我的数据集包括在多个财政年度（2013财年，14财年和15财年）以及不同地区的调查中所问问题的分数和总受访者。
我的目标是遍历<link rel="stylesheet" href="/node_modules/...">
列并确定何时针对每个区域提出每个问题。并将此信息存储在新列中。
这是可重现的样品的样子-
FY我首先创建ID列，方法是将testdf=data.frame(FY=c("FY13","FY14","FY15","FY14","FY15","FY13","FY14","FY15","FY13","FY15","FY13","FY14","FY15","FY13","FY14","FY15"),
Region=c(rep("AFRICA",5),rep("ASIA",5),rep("AMERICA",6)),
QST=c(rep("Q2",3),rep("Q5",2),rep("Q2",3),rep("Q5",2),rep("Q2",3),rep("Q5",3)),
Very.Satisfied=runif(16,min = 0, max=1),
Total.Very.Satisfied=floor(runif(16,min=10,max=120)),
Satisfied=runif(16,min = 0, max=1),
Total.Satisfied=floor(runif(16,min=10,max=120)),
Dissatisfied=runif(16,min = 0, max=1),
Total.Dissatisfied=floor(runif(16,min=10,max=120)),
Very.Dissatisfied=runif(16,min = 0, max=1),
Total.Very.Dissatisfied=floor(runif(16,min=10,max=120)))
和Region串联起来
QST我的目标
1）对于每个唯一的library(tidyr)
testdf = testdf %>%
unite(ID,c('Region','QST'),sep = "",remove = F)
，请确定是否提出了给定问题-
a）仅在一年中（13财年，14财年或15财年）
b）过去两年（仅2015和2014财年）
c）过去三年（2015财年，14财年和13财年）
d）仅在2013财年和2015财年
我的尝试
对于这个问题，我尝试创建一个ID，对于每个唯一的for loop，我首先将问题所在的每个FY的唯一出现存储在向量ID中。然后，使用IF条件语句，根据这些情况，为新创建的v列分配注释。
Tally循环似乎没有抛出错误消息，但是似乎并没有创建新的for (i in unique(testdf$ID))
{
v=unique(testdf$FY)
if(('FY15' %in% v) & ('FY14' %in% v)) {
testdf$Tally=='Asked Over The Past Two Years'
}
else if(('FY15' %in% v) & ('FY14' %in% v) & ('FY13' %in% v)) {
testdf$Tally=='Asked Over The Past Three Years'
}
else if(('FY13' %in% v) & ('FY15' %in% v)) {
testdf$Tally=='Question Asked in FY13 & FY15 Only'
}
else { testdf$Tally=='Question Asked Once Only'
}
}
列。
任何帮助，将不胜感激。
答案 0 :(得分：2)
在您的代码中，主要问题是在if-else子句中，您不是在进行赋值（使用“ <-”），而是在进行比较，使用“ ==”。我发现这是一个更优雅的解决方案，因为它不使用循环：
require(tidyverse)
testdf %>%
select(ID, FY) %>%
unique() %>%
mutate(is_true = 1) %>%
spread(key = FY, value = is_true, fill = 0) %>%
mutate(tally = case_when(
FY13 == 1 & FY14 == 1 & FY15 == 1 ~ 'Asked Over The Past Three Years',
FY14 == 1 & FY15 == 1 ~ 'Asked Over the Past Two Years',
FY13 == 1 & FY15 == 1 ~ 'Asked in FY12 & FY15 Only',
TRUE ~ 'Question Asked Once Only'
))
输出：
+------------------------------------------------------------+
| ID FY13 FY14 FY15 tally |
+------------------------------------------------------------+
| 1 AFRICAQ2 1 1 1 Asked Over The Past Three Years |
| 2 AFRICAQ5 0 1 1 Asked Over the Past Two Years |
| 3 AMERICAQ2 1 1 1 Asked Over The Past Three Years |
| 4 AMERICAQ5 1 1 1 Asked Over The Past Three Years |
| 5 ASIAQ2 1 1 1 Asked Over The Past Three Years |
| 6 ASIAQ5 1 0 1 Asked in FY12 & FY15 Only |
+------------------------------------------------------------+
答案 1 :(得分：1)
无需循环：
library(tidyverse)
result <- testdf %>%
