 公里处的第 5 段）中，第 3 行和第 4 行包含本应在第 10 列（最终时间）中结束的信息。
如何仅为这些行分隔列 (9) 并将必要的数据粘贴到现有的列 (10) 中？
我知道如何使用 tidyr::separate 函数，但不知道如何（如果）在这里应用它。任何帮助和指导将不胜感激。
structure(list(Rank = c("23", "24", "25", "26"), `Race Number` = c("13",
"11", "29", "30"), Name = c("FOSS Tobias S.", "McNULTY Brandon",
"BENNETT George", "KUKRLE Michael"), `NOC Code` = c("NOR", "USA",
"NZL", "CZE"), `Split 1 at 9.7km` = c("13:47.65(22)", "13:28.23(15)",
"14:05.46(30)", "14:05.81(32)"), `Split 2 at 15.0km` = c("19:21.16(22)",
"19:04.80(18)", "19:47.53(31)", "19:48.77(32)"), `Split 3 at 22.1km` = c("29:17.44(24)",
"29:01.94(20)", "29:58.88(28)", "29:58.09(27)"), `Split 4 at 31.8km` = c("44:06.82(24)",
"43:51.67(23)", "44:40.28(25)", "44:42.74(26)"), `Split 5 at 37.1km` = c("49:49.65(24)",
"49:40.49(23)", "50:21.82(25)1:00:28.39 (25)", "50:30.02(26)1:00:41.55 (26)"
), `Final Time` = c("59:51.68 (23)", "59:57.73 (24)", "", ""),
`Time Behind` = c("+4:47.49", "+4:53.54", "+5:24.20", "+5:37.36"
), `Average Speed` = c("44.302", "44.228", "43.854", "43.696"
)), class = "data.frame", row.names = c(NA, -4L))
正在调用 df 到您的数据框：
library(tidyr)
library(dplyr)
df %>%
separate(`Split 5 at 37.1km`, into = c("Split 5 at 37.1km","aux"), sep = "\)") %>%
mutate(`Final Time` = coalesce(if_else(`Final Time`!="",`Final Time`, NA_character_), paste0(aux, ")")),
aux = NULL,
`Split 5 at 37.1km` = paste0(`Split 5 at 37.1km`, ")"))
Rank Race Number Name NOC Code Split 1 at 9.7km Split 2 at 15.0km Split 3 at 22.1km Split 4 at 31.8km Split 5 at 37.1km Final Time
1 23 13 FOSS Tobias S. NOR 13:47.65(22) 19:21.16(22) 29:17.44(24) 44:06.82(24) 49:49.65(24) 59:51.68 (23)
2 24 11 McNULTY Brandon USA 13:28.23(15) 19:04.80(18) 29:01.94(20) 43:51.67(23) 49:40.49(23) 59:57.73 (24)
3 25 29 BENNETT George NZL 14:05.46(30) 19:47.53(31) 29:58.88(28) 44:40.28(25) 50:21.82(25) 1:00:28.39 (25)
4 26 30 KUKRLE Michael CZE 14:05.81(32) 19:48.77(32) 29:58.09(27) 44:42.74(26) 50:30.02(26) 1:00:41.55 (26)
Time Behind Average Speed
1 +4:47.49 44.302
2 +4:53.54 44.228
3 +5:24.20 43.854
4 +5:37.36 43.696
我的回答不是很花哨，但它对最后时间列中的任何数字都有效。只要末尾的括号中始终有数字，它就可以工作。
# dummy df
df <- data.frame("split" = c("49:49.65(24)", "49:40.49(23)", "50:21.82(25)1:00:28.39 (25)", "50:30.02(26)1:00:41.55 (26)"),
"final" = c("59:51.68 (23)", "59:57.73 (24)", "", ""))
# combining & splitting strings
merge_strings <- paste0(df$split, df$final)
split_strings <- strsplit(merge_strings, ")")
df$split <- paste0(unlist(lapply(split_strings, "[[", 1)),")")
df$final <- paste0(unlist(lapply(split_strings, "[[", 2)),")")
这给出：
split final
