I'm trying to build a linear regression model using eight independent variables, but when I run lm() one variable--what I anticipate being my best predictor!--keeps returning NA. I'm still new to R, and I cannot find a solution.
Here are my independent variables:
TEMPERATURE
HUMIDITY
WIND_SPEED
VISIBILITY
DEW_POINT_TEMPERATURE
SOLAR_RADIATION
RAINFALL
SNOWFALL
My df is training_set and looks like:
I'm not sure whether this matters, but training_set is 75% of my original df, and testing_set is 25%. Created thusly:
set.seed(1234)
split_bike_sharing <- sample(c(rep(0, round(0.75 * nrow(bike_sharing_df))), rep(1, round(0.25 * nrow(bike_sharing_df)))))
This gave me table(split_bike_sharing):
0
1
6349
2116
And then I did:
training_set <- bike_sharing_df[split_bike_sharing == 0, ]
testing_set <- bike_sharing_df[split_bike_sharing == 1, ]
The structure of training_set is like:
To create the model I run the code:
lm_model_weather=lm(RENTED_BIKE_COUNT ~ TEMPERATURE + HUMIDITY + WIND_SPEED + VISIBILITY + DEW_POINT_TEMPERATURE +
SOLAR_RADIATION + RAINFALL + SNOWFALL, data = training_set)
However, as you can see the resultant model returns RAINFALL as NA. Here is the resultant model:
My first thought was to check RAINFALL datatype, which is numeric with range 0-1 (because at an earlier step I performed min-max normalization). But SNOWFALL also is numeric, and I've done nothing (that I know of!) to the one but not the other. My second thought was to confirm that RAINFALL contains enough values to work, and that does not appear to be an issue: summary(training_set$RAINFALL):
So, how do I correct the NAs in RAINFALL? Truly I will be most grateful for your guidance to a solution.
UPDATE 10 MARCH 2022
I've now checked for collinearity:
X <- model.matrix(RENTED_BIKE_COUNT ~ ., data = training_set)
X2 <- caret::findLinearCombos(X)
print(X2)
This gave me:
I believe this means certain columns are jointly multicollinear. As you can see, columns 8, 13, and 38 are:
[8] is RAINFALL
[13] is SEASONS_WINTER
[38] is HOUR_23
Question: if I want to preserve RAINFALL as a predictor variable (viz., return proper values rather than NAs when I run lm()), what do I do? Remove columns [13] and [38] from the dataset?
Related
In the past few days I have been trying to find how to do Fama Macbeth regressions in R. It is advised to use the plm package with pmg, however every attempt I do returns me that I have an insufficient number of time periods.
My Dataset consists of 2828419 observations with 13 columns of variables of which I am looking to do multiple cross-sectional regressions.
My firms are specified by seriesis, I have got a variable date and want to do the following Fama Macbeth regressions:
totret ~ size
totret ~ momentum
totret ~ reversal
totret ~ volatility
totret ~ value size
totret ~ value + size + momentum
totret ~ value + size + momentum + reversal + volatility
I have been using this command:
fpmg <- pmg(totret ~ momentum, Data, index = c("date", "seriesid")
Which returns: Error in pmg(totret ~ mom, Dataset, index = c("seriesid", "datem")) : Insufficient number of time periods
I tried it with my dataset being a datatable, dataframe and pdataframe. Switching the index does not work as well.
My data contains NAs as well.
Who can fix this, or find a different way for me to do Fama Macbeth?
This is almost certainly due to having NAs in the variables in your formula. The error message is not very helpful - it is probably not a case of "too few time periods to estimate" and very likely a case of "there are firm/unit IDs that are not represented across all time periods" due to missing data being dropped.
You have two options - impute the missing data or drop observations with missing data (the latter being a quick test that the model works without missing points before deciding what you want to do that is valid for estimtation).
If the missingness in your data is truly random, you might be okay just dropping observations with missingness. Otherwise you should probably impute. A common strategy here is to impute multiple times - at least 5 - and then estimate for each of those 5 resulting data sets and average the effect together. Amelia or mice are very strong imputation packages. I like Amelia because with one call you can impute n times for that many resulting data sets and it's easy to pass in a set of variables to not impute (e.g., id variable or time period) with the idvars parameter.
EDIT: I dug into the source code to see where the error was triggered and here is what the issue is - again likely caused by missing data, but it does interact with your degrees of freedom:
...