select(3, 2, 1) %>%
mutate(Asked = 1) %>%
spread(FY, Asked)
> result
QST Region FY13 FY14 FY15
1 Q2 AFRICA 1 1 1
2 Q2 AMERICA 1 1 1
3 Q2 ASIA 1 1 1
4 Q5 AFRICA NA 1 1
5 Q5 AMERICA 1 1 1
6 Q5 ASIA 1 NA 1
一次性回答所有四个问题。
如果您真的想要一个提示栏，请按以下方式展开：
result %>%
mutate(Tally = case_when(FY13 + FY14 + FY15 == 1 ~ "Only one year",
FY13 + FY14 + FY15 == 3 ~ "Past three years",
FY14 + FY15 == 2 ~ "Past two years",
FY13 + FY15 == 2 ~ "FY13 and FY15 only",
NA ~ NA_character_))
QST Region FY13 FY14 FY15 Tally
1 Q2 AFRICA 1 1 1 Past three years
2 Q2 AMERICA 1 1 1 Past three years
3 Q2 ASIA 1 1 1 Past three years
4 Q5 AFRICA NA 1 1 Past two years
5 Q5 AMERICA 1 1 1 Past three years
6 Q5 ASIA 1 NA 1 FY13 and FY15 only
答案 2 :(得分：0)
请考虑ave，以便根据嵌套条件ifelse内的区域和 QST 进行分组计算：
testdf <- within(testdf, {
FY13 <- ifelse(FY=='FY13', 1, 0)
FY14 <- ifelse(FY=='FY14', 1, 0)
FY15 <- ifelse(FY=='FY15', 1, 0)
Tally <- ifelse(ave(FY13, Region, QST, FUN=max) + ave(FY14, Region, QST, FUN=max) + ave(FY15, Region, QST, FUN=max) == 1,
'Asked Only on One Year',
ifelse(ave(FY13, Region, QST, FUN=max) + ave(FY14, Region, QST, FUN=max) + ave(FY15, Region, QST, FUN=max) == 3,
'Asked Over the Past Three Years',
ifelse(ave(FY14, Region, QST, FUN=max) + ave(FY15, Region, QST, FUN=max) == 2,
'Asked Over the Past Two Years',
ifelse(ave(FY13, Region, QST, FUN=max) + ave(FY15, Region, QST, FUN=max) == 2,
'Asked On FY13 & FY15 Only',
NA
)
)
)
)
FY13 <- NULL; FY14 <- NULL; FY15 <- NULL
})
testdf[c("ID", "FY", "Tally")]
# Region QST FY Tally
# 1 AFRICA Q2 FY13 Asked Over the Past Three Years
# 2 AFRICA Q2 FY14 Asked Over the Past Three Years
# 3 AFRICA Q2 FY15 Asked Over the Past Three Years
# 4 AFRICA Q5 FY14 Asked Over the Past Two Years
# 5 AFRICA Q5 FY15 Asked Over the Past Two Years
# 6 ASIA Q2 FY13 Asked Over the Past Three Years
# 7 ASIA Q2 FY14 Asked Over the Past Three Years
# 8 ASIA Q2 FY15 Asked Over the Past Three Years
# 9 ASIA Q5 FY13 Asked On FY13 & FY15 Only
# 10 ASIA Q5 FY15 Asked On FY13 & FY15 Only
# 11 AMERICA Q2 FY13 Asked Over the Past Three Years
# 12 AMERICA Q2 FY14 Asked Over the Past Three Years
# 13 AMERICA Q2 FY15 Asked Over the Past Three Years
# 14 AMERICA Q5 FY13 Asked Over the Past Three Years
# 15 AMERICA Q5 FY14 Asked Over the Past Three Years
# 16 AMERICA Q5 FY15 Asked Over the Past Three Years
答案 3 :(得分：0)
有一种使用您的ID列的解决方案。 （尽管使用paste0，但使用testdf$ID <- paste0(testdf$Region, "_", testdf$QST)可以做得更好。）
我们使用dcast包testdf reshape2您的library(reshape2)
tmp <- dcast(testdf, ID ~ FY,
value.var="QST", fun.aggregate=length)
。
tmp <- cbind(tmp,
past2=as.numeric(t2[3] + t2[4] == 2 & t2[2] == 0),
past3=as.numeric(t2[2] + t2[3] + t2[4] == 3),