1 49:49.65(24) 59:51.68 (23)
2 49:40.49(23) 59:57.73 (24)
3 50:21.82(25) 1:00:28.39 (25)
4 50:30.02(26) 1:00:41.55 (26)
您可以使用 dplyr 和 stringr:
library(dplyr)
library(stringr)
data %>%
mutate(`Final Time` = ifelse(`Final Time` == "", str_remove(`Split 5 at 37.1km`, "\d+:\d+\.\d+\(\d+\)"), `Final Time`),
`Split 5 at 37.1km` = str_extract(`Split 5 at 37.1km`, "\d+:\d+\.\d+\(\d+\)"))
哪个returns
Rank Race Number Name NOC Code Split 1 at 9.7km Split 2 at 15.0km Split 3 at 22.1km Split 4 at 31.8km
1 23 13 FOSS Tobias S. NOR 13:47.65(22) 19:21.16(22) 29:17.44(24) 44:06.82(24)
2 24 11 McNULTY Brandon USA 13:28.23(15) 19:04.80(18) 29:01.94(20) 43:51.67(23)
3 25 29 BENNETT George NZL 14:05.46(30) 19:47.53(31) 29:58.88(28) 44:40.28(25)
4 26 30 KUKRLE Michael CZE 14:05.81(32) 19:48.77(32) 29:58.09(27) 44:42.74(26)
Split 5 at 37.1km Final Time Time Behind Average Speed
1 49:49.65(24) 59:51.68 (23) +4:47.49 44.302
2 49:40.49(23) 59:57.73 (24) +4:53.54 44.228
3 50:21.82(25) 1:00:28.39 (25) +5:24.20 43.854
4 50:30.02(26) 1:00:41.55 (26) +5:37.36 43.696
我喜欢使用 regex 和 stringr。虽然这里有一些次优代码，但关键步骤是 str_extract()。使用这个我们可以 select 我们想要的两个子串，第一次的和第二次的。如果任何一个时间缺失，那么我们将有一个缺失值。因此，我们可以根据缺失发生的位置填写列。
正则字符串如下^((\d+:)?\d{2}:\d{2}.\d{2}\(\d+\))\.?+((\d+:)?\d{2}:\d{2}.\d{2} \(\d+\))$。这里我们有 4 个捕获组，第一组和第三组分别捕获两个完整时间。第二个和第四个 select 包含小时的可选组（这确保完全捕获超过一个小时的时间。此外，我们检查可选的 space.
我的代码如下：
library(tidyverse)
data <- structure(list(Rank = c("23", "24", "25", "26"), `Race Number` = c("13",
"11", "29", "30"), Name = c("FOSS Tobias S.", "McNULTY Brandon",
"BENNETT George", "KUKRLE Michael"), `NOC Code` = c("NOR", "USA",
"NZL", "CZE"), `Split 1 at 9.7km` = c("13:47.65(22)", "13:28.23(15)",
"14:05.46(30)", "14:05.81(32)"), `Split 2 at 15.0km` = c("19:21.16(22)",
"19:04.80(18)", "19:47.53(31)", "19:48.77(32)"), `Split 3 at 22.1km` = c("29:17.44(24)",
"29:01.94(20)", "29:58.88(28)", "29:58.09(27)"), `Split 4 at 31.8km` = c("44:06.82(24)",
"43:51.67(23)", "44:40.28(25)", "44:42.74(26)"), `Split 5 at 37.1km` = c("49:49.65(24)",
"49:40.49(23)", "50:21.82(25)1:00:28.39 (25)", "50:30.02(26)1:00:41.55 (26)"
), `Final Time` = c("59:51.68 (23)", "59:57.73 (24)", "", ""),
`Time Behind` = c("+4:47.49", "+4:53.54", "+5:24.20", "+5:37.36"
), `Average Speed` = c("44.302", "44.228", "43.854", "43.696"
)), class = "data.frame", row.names = c(NA, -4L))
# Take data and use a matching string to the regex pattern
data |>
mutate(match = map(`Split 5 at 37.1km`, ~unlist(str_match(., "^((\d+:)?\d{2}:\d{2}.\d{2}\(\d+\))((\d+:)?\d{2}:\d{2}.\d{2} ?\(\d+\))$")))) |>
