來自 Qualtrics 的調查結果對問題選擇進行了編碼，其中可以記錄多個回應，例如種族/民族人口統計資料（如下所示），我無法想出一個簡單的解決方案來進行分析。它記錄每個選項下每行的選定復選框（在其自己的列中），未選擇的選項保持空白。我已經決定，一個好的開始是計算每個選擇的非“NA”。然而，它并沒有按照我計劃的方式運行，并且對可用解決方案的嚴格篩選并沒有多大用處。我想出了一種使用 apply 獲取列數的方法，但處理輸出仍然有點笨拙。我有一個包含許多列的資料框，需要以這種方式進行分析，因此我使用 grep 函式來選擇需要選擇計數的相關列。
資料：
structure(list(race_White = c("White", NA, NA, "White", NA, NA,
"White", "White", NA, "White", "White", "White", "White", "White",
"White", "White", "White", "White", "White", "White", NA, "White",
"White", "White", NA), `race_Black or African American` = c(NA,
NA, "Black or African American", NA, NA, "Black or African American",
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, "Black or African American"), `race_American Indian or Alaska Native` = c(NA,
NA, NA, NA, NA, NA, NA, NA, "American Indian or Alaska Native",
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA
), race_Asian = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, "Asian", NA, NA, NA, NA),
`race_Middle Eastern or North African` = c(NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_
), `race_Hispanic, Latino or Spanish` = c(NA, "Hispanic, Latino or Spanish",
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA), `race_Native Hawaiian or Pacific Islander` = c(NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_
), `race_ Prefer not to share` = c(NA, NA, NA, NA, "Prefer not to share",
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA), race_Other = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_), education_level = structure(c(3L,
2L, 5L, 4L, 6L, 3L, 6L, 2L, 3L, 3L, 5L, 2L, 5L, 5L, 3L, 3L,
5L, 2L, 5L, 5L, 5L, 3L, 3L, 3L, 5L), .Label = c("Less than high school degree",
"High school graduate (high school diploma or equivalent)",
"Some college but no degree", "Associate's degree (2-year)",
"Bachelor's degree (4-year)", "Master's degree", "Doctoral/Professional degree (PhD, MD, JD)",
"Other/Prefer not to share"), class = "factor"), age = c(74,
43, NA, 37, 61, 64, NA, NA, 45, NA, NA, 21, NA, NA, 52, 43,
43, NA, 65, 42, NA, 27, 35, NA, 46)), row.names = c(NA, -25L
), class = c("tbl_df", "tbl", "data.frame"))
我已經使用 grep 來選擇我想通過使用以下內容來計算選擇的列號：
race<-c(grep("race", colnames(data)))
然后，我還使用了列名，以防公式需要名稱而不是數字
racenames<-colnames(data[race])
創建這些選擇后，我嘗試使用以下內容獲取不等于“”的行計數表（沒有成功）
racecounts <- sapply(data[race],FUN = function(x){length(x[x!=""])})
racecounts
這基本上總結了列中的每一行，而不是我希望的非空行。所以我嘗試了一個簡單的應用函式，它確實有效：
racecounts2 <- apply(data[race], 2, table)
racecounts2
這有效，然后我必須將其轉換為 prop.table 以獲得與 kable 一起使用的比例
racecounts2<-prop.table(racecounts2)
racecounts2%>%
kbl() %>%
kable_material_dark()
我只是好奇是否有人找到了替代/更好的方法來處理這種資料格式？我愿意嘗試任何不同的東西，這個看起來很笨拙，它的輸出有點讓人想像。如果能找到一種處理這些資料的方法，讓排名/繪圖等作業更容易進行，那就太好了。
所以我很好奇社區會如何做到這一點。
uj5u.com熱心網友回復：
您可以!is.na像這樣計算用于比賽列的非 NA 值的數量：
colSums(!is.na(data[race]))
或者，使用dplyr語法并tidyr::pivot_longer使其更像一張表格：
data %>% select(starts_with("race")) %>%
summarise(across(everything(), ~sum(!is.na(.x)))) %>%
pivot_longer(cols=everything(), names_to = "race", values_to = "count",
names_transform = list(race = \(x) str_remove(x, "race_")))
# A tibble: 9 x 2
race count
<chr> <int>
1 "White" 18
2 "Black or African American" 3
3 "American Indian or Alaska Native" 1
4 "Asian" 1
5 "Middle Eastern or North African" 0
6 "Hispanic, Latino or Spanish" 1
7 "Native Hawaiian or Pacific Islander" 0
8 " Prefer not to share" 1
9 "Other" 0
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