y13_15=as.numeric(t2[2] + t2[4] == 2 & t2[3] == 0))
现在我们已经知道问题是否在不同年份提出。为了回答其他问题，我们将做一些数学运算。
Tally 5：7列中的序列包含我们可以挤奶的所需tmp$Tally <- apply(tmp, 1, function(x) paste0(x[5:7], collapse=""))
信息
tmp$Tally <- factor(tmp$Tally, labels=c('Question Asked Once Only',
'Question Asked in FY13 & FY15 Only',
'Asked Over The Past Three Years',
'Asked Over The Past Two Years'))
按因子水平翻译成人类语言，
> merge(testdf, t3[c(1, 8)])
ID FY Region QST Tally
1 AFRICA_Q2 FY13 AFRICA Q2 Asked Over The Past Three Years
2 AFRICA_Q2 FY14 AFRICA Q2 Asked Over The Past Three Years
3 AFRICA_Q2 FY15 AFRICA Q2 Asked Over The Past Three Years
4 AFRICA_Q5 FY14 AFRICA Q5 Asked Over The Past Two Years
5 AFRICA_Q5 FY15 AFRICA Q5 Asked Over The Past Two Years
6 AMERICA_Q2 FY13 AMERICA Q2 Asked Over The Past Three Years
7 AMERICA_Q2 FY14 AMERICA Q2 Asked Over The Past Three Years
8 AMERICA_Q2 FY15 AMERICA Q2 Asked Over The Past Three Years
9 AMERICA_Q5 FY13 AMERICA Q5 Asked Over The Past Three Years
10 AMERICA_Q5 FY14 AMERICA Q5 Asked Over The Past Three Years
11 AMERICA_Q5 FY15 AMERICA Q5 Asked Over The Past Three Years
12 ANTH.CTRY_Q2 FY15 ANTH.CTRY Q2 Question Asked Once Only
13 ASIA_Q2 FY13 ASIA Q2 Asked Over The Past Three Years
14 ASIA_Q2 FY14 ASIA Q2 Asked Over The Past Three Years
15 ASIA_Q2 FY15 ASIA Q2 Asked Over The Past Three Years
16 ASIA_Q5 FY13 ASIA Q5 Question Asked in FY13 & FY15 Only
17 ASIA_Q5 FY15 ASIA Q5 Question Asked in FY13 & FY15 Only
并与原始数据帧合并以获得所需的结果。
testdf <- structure(list(FY = c("FY13", "FY14", "FY15", "FY14", "FY15",
"FY13", "FY14", "FY15", "FY13", "FY15", "FY13", "FY14", "FY15",
"FY13", "FY14", "FY15", "FY15"), Region = c("AFRICA", "AFRICA",
"AFRICA", "AFRICA", "AFRICA", "ASIA", "ASIA", "ASIA", "ASIA",
"ASIA", "AMERICA", "AMERICA", "AMERICA", "AMERICA", "AMERICA",
"AMERICA", "ANTH.CTRY"), QST = c("Q2", "Q2", "Q2", "Q5", "Q5",
"Q2", "Q2", "Q2", "Q5", "Q5", "Q2", "Q2", "Q2", "Q5", "Q5", "Q5",
"Q2")), row.names = c(NA, 17L), class = "data.frame")
HTML:
//webgazer.js Library
<script type="text/js" src="libraries/WebGazer/webgazer.js"></script>
//Three.js
<script type="text/javascript" src="libraries/Three.js/Three.js"></script>
<script type="text/javascript" src="libraries/Three.js/three.min.js"></script>
<script type="text/javascript" src="libraries/Three.js/OrbitControls.js"></script>
<!-- MY SCRIPT -->
<script type="text/javascript" src="scripts/setupPage.js"></script>
<script type="text/javascript" src="scripts/3DCanvas.js"></script>
<script type="text/javascript" src="scripts/faceFeatures.js"></script>
<script type="text/javascript" src="scripts/modals.js"></script>
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