# Grab the strings that match the whole first and second/final times
mutate(match1 = map(match, ~.[[2]]), match2 = map(match, ~.[[4]]), .keep = "unused") |>
# Check where the NAs are and put into the dataframe accordingly
mutate(`Split 5 at 37.1km`= ifelse(is.na(match1), `Split 5 at 37.1km`, match1),
`Final Time` = ifelse(is.na(match2), `Final Time`, match2), .keep = "unused")
#> Rank Race Number Name NOC Code Split 1 at 9.7km Split 2 at 15.0km
#> 1 23 13 FOSS Tobias S. NOR 13:47.65(22) 19:21.16(22)
#> 2 24 11 McNULTY Brandon USA 13:28.23(15) 19:04.80(18)
#> 3 25 29 BENNETT George NZL 14:05.46(30) 19:47.53(31)
#> 4 26 30 KUKRLE Michael CZE 14:05.81(32) 19:48.77(32)
#> Split 3 at 22.1km Split 4 at 31.8km Split 5 at 37.1km Final Time
#> 1 29:17.44(24) 44:06.82(24) 49:49.65(24) 59:51.68 (23)
#> 2 29:01.94(20) 43:51.67(23) 49:40.49(23) 59:57.73 (24)
#> 3 29:58.88(28) 44:40.28(25) 50:21.82(25) 1:00:28.39 (25)
#> 4 29:58.09(27) 44:42.74(26) 50:30.02(26) 1:00:41.55 (26)
#> Time Behind Average Speed
#> 1 +4:47.49 44.302
#> 2 +4:53.54 44.228
#> 3 +5:24.20 43.854
#> 4 +5:37.36 43.696
由 reprex package (v2.0.0)
于 2021-07-28 创建
注意上面我使用的是 R 4.1 以上的基管 |> 如果您使用的是较早的 R 版本，这可以简单地用 magrittr 管替换 %>%。
我正在整理我使用 tabulizer 从 PDF 读入 R 的一些数据。不幸的是，有些单元格没有被正确读取。在第 9 列（37.1 公里处的第 5 段）中，第 3 行和第 4 行包含本应在第 10 列（最终时间）中结束的信息。
如何仅为这些行分隔列 (9) 并将必要的数据粘贴到现有的列 (10) 中？
我知道如何使用 tidyr::separate 函数，但不知道如何（如果）在这里应用它。任何帮助和指导将不胜感激。
structure(list(Rank = c("23", "24", "25", "26"), `Race Number` = c("13",
"11", "29", "30"), Name = c("FOSS Tobias S.", "McNULTY Brandon",
"BENNETT George", "KUKRLE Michael"), `NOC Code` = c("NOR", "USA",
"NZL", "CZE"), `Split 1 at 9.7km` = c("13:47.65(22)", "13:28.23(15)",
"14:05.46(30)", "14:05.81(32)"), `Split 2 at 15.0km` = c("19:21.16(22)",
"19:04.80(18)", "19:47.53(31)", "19:48.77(32)"), `Split 3 at 22.1km` = c("29:17.44(24)",
"29:01.94(20)", "29:58.88(28)", "29:58.09(27)"), `Split 4 at 31.8km` = c("44:06.82(24)",
"43:51.67(23)", "44:40.28(25)", "44:42.74(26)"), `Split 5 at 37.1km` = c("49:49.65(24)",
"49:40.49(23)", "50:21.82(25)1:00:28.39 (25)", "50:30.02(26)1:00:41.55 (26)"
), `Final Time` = c("59:51.68 (23)", "59:57.73 (24)", "", ""),
`Time Behind` = c("+4:47.49", "+4:53.54", "+5:24.20", "+5:37.36"
), `Average Speed` = c("44.302", "44.228", "43.854", "43.696"
)), class = "data.frame", row.names = c(NA, -4L))
正在调用 df 到您的数据框：
library(tidyr)
library(dplyr)
df %>%
separate(`Split 5 at 37.1km`, into = c("Split 5 at 37.1km","aux"), sep = "\)") %>%
mutate(`Final Time` = coalesce(if_else(`Final Time`!="",`Final Time`, NA_character_), paste0(aux, ")")),
aux = NULL,
