############ uncoded data
x10<- structure(c(0L, 0L, 0L, 0L, 1L, 1L, 1L, 5L, 8L, 9L, 31L, 1L,
0L, 0L, 0L, 1L, 0L, 1L, 2L, 7L, 2L, 10L, 0L, 2L, 0L, 2L, 2L,
5L, 2L, 4L, 6L, 8L, 4L, 1L, 1L, 3L, 2L, 2L, 6L, 1L, 12L, 18L,
7L, 29L, 8L, 4L, 6L, 8L, 6L, 19L, 3L, 9L, 12L, 3L, 12L, 14L,
1L, 2L, 1L, 3L, 1L, 0L, 4L, 6L, 3L, 11L, 0L, 0L, 0L, 1L, 3L,
7L, 5L, 8L, 21L, 26L, 51L, 0L, 1L, 0L, 3L, 5L, 10L, 9L, 29L,
55L, 60L, 125L, 3L, 0L, 1L, 1L, 3L, 10L, 1L, 6L, 18L, 17L, 13L,
6L, 3L, 4L, 13L, 6L, 33L, 17L, 48L, 84L, 54L, 103L, 34L, 11L,
20L, 27L, 26L, 50L, 29L, 30L, 54L, 28L, 34L, 31L, 5L, 7L, 3L,
4L, 20L, 8L, 16L, 16L, 8L, 41L, 1L, 0L, 0L, 3L, 1L, 3L, 3L, 11L,
19L, 16L, 56L, 0L, 0L, 0L, 0L, 3L, 11L, 3L, 18L, 25L, 21L, 62L,
3L, 0L, 1L, 4L, 2L, 7L, 8L, 15L, 22L, 12L, 19L, 5L, 2L, 8L, 9L,
9L, 42L, 18L, 51L, 70L, 45L, 103L, 29L, 15L, 23L, 34L, 25L, 57L,
23L, 38L, 55L, 30L, 33L, 36L, 5L, 5L, 6L, 6L, 16L, 6L, 10L, 17L,
9L, 35L, 2L, 0L, 1L, 1L, 2L, 4L, 6L, 8L, 22L, 33L, 73L, 0L, 0L,
0L, 1L, 2L, 7L, 7L, 15L, 27L, 21L, 56L, 1L, 2L, 2L, 0L, 2L, 9L,
4L, 8L, 24L, 13L, 17L, 14L, 2L, 8L, 10L, 16L, 51L, 16L, 51L,
69L, 29L, 99L, 44L, 18L, 25L, 34L, 19L, 49L, 26L, 43L, 63L, 15L,
30L, 42L, 9L, 17L, 7L, 3L, 16L, 8L, 13L, 22L, 18L, 45L, 0L, 0L,
1L, 3L, 0L, 7L, 4L, 14L, 15L, 20L, 47L, 0L, 1L, 0L, 1L, 1L, 3L,
3L, 5L, 6L, 11L, 21L, 1L, 0L, 0L, 4L, 2L, 3L, 8L, 7L, 17L, 3L,
13L, 5L, 2L, 6L, 13L, 15L, 34L, 19L, 42L, 62L, 37L, 83L, 52L,
16L, 26L, 26L, 29L, 53L, 28L, 45L, 45L, 15L, 22L, 26L, 8L, 12L,
11L, 5L, 12L, 5L, 7L, 17L, 10L, 28L), .Dim = c(11L, 6L, 5L), .Dimnames = structure(list(
c("0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "10"),
c("I've changed for work/ a new job/ gone on a work plan",
"I want a phone that doesn't offer", "I want Best Mates/ Favourites",
"I was offered or saw a better offer on another network",
"Issues with the network (poor coverage)", "Other"
), YearQuarter = c("2011-09-01", "2011-12-01", "2012-03-01",
"2012-06-01", "2012-09-01")), .Names = c("", "", "YearQuarter"
)), class = "table")
############ recoded data
x10 <- structure(c(40L, 3L, 13L, 12L, 3L, 9L, 12L, 13L, 10L, 36L, 16L,
30L, 15L, 54L, 21L, 14L, 22L, 10L, 77L, 16L, 29L, 185L, 28L,
84L, 30L, 19L, 24L, 157L, 82L, 132L, 62L, 197L, 84L, 49L, 78L,
32L, 72L, 11L, 30L, 83L, 17L, 43L, 31L, 25L, 37L, 148L, 93L,
121L, 63L, 206L, 93L, 44L, 80L, 27L, 106L, 16L, 30L, 77L, 17L,
