tapply - 創建NA？

Question

我正試圖計算每人獨特水果的平均數量（我通常的練習數據）。這與這兩行代碼很好地工作：

with(df, tapply(fruit, names, FUN = function(x) length(unique(x))))->uniques 
sum(uniques)/length(unique(df$names)) 

aggregate(df[,"fruit"], by=list(id=names), FUN = function(x) length(unique(x)))->d1 
sum(d1$x)/length(unique(df$names)) 

我的問題是，當我在我的真實數據上使用代碼它不起作用。我的真實數據是處方數據，我想要每個人平均數量的獨特藥物。使用tapply代碼，它似乎創建了原始df中不存在的全新患者id。它也已經給出了1000的NA值。有我的ID列沒有缺失值並沒有在drug_code列要麼

with(dt3, tapply(drug_code, id, FUN = function(x) length(unique(x))))->uniques  

head(uniques) 
        uniques 
Patient HAI0000001  NA 
Patient HAI0000003  NA 
Patient HAI0000008  NA 
Patient HAI0000010  NA 
Patient HAI0000014  NA 
Patient HAI0000020  NA 

table(dt3$id=="Patient HAI0000001") ##checking to see if HA10000001 occurs in original df. the dim of df are 228954 rows and 5 cols 

FALSE 
228954 

對於聚集代碼我得到一個錯誤：

aggregate(dt3[,"drug_code"], by=list(id=id), FUN = function(x) length(unique(x)))->d1 

Error in aggregate.data.frame(as.data.frame(x), ...) : 
    arguments must have same length 

我不明白髮生了什麼。我的真實數據與我的練習數據相似，因爲它有一個ID欄，並有一個藥物/水果欄。 df中沒有丟失的數據。我知道lapply對數據框更好，但我不一定需要df。在任何情況下，tapply代碼都是針對df的練習數據。有沒有人知道這裏發生了什麼？

實踐DF：真實數據的

names<-as.character(c("john", "john", "john", "john", "john", "mary", "mary","mary","mary","mary", "jim", "sylvia","ted","ted","mary", "sylvia", "jim", "ted", "john", "ted")) 
dates<-as.Date(c("2010-07-01", "2010-09-01", "2010-11-01", "2010-12-01", "2011-01-01", "2010-08-12", "2010-11-11", "2010-05-12", "2010-12-03", "2010-07-12", "2010-12-21", "2010-02-18", "2010-10-29", "2010-08-13", "2010-11-11", "2010-05-12", "2010-04-01", "2010-05-06", "2010-09-28", "2010-11-28")) 
fruit<-as.character(c("kiwi","apple","banana","orange","apple","orange","apple","orange", "apple", "apple", "pineapple", "peach", "nectarine", "grape", "melon", "apricot", "plum", "lychee", "watermelon", "apple")) 
df<-data.frame(names,dates,fruit) 

例如：

head(dt3) 
     id   quantity date_of_claim drug_code index 
1 Patient HAI0000560  1 2009-10-15 R03AC02 2010-04-06 
2 Patient HAI0000560  1 2009-10-15 R03AK06 2010-04-06 
3 Patient HAI0000560  30 2009-10-15 R03BB04 2010-04-06 
4 Patient HAI0000560  30 2009-10-15 A02BC01 2010-04-06 
5 Patient HAI0000560  50 2009-10-15 M02AA15 2010-04-06 
6 Patient HAI0000560  30 2009-10-15 N02BE51 2010-04-06 

Answer 1

在你的情況你問杉木單號：一個特定的載體(unique(fruits))的所有單個長度內的平均患者ID。這shws你先蹦牀網上單人獨特的計數，然後平均函數結果：

> with(df, tapply(fruit, names, function(x) length(unique(x)))) 
    jim john mary sylvia ted 
    2  5  3  2  4 
> mean (with(df, tapply(fruit, names, function(x) length(unique(x))))) 
[1] 3.2 

我會評論說，你上面的代碼中的一個特定值的遏制測試過，可能已導致問題尾隨空間。 "string "將不等於"string"。我在我的.Rprofile文件中放入了一個使用修剪功能pkg::gdata的副本，以便我更輕鬆地處理這種可能性。

Answer 2

我可能會錯過一些東西，但是不會在這裏簡單地工作tapply？下面的線計算每人

x=tapply(df$fruit,df$names,function(x){length(unique(x))}) 

不同的水果的數量，然後mean(x)會給你整個人平均？