如何根據多個條件填充單元格？
這個游戲有很多玩家（列），但為了這個例子，我只包括了 2 個。我想回圈很多玩家。
每行代表一個游戲回合。
條件：
如果 player00[i] 得分 = 0 &
IF lossallowed00[i] = "否"
然后用“FLAG”填充flag00 [i]
df <-data.frame(player001 = c(1,0,3),
player002 = c(1,0,5),
lossallowed001 = c("no", "yes", "no"),
lossallowed002 = c("no", "no", "yes"),
flag001 = NA, flag002 = NA)
#desired output:
#player001 player002 lossallowed001 lossallowed002 flag001 flag002
# 1 1 no no NA NA
# 0 0 yes no NA FLAG
# 3 5 no yes NA NA
uj5u.com熱心網友回復：
如果您使用一種重塑為長格式的方法，根據列名的模式拆分 ID，列名是由字母組成的變數，ID 是由數字組成的，您可以在幾行中一次完成所有操作，然后重新調整為寬的。使用正則運算式意味著您不受玩家數量或列名稱的約束。我為游戲添加了一個 ID 列來區分行；之后你可以放棄它。
重塑本身已經被廣泛涵蓋（例如，將多組測量列（寬格式）重塑為單列（長格式）），但對于需要像這樣擴展的問題很有用。
library(dplyr)
df %>%
tibble::rowid_to_column(var = "game") %>%
tidyr::pivot_longer(-game, names_to = c(".value", "num"),
names_pattern = "(^[a-z] )(\\d $)") %>%
mutate(flag = ifelse(player == 0 & lossallowed == "no", "FLAG", NA_character_)) %>%
tidyr::pivot_wider(id_cols = game, names_from = num, values_from = player:flag,
names_glue = "{.value}{num}")
#> # A tibble: 3 × 7
#> game player001 player002 lossallowed001 lossallowed002 flag001 flag002
#> <int> <dbl> <dbl> <chr> <chr> <chr> <chr>
#> 1 1 1 1 no no <NA> <NA>
#> 2 2 0 0 yes no <NA> FLAG
#> 3 3 3 5 no yes <NA> <NA>
uj5u.com熱心網友回復：
一個可能的解決方案：
library(tidyverse)
df <-data.frame(player001 = c(1,0,3), player002 = c(1,0,5),lossallowed001 = c("no", "yes", "no"), loseallowed002 = c("no", "no", "yes"),flag001 = NA, flag002 = NA)
df %>%
rownames_to_column("id") %>%
mutate(across(where(is.numeric), as.character)) %>%
pivot_longer(cols = -id) %>%
group_by(str_extract(name, "\\d{3}$"), id) %>%
mutate(value = if_else(row_number() == 3 & first(value) == "0" &
nth(value, 2) == "no", "FLAG", value)) %>%
ungroup %>% select(name, value) %>%
pivot_wider(names_from = name, values_from = value, values_fn = list) %>%
unnest(cols = everything()) %>% type.convert(as.is = TRUE)
#> # A tibble: 3 × 6
#> player001 player002 lossallowed001 loseallowed002 flag001 flag002
#> <int> <int> <chr> <chr> <lgl> <chr>
#> 1 1 1 no no NA <NA>
#> 2 0 0 yes no NA FLAG
#> 3 3 5 no yes NA <NA>
uj5u.com熱心網友回復：
你可以這樣做。首先重塑資料，然后添加列。使用bind_cols，如果你想將資料合并回來。
library(purrr)
library(dplyr)
map(set_names(paste0("00", 1:2)), ~ select(df, ends_with(.x))) %>%
map(., ~ mutate(., newcol = ifelse(.[[1]] == 0 & .[[2]] == "no", "FLAG", NA)))
$`001`
player001 lossallowed001 flag001 newcol
1 1 no NA NA
2 0 yes NA NA
3 3 no NA NA
$`002`
player002 loseallowed002 flag002 newcol
1 1 no NA <NA>
2 0 no NA FLAG
3 5 yes NA <NA>
uj5u.com熱心網友回復：
這是tidyverse. 雖然我在這個解決方案抵達獨立，這是有可能的副本@camille的解決方案在這里，這是我前不久公布。
library(tidyverse)
# ...
# Code to generate 'df'.
# ...
df %>%
# Index the matches.
mutate(match_id = row_number()) %>%
# Pivot to get a row for each player {001, 002, ...} and match.
pivot_longer(
# Target columns whose names end with a separate suffix of 3 digits.
matches("^(.*\\D)(\\d{3,})$"),
names_pattern = "^(.*\\D)(\\d{3,})$",
# Index the players by their suffixes; and give each the following three columns:
# 'player' (score), 'lossallowed', and 'flag'.
names_to = c(".value", "player_id")
) %>%
# Flag the appropriate cases.
mutate(
flag = if_else(player == 0 & lossallowed == "no", "FLAG", NA_character_)
) %>%
# Return to original, wide format.
pivot_wider(
names_from = player_id,
values_from = !c(match_id, player_id),
names_glue = "{.value}{player_id}"
) %>%
arrange(match_id) %>% select(!match_id)
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