I have a dataframe movielens:
str(u.data)
'data.frame': 100000 obs. of 4 variables:
$ userID : int 196 186 22 244 166 298 115 253 305 6 ...
$ movieID : int 242 302 377 51 346 474 265 465 451 86 ...
$ rating : int 3 3 1 2 1 4 2 5 3 3 ...
$ timestamp: int 881250949 891717742 878887116 880606923 886397596 884182806 881171488 891628467 886324817 883603013 ...
and
str(u.item)
'data.frame': 1681 obs. of 20 variables:
$ unknown : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Action : Factor w/ 2 levels "0","1": 1 2 1 2 1 1 1 1 1 1 ...
$ Adventure : Factor w/ 2 levels "0","1": 1 2 1 1 1 1 1 1 1 1 ...
$ Animation : Factor w/ 2 levels "0","1": 2 1 1 1 1 1 1 1 1 1 ...
$ Childrens : Factor w/ 2 levels "0","1": 2 1 1 1 1 1 1 2 1 1 ...
$ Comedy : Factor w/ 2 levels "0","1": 2 1 1 2 1 1 1 2 1 1 ...
$ Crime : Factor w/ 2 levels "0","1": 1 1 1 1 2 1 1 1 1 1 ...
$ Documentary: Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Drama : Factor w/ 2 levels "0","1": 1 1 1 2 2 2 2 2 2 2 ...
$ Fantasy : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Film-Noir : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Horror : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Musical : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Mystery : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Romance : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Sci-Fi : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 2 1 1 1 ...
$ Thriller : Factor w/ 2 levels "0","1": 1 2 2 1 2 1 1 1 1 1 ...
$ War : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 2 ...
$ Western : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ movieID : int 1 2 3 4 5 6 7 8 9 10 ...
The number of row of u.data is 100.000
nrow(u.data)
100000
And
nrow(u.item)
[1] 1681
Then, I want to merge them:
all_data = u.data
all_data = merge(all_data, u.item, by = "movieID")
But the merged data has only 99.999 rows
nrow(all_data)
[1] 99999
Did I did something wrong while merging these two data frames?
This happens if min(u.data$movieID) < min(u.item$movieID) or if max(u.data$movieID) > max(u.item$movieID). Example for the latter:
# max(u.data$movieID) = 10
u.data <- data.frame(movieID = 1:10, NAME = LETTERS[1:10])
dim(u.data)
# [1] 10 2
# max(u.item$movieID) = 11
u.item <- data.frame(movieID = c(1:9,11), name = letters[c(1:9,11)])
dim(u.item)
# [1] 10 2
out <- merge(u.data, u.item, by = "movieID")
dim(out)
# [1] 9 3
# check if all elements of u.item$movieID exist in u.data$movieID
is.element(u.data$movieID, u.item$movieID)
# [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
Suggested by Batanichek:
out <- merge(u.data, u.item, by = "movieID", all.x = TRUE)
dim(out)
# [1] 10 3
Related
I am trying to do logistic regression by using the code:
model <- glm (Participation ~ Gender + Race + Ethnicity + Education + Comorbidities + WLProgram + LoseWeight + EverLoseWeight + PastYearLW + Age + BMI, data = LogisticData, family = binomial)
summary(model)
I keep getting the error:
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels
Upon checking the forums I checked to see which variables were factors:
str(LogisticData)
'data.frame': 994 obs. of 13 variables:
$ outcome : Factor w/ 2 levels "No","Yes": 1 1 2 2 1 2 2 1 2 2 ...
$ Gender : Factor w/ 3 levels "Male","Female",..: 1 2 2 1 2 1 1 1 1
$ Race : Factor w/ 3 levels "White","Black",..: 1 1 1 3 1 1 1 1 1 1
$ Ethnicity : Factor w/ 2 levels "Hispanic/Latino",..: 2 2 2 2 2 2 2 2 2
$ Education : Factor w/ 2 levels "Below Bachelors",..: 1 1 1 2 1 1 1 2 1
$ Comorbidities : Factor w/ 2 levels "No","Yes": 1 1 2 1 1 1 2 2 1 1 ...