42L, 30L, 20L, 32L, 128L, 117L, 120L, 45L, 215L, 106L, 63L, 102L,
35L, 67L, 15L, 29L, 32L, 9L, 11L, 16L, 18L, 24L, 120L, 94L, 104L,
37L, 230L, 90L, 38L, 79L, 24L), .Dim = c(3L, 6L, 5L), .Dimnames = structure(list(
c("Promoters", "Detractors", "Passive"), c("I've changed for work/ a new job/ gone on a work plan",
"I want a phone that doesn't offer", "I want Best Mates/ Favourites",
"I was offered or saw a better offer on another network",
"Issues with the network (poor coverage)", "Other"
), YearQuarter = c("2011-09-01", "2011-12-01", "2012-03-01",
"2012-06-01", "2012-09-01")), .Names = c("", "", "YearQuarter"
)), class = "table")
x10.p <- round(prop.table(x10,c(3,2)),2)*100
你好
净促销分数是一个问题，要求消费者评价推荐产品或服务的可能性。在零到十的比例。据报道有10人和9人的人被称为“推动者”，被评为8和7的人被视为“被动”，被报道少于6人被视为诋毁者。净促销分数是“促销员”的百分比之间的差异。减去＆＃39; Detractors＆＃39;。
的百分比我将2011年9月至2012年9月的问题答案重新编入表格x10中。这些数字是每个群组的实际人数（发起人，擅离者和被动者）。为三维表格道歉，我对每个原因的净推动者得分感兴趣（即，促销员和批评者之间的差异是什么？＃34;我改变了工作/新工作/ / 2012年9月的工作计划＃34;
在我可以绘制之前的净促销分数需要一点操作。我想知道是否有人知道它是怎么做的？
干杯
答案 0 :(得分：1)
首先，在你完成所有计算之前不要进行舍入（否则你的百分比不会增加到1）
x10.p <- prop.table(x10,c(3,2))*100
# get the total promoters
promoters <- apply(x10.p, 2:3, function(x) sum(tail(x,2)))
# and detractors
detractors <- apply(x10.p, 2:3, function(x) sum(head(x,7)))
# passive is everything else
passive <- passive <- 100 - (detractors +promoters)
# the net score
net <- promoters - detractors
net
YearQuarter
2011-09-01 2011-12-01 2012-03-01 2012-06-01 2012-09-01
I've changed for work/ a new job/ gone on a work plan 66.071429 50.00000 53.982301 59.210526 46.846847
I want a phone that doesn't offer 37.500000 52.86195 46.153846 44.117647 44.230769
I want Best Mates/ Favourites -2.857143 15.06849 6.451613 12.195122 -3.448276
I was offered or saw a better offer on another network 24.390244 20.21563 15.193370 3.013699 8.176101
Issues with the network (poor coverage) -43.333333 -39.35860 -39.502762 -46.448087 -54.061625
Other -17.391304 -18.23899 -23.841060 -19.500000 -29.078014
您希望2012年9月，只选择该列，drop = FALSE以确保它仍然是包含1列的矩阵。
net[,'2012-09-01', drop = FALSE]
YearQuarter
2012-09-01
I've changed for work/ a new job/ gone on a work plan 46.846847
I want a phone that doesn't offer 44.230769
I want Best Mates/ Favourites -3.448276
I was offered or saw a better offer on another network 8.176101
Issues with the network (poor coverage) -54.061625
Other -29.078014