`Split 5 at 37.1km` = paste0(`Split 5 at 37.1km`, ")"))
Rank Race Number Name NOC Code Split 1 at 9.7km Split 2 at 15.0km Split 3 at 22.1km Split 4 at 31.8km Split 5 at 37.1km Final Time
1 23 13 FOSS Tobias S. NOR 13:47.65(22) 19:21.16(22) 29:17.44(24) 44:06.82(24) 49:49.65(24) 59:51.68 (23)
2 24 11 McNULTY Brandon USA 13:28.23(15) 19:04.80(18) 29:01.94(20) 43:51.67(23) 49:40.49(23) 59:57.73 (24)
3 25 29 BENNETT George NZL 14:05.46(30) 19:47.53(31) 29:58.88(28) 44:40.28(25) 50:21.82(25) 1:00:28.39 (25)
4 26 30 KUKRLE Michael CZE 14:05.81(32) 19:48.77(32) 29:58.09(27) 44:42.74(26) 50:30.02(26) 1:00:41.55 (26)
Time Behind Average Speed
1 +4:47.49 44.302
2 +4:53.54 44.228
3 +5:24.20 43.854
4 +5:37.36 43.696
我的回答不是很花哨，但它对最后时间列中的任何数字都有效。只要末尾的括号中始终有数字，它就可以工作。
# dummy df
df <- data.frame("split" = c("49:49.65(24)", "49:40.49(23)", "50:21.82(25)1:00:28.39 (25)", "50:30.02(26)1:00:41.55 (26)"),
"final" = c("59:51.68 (23)", "59:57.73 (24)", "", ""))
# combining & splitting strings
merge_strings <- paste0(df$split, df$final)
split_strings <- strsplit(merge_strings, ")")
df$split <- paste0(unlist(lapply(split_strings, "[[", 1)),")")
df$final <- paste0(unlist(lapply(split_strings, "[[", 2)),")")
这给出：
split final
1 49:49.65(24) 59:51.68 (23)
2 49:40.49(23) 59:57.73 (24)
3 50:21.82(25) 1:00:28.39 (25)
4 50:30.02(26) 1:00:41.55 (26)
您可以使用 dplyr 和 stringr:
library(dplyr)
library(stringr)
data %>%
mutate(`Final Time` = ifelse(`Final Time` == "", str_remove(`Split 5 at 37.1km`, "\d+:\d+\.\d+\(\d+\)"), `Final Time`),
`Split 5 at 37.1km` = str_extract(`Split 5 at 37.1km`, "\d+:\d+\.\d+\(\d+\)"))
哪个returns
Rank Race Number Name NOC Code Split 1 at 9.7km Split 2 at 15.0km Split 3 at 22.1km Split 4 at 31.8km
1 23 13 FOSS Tobias S. NOR 13:47.65(22) 19:21.16(22) 29:17.44(24) 44:06.82(24)
2 24 11 McNULTY Brandon USA 13:28.23(15) 19:04.80(18) 29:01.94(20) 43:51.67(23)
3 25 29 BENNETT George NZL 14:05.46(30) 19:47.53(31) 29:58.88(28) 44:40.28(25)
4 26 30 KUKRLE Michael CZE 14:05.81(32) 19:48.77(32) 29:58.09(27) 44:42.74(26)
Split 5 at 37.1km Final Time Time Behind Average Speed
1 49:49.65(24) 59:51.68 (23) +4:47.49 44.302
2 49:40.49(23) 59:57.73 (24) +4:53.54 44.228
3 50:21.82(25) 1:00:28.39 (25) +5:24.20 43.854
4 50:30.02(26) 1:00:41.55 (26) +5:37.36 43.696
我喜欢使用 regex 和 stringr。虽然这里有一些次优代码，但关键步骤是 str_extract()。使用这个我们可以 select 我们想要的两个子串，第一次的和第二次的。如果任何一个时间缺失，那么我们将有一个缺失值。因此，我们可以根据缺失发生的位置填写列。
正则字符串如下^((\d+:)?\d{2}:\d{2}.\d{2}\(\d+\))\.?+((\d+:)?\d{2}:\d{2}.\d{2} \(\d+\))$。这里我们有 4 个捕获组，第一组和第三组分别捕获两个完整时间。第二个和第四个 select 包含小时的可选组（这确保完全捕获超过一个小时的时间。此外，我们检查可选的 space.