$ WLProgram : Factor w/ 2 levels "No","Yes": NA 1 2 2 1 1 1 NA 1 1 ...
$ LoseWeight : Factor w/ 2 levels "Yes","No": 2 1 1 1 1 1 1 2 1 1 ...
$ PastYearLW : Factor w/ 2 levels "Yes","No": NA 2 1 1 1 2 1 NA 1 1 ...
$ EverLoseWeight: Factor w/ 2 levels "Yes","No": 2 1 1 1 1 1 1 2 1 1 ...
$ Age : int 29 35 69 32 21 45 40 62 59 58 ...
$ Participation : Factor w/ 2 levels "Yes","No": 2 2 1 1 1 1 1 2 1 2 ...
$ BMI : num 25.7 33.8 26.4 32.3 27.5 ...
All factors appear to have 2 or more levels.
I also tried to omit NA's which still gave me this error.
I want all variables in the regression, and can't figure out why it won't run.
When performing :
newdata <- droplevels(na.omit(LogisticData))
> str(newdata)
'data.frame': 840 obs. of 13 variables:
$ outcome : Factor w/ 2 levels "No","Yes": 1 2 2 1 2 2 2 2 2 2 ...
$ Gender : Factor w/ 3 levels "Male","Female",..: 2 2 1 2 1 1 1 2 1
$ Race : Factor w/ 3 levels "White","Black",..: 1 1 3 1 1 1 1 1 3
$ Ethnicity : Factor w/ 2 levels "Hispanic/Latino",..: 2 2 2 2 2 2 2 2
$ Education : Factor w/ 2 levels "Below Bachelors",..: 1 1 2 1 1 1 1 1
$ Comorbidities : Factor w/ 2 levels "No","Yes": 1 2 1 1 1 2 1 1 1 2 ...
$ WLProgram : Factor w/ 2 levels "No","Yes": 1 2 2 1 1 1 1 1 1 1 ...
$ LoseWeight : Factor w/ 1 level "Yes": 1 1 1 1 1 1 1 1 1 1 ...
$ PastYearLW : Factor w/ 2 levels "Yes","No": 2 1 1 1 2 1 1 1 1 2 ...
$ EverLoseWeight: Factor w/ 1 level "Yes": 1 1 1 1 1 1 1 1 1 1 ...
$ Age : int 35 69 32 21 45 40 59 58 23 32 ...
$ Participation : Factor w/ 2 levels "Yes","No": 2 1 1 1 1 1 1 2 2 1 ...
$ BMI : num 33.8 26.4 32.3 27.5 45.4 ...
- attr(*, "na.action")=Class 'omit' Named int [1:154] 1 8 13 14 21 24 25
46 55 58 ...
.. ..- attr(*, "names")= chr [1:154] "1" "8" "13" "14" ...
This doesn't make sense to me because you can see in the first str(Logisitic Data) that there is obviously 2 levels in EverLoseWeight as you can see both the Yes and the No and the 1 and 2? How do I fix this anomaly?
Given your update, it looks like you have at least two possibilities.
1: Remove the factors that are left with only a single level after removing the NAs (i.e. LoseWeight and EverLoseWeight).
2: Treat the NAs as an extra level. Something along the lines of
a = as.factor(c(1,1,NA,2))
b = as.factor(c(1,1,2,1))
# 0 is an unused factor level for a
x = data.frame(a, b)
levels(x$a) = c(levels(x$a), 0)
x$a[is.na(x$a)] = 0
But this might not deal with any singularity issues that also resulted in having single-level factors.
Try doing summary on your raw data and make sure that all of the levels have values. I would put this in a comment, but I don't have the reputation points :(
I'm trying to convert a triple nested list into a dataframe. This question has helped, but I can't get the dataframe I'd like.
