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我的任務是創建頻率表與統計摘要。我的目標是創建一個可以簡單導出爲excel的數據框。 其中大部分可以使用存儲過程在SQL中，但我決定在R中執行此操作。我正在學習R，所以我可能會做很長的路要走。這是問題的一個後續從getting-r-frequency-counts-for-all-possible-answers一個簡單的方法來實現頻率計數與平均值，總和，長度和SD在R
鑑於
Id <- c(1,2,3,4,5,6,7,8,9,10)
ClassA <- c(1,NA,3,1,1,2,1,4,5,3)
ClassB <- c(2,1,1,3,3,2,1,1,3,3)
R <- c(1,2,3,NA,9,2,4,5,6,7)
S <- c(3,7,NA,9,5,8,7,NA,7,6)
df <- data.frame(Id,ClassA,ClassB,R,S)
ZeroTenNAScale <- c(0:10,NA);
R.freq <- setNames(nm=c('answer','value'),data.frame(table(factor(df$R,levels=ZeroTenNAScale,exclude=NULL))));
R.freq[, 1] <- as.numeric(as.character(R.freq[, 1]))
R.freq <- cbind(question='R',R.freq)
S.freq <- setNames(nm=c('answer','value'),data.frame(table(factor(df$S,levels=ZeroTenNAScale,exclude=NULL))));
S.freq[, 1] <- as.numeric(as.character(S.freq[, 1]))
S.freq <- cbind(question='S',S.freq)
R.mean = mean(df$R, na.rm = TRUE)
R.length = sum(!is.na(df$R))
R.sd = sd(df$R, na.rm = TRUE)
R.sum = sum(df$R, na.rm = TRUE)
S.mean = mean(df$S, na.rm = TRUE)
S.length = sum(!is.na(df$S))
S.sd = sd(df$S, na.rm = TRUE)
S.sum = sum(df$S, na.rm = TRUE)
S.row <- cbind('S','sum',as.numeric(S.sum))
S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
S.freq = rbind(S.freq, S.row)
S.row <- cbind('S','length',as.numeric(S.length))
S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
S.freq = rbind(S.freq, S.row)
S.row <- cbind('S','mean',as.numeric(S.mean))
S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
S.freq = rbind(S.freq, S.row)
S.row <- cbind('S','sd',as.numeric(S.sd))
S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
S.freq = rbind(S.freq, S.row)
R.row <- cbind('R','sum',as.numeric(R.sum))
R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
R.freq = rbind(R.freq, R.row)
R.row <- cbind('R','length',as.numeric(R.length))
R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
R.freq = rbind(R.freq, R.row)
R.row <- cbind('R','mean',as.numeric(R.mean))
R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
R.freq = rbind(R.freq, R.row)
R.row <- cbind('R','sd',as.numeric(R.sd))
R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
R.freq = rbind(R.freq, R.row)
result <- rbind(R.freq,S.freq)
result <- cbind(filter='None',result)
result
我得到
filter question answer value
1 None R 0 0
2 None R 1 1
3 None R 2 2
4 None R 3 1
5 None R 4 1
6 None R 5 1
7 None R 6 1
8 None R 7 1
9 None R 8 0
10 None R 9 1
11 None R 10 0
12 None R <NA> 1
13 None R sum 39
14 None R length 9
15 None R mean 4.33333333333333
16 None R sd 2.64575131106459
17 None S 0 0
18 None S 1 0
19 None S 2 0
20 None S 3 1
21 None S 4 0
22 None S 5 1
23 None S 6 1
24 None S 7 3
25 None S 8 1
26 None S 9 1
27 None S 10 0
28 None S <NA> 2
29 None S sum 52
30 None S length 8
31 None S mean 6.5
32 None S sd 1.8516401995451
這是差不多就是我要找的。下一步，我看到它開始在一些函數中包裝，以簡化代碼，然後再開始添加類ClassA = 1，ClassA = n + 1 ... ClassA = NA，ClassB = 1，ClassB = 2 ... ClassB = NA。有沒有更簡單的方法來做到這一點？
# https://stackoverflow.com/questions/36790376/a-simpler-way-to-achieve-a-frequency-count-with-mean-sum-length-and-sd-in-r/36794422#36794422
# create the summary function
summaryStatistics <- function(x) {
xx <- na.omit(x)
c(table(factor(x, levels=0:10), useNA='always', exclude=NULL),
sum=sum(xx), length=length(x), mean=mean(xx), sd=sqrt(var(xx)))
}
# create the test data frame
Id <- c(1,2,3,4,5,6,7,8,9,10)
ClassA <- c(1,NA,3,1,1,2,1,4,5,3)
ClassB <- c(2,1,1,3,3,2,1,1,3,3)
R <- c(1,2,3,NA,9,2,4,5,6,7)
S <- c(3,7,NA,9,5,8,7,NA,7,6)
df <- data.frame(Id,ClassA,ClassB,R,S)
# create the result
result <- setNames(
nm=c('answer','question','value'),
as.data.frame(
as.table(
simplify2array(
lapply(
df[c('R', 'S')],
summaryStatistics
)
)
)
)
)
# change the order to question, answer, value
result <- result[, c(2, 1, 3)]
# add the filter
result <- cbind(filter='None',result)
# return the result
result
這是更簡單，讓我的訓練我們的隊伍更加簡單等任務。感謝Ernest A和Imo。
關於我的R理解接下來的問題是Using vectors in R to change the output of a function
可以將水平傳遞給函數。我有一些問題有不同的尺度。例如，一些問題的等級= 1：3。 –
謝謝@ErnestA。乾杯。 –