正常化數據R

Question

你好，我有以下data.frame（追加）。我想添加一個標準化計數的額外列N = N/sum(N)。我有沒有日期列前一個data.frame，並能夠做到這一點使用

oo[, N.norm := N/sum(N), by=Operator]

我試圖通過功能

oo[, N.norm := N/sum(N), by=Operator,Date] 

到日期添加到，但收到一條錯誤消息

Error in `[.data.frame`(oo, , `:=`(N.norm, N/sum(N)), by = Operator, Date) : 
    unused argument(s) (by = Operator) 

例如，對於運營商「A」在月「2013年1月」，我有每個計數N數量= c（「好」，「好」，「差」，「廢話」）。我想總結n該組合（A和2013年1月）和sum(N)

劃分數N在另一方面，任何人都可以給我提供一個體面的介紹操縱data.frames R中

structure(list(Operator = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), .Label = c("A", 
"D", "J", "L", "M"), class = "factor"), ROI_Score = structure(c(1L, 
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 
4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 
3L, 3L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 1L, 
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 
4L, 4L, 4L), .Label = c("Crap", "Good", "OK", "Poor"), class = "factor"), 
    Date = c("Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013", "Apr 2013", "Feb 2013", "Jan 2013", "Mar 2013", 
    "May 2013"), N = c(0, 0, 0, 0, 0, 1, 2, 15, 1, 5, 3, 2, 3, 
    1, 0, 3, 0, 5, 5, 1, 0, 0, 0, 1, 0, 14, 17, 16, 8, 7, 5, 
    10, 6, 1, 5, 24, 27, 31, 16, 15, 0, 0, 0, 0, 0, 26, 24, 20, 
    11, 18, 3, 4, 17, 3, 2, 20, 36, 12, 21, 9, 0, 0, 0, 0, 0, 
    3, 12, 5, 12, 4, 0, 0, 3, 4, 0, 29, 37, 41, 25, 10, 0, 0, 
    0, 0, 0, 9, 9, 15, 17, 3, 6, 4, 5, 4, 1, 14, 13, 9, 15, 9 
    )), .Names = c("Operator", "ROI_Score", "Date", "N"), row.names = c(NA, 
100L), class = "data.frame") 

---

我不確定數據是以data.frame還是data.table格式。這裏是我的代碼，改編自阿倫（reshape/remould data frame to create normalized bar chart and pie chart）給出解決辦法

df <- data.frame(read.csv("/misc/jaguar_data/report/system/db_fs/roi_scores.csv")) 
#Get date into nice structure for faceting 
df$Date = strftime(strptime(df$Date,f="%d/%m/%Y"), "%b %Y") 
dt <- data.table(df) 
ops <- as.character(unique(dt$Operator)) 
scr <- as.character(unique(dt$ROI_Score)) 
dts <- unique(dt$Date) 

oo <- setkey(dt[, .N, by="Operator,ROI_Score,Date"], Operator, 
ROI_Score,Date)[CJ(ops, scr,dts)][is.na(N), N:= 0L] 

oo[, N.norm := N/sum(N), by=Operator] 

Answer 1

你的代碼是（差不多）完美。兩個輕微的問題。

1：您正在使用data.table語法，所以不是oo是一個data.frame它應該是一個data.table。只需使用：

library(data.table) 
oo <- data.table(oo) 

2：當使用by有多個列，請務必將列list(..)或作爲一個單獨的逗號分隔的字符串。例如

oo[, N.norm := N/sum(N), by=list(Operator,Date)] 

# - or - # 
oo[, N.norm := N/sum(N), by="Operator,Date"] 

編輯：如果你希望每個總對每個Operator劃分 - Date組，那麼你的代碼應該是以上。相反，如果你想總的整個數據來劃分，然後用

oo[, N.norm := N/sum(DT$N), by=list(Operator,Date)] 

---

固定這兩件事情，並使用一切正是因爲你知道了：

 Operator ROI_Score  Date N N.norm 
    1:  A  Crap Apr 2013 0 0.0000000 
    2:  A  Crap Feb 2013 0 0.0000000 
    3:  A  Crap Jan 2013 0 0.0000000 
    4:  A  Crap Mar 2013 0 0.0000000 
    5:  A  Crap May 2013 0 0.0000000 
---           
96:  M  Poor Apr 2013 14 0.4827586 
97:  M  Poor Feb 2013 13 0.5000000 
98:  M  Poor Jan 2013 9 0.3103448 
99:  M  Poor Mar 2013 15 0.4166667 
100:  M  Poor May 2013 9 0.6923077 

---

編輯2：

只是一個說明。一般來說，如果您使用[括號]中的表達式，尤其是參考賦值運算符:=，那麼您的對象應該是data.table。

如果你看到一個錯誤，如

Error in `[.data.frame`(_<your object name>_, ... 

那麼這可能是由於這樣的事實，或者是（a）你的對象不是data.table或（b）你忘了加載數據。表package。

Answer 2

我不認爲你可以做你想做這個數據集的內容。這裏的原因：

install.packages("plyr") 
library("plyr") 
str(tmp) # this is your data 
count(tmp, vars = c("Operator", "ROI_Score")) 

給出了這樣的：

Operator ROI_Score freq 
1   A  Crap 5 
2   A  Good 5 
3   A  OK 5 
4   A  Poor 5 
5   D  Crap 5 
6   D  Good 5 
7   D  OK 5 
8   D  Poor 5 
9   J  Crap 5 
10  J  Good 5 
11  J  OK 5 
12  J  Poor 5 
13  L  Crap 5 
14  L  Good 5 
15  L  OK 5 
16  L  Poor 5 
17  M  Crap 5 
18  M  Good 5 
19  M  OK 5 
20  M  Poor 5 

而且包括Date使每一個獨特的價值，所以都具有1

使用data.frame計數，你要能在什麼原理獲得者：

ans <- aggregate(N ~ Operator + ROI_Score + Date, data = tmp, FUN = sum) 

然後改變函數做你想要的東西（除以100，條目數？）。但我不確定這是你想要的。

編輯

由於要通過運營商和日期各評級類別的百分比，我會第一子集，然後彙總：

tmp2 <- subset(tmp, Operator == "A") 
ans2 <- aggregate(N ~ ROI_Score, data = tmp2, FUN = sum) 
ans2$N.norm <- ans2$N/sum(ans2$N) 

給出：

ROI_Score N N.norm 
1  Crap 0 0.0000000 
2  Good 24 0.5106383 
3  OK 9 0.1914894 
4  Poor 14 0.2978723