The list is an options chain obtained from IBrokers, a summary is shown below. I've uploaded the actual chain here which is more detailed.
Chain <-
list(
list(
list(
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180621",strike="25")),
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180621",strike="26"))
),
list(
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180730",strike="25")),
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180730",strike="26"))
)
),
list(
list(
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180621",strike="65")),
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180621",strike="64"))
),
list(
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180730",strike="65")),
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180730",strike="64"))
)
)
)
I'd like to convert the list into a dataframe like this:
Contracts <- data.frame(symbol=c("BHP","BHP","BHP","BHP","CBA","CBA","CBA","CBA"),
right=c("C","C","C","C","C","C","C","C"),
expiry=c("20180621","20180621","20180730","20180730","20180621","20180621","20180730","20180730"),
strike=c("25","26","25","26","65","64","65","64"))
I tried this code, but it didn't give me the dataframe I wanted.
X <- lapply(Chain,function(x) as.data.frame.list(lapply(x,as.data.frame.list)))
dfx <- do.call(rbind,X)
Any suggestions please?
How about the following?
df <- as.data.frame(matrix(unlist(Chain, recursive = T), ncol = 5, byrow = T)[, -1]);
colnames(df) <- c("symbol", "right", "expiry", "strike");
# symbol right expiry strike
#1 BHP C 20180621 25
#2 BHP C 20180621 26
#3 BHP C 20180730 25
#4 BHP C 20180730 26
#5 CBA C 20180621 65
#6 CBA C 20180621 64
#7 CBA C 20180730 65
#8 CBA C 20180730 64
Explanation: Recursively unlist the nested Chain, then recast as matrix, remove column version and convert to data.frame. The only minor down-side is that we have to manually add column names.
Update
Since your actual data is quite different, here is a possibility.
Note: I assume the structure from the Gist is stored in tbl.
tbl;
#Source: local data frame [2 x 6]
#Groups: <by row>
#
## A tibble: 2 x 6
# symbol sectype exch currency multiplier Chain
# <fct> <fct> <fct> <fct> <fct> <list>
#1 BHP OPT ASX AUD 100 <list [1,241]>
#2 CBA OPT ASX AUD 100 <list [1,204]>
The following list contains two data.frames, one for each row from tbl.
lst <- lapply(tbl$Chain, function(x)
do.call(rbind.data.frame, lapply(x, function(y) as.data.frame(unclass(y$contract)))))
#List of 2
# $ :'data.frame': 1241 obs. of 16 variables:
# ..$ conId : Factor w/ 1241 levels "198440202","198440207",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ symbol : Factor w/ 1 level "BHP": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ sectype : Factor w/ 1 level "OPT": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ exch : Factor w/ 1 level "ASX": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ primary : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ expiry : Factor w/ 18 levels "20180628","20181220",..: 1 1 1 1 1 1 1 1 1 1 ...
# ..$ strike : Factor w/ 118 levels "25","26","27",..: 1 1 2 2 3 3 4 4 5 5 ...
# ..$ currency : Factor w/ 1 level "AUD": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ right : Factor w/ 2 levels "C","P": 1 2 1 2 1 2 1 2 1 2 ...
# ..$ local : Factor w/ 1241 levels "BHPV78","BHPV88",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ multiplier : Factor w/ 1 level "100": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ combo_legs_desc: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ comboleg : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ include_expired: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secIdType : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secId : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# $ :'data.frame': 1204 obs. of 16 variables:
# ..$ conId : Factor w/ 1204 levels "198447027","198447030",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ symbol : Factor w/ 1 level "CBA": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ sectype : Factor w/ 1 level "OPT": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ exch : Factor w/ 1 level "ASX": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ primary : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ expiry : Factor w/ 18 levels "20180628","20181220",..: 1 1 1 1 1 1 1 1 1 1 ...