我的代码如下：
library(tidyverse)
data <- structure(list(Rank = c("23", "24", "25", "26"), `Race Number` = c("13",
"11", "29", "30"), Name = c("FOSS Tobias S.", "McNULTY Brandon",
"BENNETT George", "KUKRLE Michael"), `NOC Code` = c("NOR", "USA",
"NZL", "CZE"), `Split 1 at 9.7km` = c("13:47.65(22)", "13:28.23(15)",
"14:05.46(30)", "14:05.81(32)"), `Split 2 at 15.0km` = c("19:21.16(22)",
"19:04.80(18)", "19:47.53(31)", "19:48.77(32)"), `Split 3 at 22.1km` = c("29:17.44(24)",
"29:01.94(20)", "29:58.88(28)", "29:58.09(27)"), `Split 4 at 31.8km` = c("44:06.82(24)",
"43:51.67(23)", "44:40.28(25)", "44:42.74(26)"), `Split 5 at 37.1km` = c("49:49.65(24)",
"49:40.49(23)", "50:21.82(25)1:00:28.39 (25)", "50:30.02(26)1:00:41.55 (26)"
), `Final Time` = c("59:51.68 (23)", "59:57.73 (24)", "", ""),
`Time Behind` = c("+4:47.49", "+4:53.54", "+5:24.20", "+5:37.36"
), `Average Speed` = c("44.302", "44.228", "43.854", "43.696"
)), class = "data.frame", row.names = c(NA, -4L))
# Take data and use a matching string to the regex pattern
data |>
mutate(match = map(`Split 5 at 37.1km`, ~unlist(str_match(., "^((\d+:)?\d{2}:\d{2}.\d{2}\(\d+\))((\d+:)?\d{2}:\d{2}.\d{2} ?\(\d+\))$")))) |>
# Grab the strings that match the whole first and second/final times
mutate(match1 = map(match, ~.[[2]]), match2 = map(match, ~.[[4]]), .keep = "unused") |>
# Check where the NAs are and put into the dataframe accordingly
mutate(`Split 5 at 37.1km`= ifelse(is.na(match1), `Split 5 at 37.1km`, match1),
`Final Time` = ifelse(is.na(match2), `Final Time`, match2), .keep = "unused")
#> Rank Race Number Name NOC Code Split 1 at 9.7km Split 2 at 15.0km
#> 1 23 13 FOSS Tobias S. NOR 13:47.65(22) 19:21.16(22)
#> 2 24 11 McNULTY Brandon USA 13:28.23(15) 19:04.80(18)
#> 3 25 29 BENNETT George NZL 14:05.46(30) 19:47.53(31)
#> 4 26 30 KUKRLE Michael CZE 14:05.81(32) 19:48.77(32)
#> Split 3 at 22.1km Split 4 at 31.8km Split 5 at 37.1km Final Time
#> 1 29:17.44(24) 44:06.82(24) 49:49.65(24) 59:51.68 (23)
#> 2 29:01.94(20) 43:51.67(23) 49:40.49(23) 59:57.73 (24)
#> 3 29:58.88(28) 44:40.28(25) 50:21.82(25) 1:00:28.39 (25)
#> 4 29:58.09(27) 44:42.74(26) 50:30.02(26) 1:00:41.55 (26)
#> Time Behind Average Speed
#> 1 +4:47.49 44.302
#> 2 +4:53.54 44.228
#> 3 +5:24.20 43.854
#> 4 +5:37.36 43.696
由 reprex package (v2.0.0)
于 2021-07-28 创建注意上面我使用的是 R 4.1 以上的基管 |> 如果您使用的是较早的 R 版本，这可以简单地用 magrittr 管替换 %>%。