# ..$ strike : Factor w/ 179 levels "79.68","81.68",..: 1 1 2 2 3 3 4 4 5 5 ...
# ..$ currency : Factor w/ 1 level "AUD": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ right : Factor w/ 2 levels "C","P": 1 2 1 2 1 2 1 2 1 2 ...
# ..$ local : Factor w/ 1204 levels "CBAKT9","CBAKU9",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ multiplier : Factor w/ 1 level "100": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ combo_legs_desc: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ comboleg : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ include_expired: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secIdType : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secId : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
You can use unstack
unstack(data.frame(d<-unlist(Chain),names(d)))
contract.expiry contract.right contract.strike contract.symbol version
1 20180621 C 25 BHP 8
2 20180621 C 26 BHP 8
3 20180730 C 25 BHP 8
4 20180730 C 26 BHP 8
5 20180621 C 65 CBA 8
6 20180621 C 64 CBA 8
7 20180730 C 65 CBA 8
8 20180730 C 64 CBA 8
If you want you can delete the word contract.
unstack(data.frame(d<-unlist(Chain),sub(".*[.]","",names(d))))
expiry right strike symbol version
1 20180621 C 25 BHP 8
2 20180621 C 26 BHP 8
3 20180730 C 25 BHP 8
4 20180730 C 26 BHP 8
5 20180621 C 65 CBA 8
6 20180621 C 64 CBA 8
7 20180730 C 65 CBA 8
8 20180730 C 64 CBA 8
This can also be written as unstack(data.frame(d<-unlist(Chain),sub("contract[.]","",names(d)))) Although I would prefer to maintain the name contract in order to know which columns indeed form the contract dataframe needed
Or even you can change the names After unstacking.
With the new data:
a=readLines("https://raw.githubusercontent.com/hughandersen/OptionsTrading/master/Stocks_option_chain")
b=eval(parse(text=paste(a,collapse="")))
s=unstack(data.frame(d<-unlist(b[6]),names(d)))
I'm trying to perform a Hierarchical cluster analysis with RStudio, by using the package daisy. This is my dataset:
data.frame':341 obs. of 28 variables:
$ Impo_Env : Ord.factor w/ 3 levels "Low"<"Med"<"High": 3 2 3 2 3 2 3 3 2 3 ...
$ ComparativePriority_IAS: Ord.factor w/ 3 levels "Low"<"Med"<"High": 3 1 3 2 3 2 3 2 3 2 ...
$ Strategy_Eradication: Ord.factor w/ 3 levels "No intervention"<..: 3 2 3 2 3 2 3 2 2 3 ...
$ Knowl_BiodivLoss: Factor w/ 2 levels "0","1": 2 1 2 2 2 1 2 2 2 2 ...
$ Control_Trade: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
$ Engagement_Retail: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
$ Knowl_PastProj: Factor w/ 2 levels "0","1": 1 1 1 1 2 1 1 1 2 1 ...
$ Priority_IAS: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
$ Knowl_Eradic: Factor w/ 2 levels "0","1": 2 1 2 1 2 2 1 2 2 1 ...
$ Alert_CFS: Factor w/ 2 levels "0","1": 1 2 1 2 1 2 2 1 2 1 ...
$ Alert_Municipality: Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Alert_Park: Factor w/ 2 levels "0","1": 2 1 2 1 2 1 1 2 1 1 ...
$ Alert_Police: Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ Alert_Firemen: Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 2 ...
$ Supp_AuthorityIAS: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
$ Knowl_Env: Factor w/ 2 levels "0","1": 2 1 2 2 2 2 2 2 2 2 ...
$ Info_Tv: Factor w/ 2 levels "0","1": 2 2 1 2 2 2 2 1 2 1 ...
$ Info_Web: Factor w/ 2 levels "0","1": 2 1 2 2 2 1 2 1 2 2 ...
$ Info_Radio: Factor w/ 2 levels "0","1": 1 1 1 1 1 2 1 2 1 1 ...
$ Info_Magazines: Factor w/ 2 levels "0","1": 1 1 2 1 2 1 1 2 1 1 ...
$ Info_School: Factor w/ 2 levels "0","1": 1 1 2 1 1 1 1 1 1 2 ...
$ Blacklist: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
$ Workshop: Factor w/ 2 levels "0","1": 1 1 2 1 2 1 2 2 1 1 ...
$ SuppFin_FutProj: Factor w/ 2 levels "0","1": 2 1 2 1 2 2 2 2 2 2 ...
$ Tourist_dummy: Factor w/ 2 levels "0","1": 1 1 1 2 2 1 1 1 2 1 ...
$ Gender: Factor w/ 2 levels "Female","Male": 1 2 1 2 1 1 2 2 2 1 ...
$ logIASknown: num 2.89 2.94 2.89 2.56 3.14 ...
$ Age: int 20 41 14 10 26 33 19 59 23 16 ...
I would like to use the Euclidean distance with daisy, however when I run
daisy(fuu, metric = c("euclidean"), type=list(ordratio = c(1,2,3), asymm=c(4:24), symm=c(25,26)))
The output is not fine. Gower's distance is used instead of Euclidean distance:
Warning message:In daisy(fuu, metric = c("euclidean"), type = list(ordratio = c(1,:with mixed variables, metric "gower" is used automatically
How can I fix it?
As described in "Details" section within the documentation of the daisy function contained in the cluster package:
The handling of nominal, ordinal, and (a)symmetric binary data is
achieved by using the general dissimilarity coefficient of Gower
(1971). If x contains any columns of these data-types, both arguments
metric and stand will be ignored and Gower's coefficient will be used
as the metric.
In other words, for euclidean metrics (distances as root sum-of-squares of differences) to be computed, input columns must be numeric (mode) variables (i.e. all columns when x is a matrix) and thus recognised as interval scaled variables, as opposed to nominal (columns of class factor) variables or ordinal (columns of class ordered) variables. Specifying variable type within the type argument does not change this fact.
Under these premises, and supposing it makes sense for all of your 28 variables despite some being qualitative binary, you might try converting them with as.numeric and proceed then, reason being: with mixed variables metric "gower" overrides being automatically used.
I have 50 year-classes, and age and length data on individuals within each year class.
Without inputting a different data set for each year class I'm trying to plot average age for each year class.
As in year class along the x axis and average age (or length) (for each year class) along the y.
This is my data frame
'data.frame': 236628 obs. of 7 variables:
$ maturity: Ord.factor w/ 2 levels "0"<"1": 1 1 1 1 1 1 1 1 1 1 ...
$ year : Factor w/ 50 levels "1966","1967",..: 1 1 1 1 1 1 1 1 1 1 ...
$ quarter : Factor w/ 4 levels "1","2","3","4": 1 1 1 1 1 1 1 1 1 1 ...
$ area : Factor w/ 10 levels "1","2","3","4",..: 2 2 2 2 2 2 2 2 2 2 ...
$ lngth : int 145 145 145 150 150 150 150 150 155 155 ...
$ age : int 1 1 1 1 1 1 1 1 1 1 ...
$ sex : Ord.factor w/ 2 levels "0"<"1": 1 2 2 2 2 2 2 2 1 2 ...
Cheers
I'm looking for an elegant way to change multiple vectors' datatypes in R.
I'm working with an educational dataset: 426 students' answers to eight multiple choice questions (1 = correct, 0 = incorrect), plus a column indicating which instructor (1, 2, or 3) taught their course.
As it stands, my data is sitting pretty in data.df, like this:
str(data.df)
'data.frame': 426 obs. of 9 variables:
$ ques01: int 1 1 1 1 1 1 0 0 0 1 ...
$ ques02: int 0 0 1 1 1 1 1 1 1 1 ...
$ ques03: int 0 0 1 1 0 0 1 1 0 1 ...
$ ques04: int 1 0 1 1 1 1 1 1 1 1 ...
$ ques05: int 0 0 0 0 1 0 0 0 0 0 ...
$ ques06: int 1 0 1 1 0 1 1 1 1 1 ...
$ ques07: int 0 0 1 1 0 1 1 0 0 1 ...
$ ques08: int 0 0 1 1 1 0 1 1 0 1 ...
$ inst : num 1 1 1 1 1 1 1 1 1 1 ...
But those ques0x values aren't really integers. Rather, I think it's better to have R treat them as experimental factors. Same goes for the "inst" values.
I'd love to turn all those ints and nums into factors
Ideally, an elegant solution should produce a dataframe—I call it factorData.df—that looks like this:
str(factorData.df)
'data.frame': 426 obs. of 9 variables:
$ ques01: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 1 1 1 2 ...
$ ques02: Factor w/ 2 levels "0","1": 1 1 2 2 2 2 2 2 2 2 ...
$ ques03: Factor w/ 2 levels "0","1": 1 1 2 2 1 1 2 2 1 2 ...
$ ques04: Factor w/ 2 levels "0","1": 2 1 2 2 2 2 2 2 2 2 ...
$ ques05: Factor w/ 2 levels "0","1": 1 1 1 1 2 1 1 1 1 1 ...
$ ques06: Factor w/ 2 levels "0","1": 2 1 2 2 1 2 2 2 2 2 ...
$ ques07: Factor w/ 2 levels "0","1": 1 1 2 2 1 2 2 1 1 2 ...
$ ques08: Factor w/ 2 levels "0","1": 1 1 2 2 2 1 2 2 1 2 ...
$ inst : Factor w/ 3 levels "1","2","3": 1 1 1 1 1 1 1 1 1 1 ...
I'm fairly certain that whatever solution you folks come up with, it ought to be easy to generalize to any n number of variables that'd need to get reclassified, and would work across most common conversions (int -> factor and num -> int, for example).
No matter what solution you folks generate, it's bound to be more elegant than mine
Because my current clunky code is just 9 separate factor() statements, one for each variable, like this
factorData.df$ques01
I'm brand-new to R, programming, and stackoverflow. Please be gentle, and thanks in advance for your help!
This was also answered in R-Help.
I imagine that there's a better way to do it, but here are two options:
# use a sample data set
> str(cars)
'data.frame': 50 obs. of 2 variables:
$ speed: num 4 4 7 7 8 9 10 10 10 11 ...
$ dist : num 2 10 4 22 16 10 18 26 34 17 ...
> data.df <- cars
You can use lapply:
> data.df <- data.frame(lapply(data.df, factor))
Or a for statement:
> for(i in 1:ncol(data.df)) data.df[,i] <- as.factor(data.df[,i])
In either case, you end up with what you want:
> str(data.df)
'data.frame': 50 obs. of 2 variables:
$ speed: Factor w/ 19 levels "4","7","8","9",..: 1 1 2 2 3 4 5 5 5 6 ...
$ dist : Factor w/ 35 levels "2","4","10","14",..: 1 3 2 9 5 3 7 11 14 6 ...
I found an alternative solution in the plyr package:
# load the package and data
> library(plyr)
> data.df <- cars
Use the colwise function:
> data.df <- colwise(factor)(data.df)
> str(data.df)
'data.frame': 50 obs. of 2 variables:
$ speed: Factor w/ 19 levels "4","7","8","9",..: 1 1 2 2 3 4 5 5 5 6 ...
$ dist : Factor w/ 35 levels "2","4","10","14",..: 1 3 2 9 5 3 7 11 14 6 ...
Incidentally, if you look inside the colwise function, it just uses lapply:
df <- as.data.frame(lapply(filtered, .fun, ...))