# part of the code where error is triggered below, here is context:
# X = matrix of the RHS of your model including intercept, so X[,1] is all 1s
# k = number of coefficients used determined by length(coef(plm.model))
# ind = vector of ID values
# so t here is the minimum value from a count of occurrences for each unique ID
t <- min(tapply(X[,1], ind, length))
# then if the minimum number of times a single ID appears across time is
# less than the number of coefficients + 1, you do not have enough time
# points (for that ID/those IDs) to estimate.
if (t < (k + 1))
stop("Insufficient number of time periods")
That is what is triggering your error. So imputation is definitely a solution, but there might be a single offender in your data and importantly, once this condition is satisfied your model will run just fine with missing data.
Lately, I fixed the Fama Macbeth regression in R.
From a Data Table with all of the characteristics within the rows, the following works and gives the opportunity to equally weight or apply weights to the regression (remove the ",weights = marketcap" for equally weighted). totret is a total return variable, logmarket is the logarithm of market capitalization.
logmarket<- df %>%
group_by(date) %>%
summarise(constant = summary(lm(totret~logmarket, weights = marketcap))$coefficient[1], rsquared = summary(lm(totret~logmarket*, weights = marketcap*))$r.squared, beta= summary(lm(totret~logmarket, weights = marketcap))$coefficient[2])
You obtain a DataFrame with monthly alphas (constant), betas (beta), the R squared (rsquared).
To retrieve coefficients with t-statistics in a dataframe:
Summarystatistics <- as.data.frame(matrix(data=NA, nrow=6, ncol=1)
names(Summarystatistics) <- "logmarket"
row.names(Summarystatistics) <- c("constant","t-stat", "beta", "tstat", "R^2", "observations")
Summarystatistics[1,1] <- mean(logmarket$constant)
Summarystatistics[2,1] <- coeftest(lm(logmarket$constant~1))[1,3]
Summarystatistics[3,1] <- mean(logmarket$beta)
Summarystatistics[4,1] <- coeftest(lm(logmarket$beta~1))[1,3]
Summarystatistics[5,1] <- mean(logmarket$rsquared)
Summarystatistics[6,1] <- nrow(subset(df, !is.na(logmarket)))
There are some entries of "seriesid" with only one entry. Therefore the pmg gives the error. If you do something like this (with variable names you use), it will stop the error:
try2 <- try2 %>%
group_by(cusip) %>%
mutate(flag = (if (length(cusip)==1) {1} else {0})) %>%
ungroup() %>%
filter(flag == 0)
I am trying to simulate monthly panels of data where one variable depends on lagged values of that variable in R. My solution is extremely slow. I need around 1000 samples of 2545 individuals, each of whom is observed monthly over many years, but the first sample took my computer 8.5 hours to construct. How can I make this faster?
I start by creating an unbalanced panel of people with different birth dates, monthly ages, and variables xbsmall and error that will be compared to determine the Outcome. All of the code in the first block is just data setup.
# Setup:
library(plyr)
# Would like to have 2545 people (nPerson).
#Instead use 4 for testing.
nPerson = 4
# Minimum and maximum possible ages and birth dates
AgeMin = 10
AgeMax = 50
BornMin = 1950
BornMax = 1963
# Person-specific characteristics
ind =
data.frame(
id = 1:nPerson,
BornYear = floor(runif(length(1:nPerson), min=BornMin, max=BornMax+1)),
BornMonth = ceiling(runif(length(1:nPerson), min=0, max=12))
)
# Make an unbalanced panel of people over age 10 up to year 1986
# panel = ddply(ind, ~id, transform, AgeMonths = BornMonth)
panel = ddply(ind, ~id, transform, AgeMonths = (AgeMin*12):((1986-BornYear)*12 + 12-BornMonth))
# Set up some random variables to approximate the data generating process
panel$xbsmall = rnorm(dim(panel)[1], mean=-.3, sd=.45)
# Standard normal error for probit
panel$error = rnorm(dim(panel)[1])
# Placeholders
panel$xb = rep(0, dim(panel)[1])
panel$Outcome = rep(0, dim(panel)[1])
Now that we have data, here is the part that is slow (around a second on my computer for only 4 observations but hours for thousands of observations). Each month, a person gets two draws (xbsmall and error) from two different normal distributions (these were done above), and Outcome == 1 if xbsmall > error. However, if Outcome equals 1 in the previous month, then Outcome in the current month equals 1 if xbsmall + 4.47 > error. I use xb = xbsmall+4.47 in the code below (xb is the "linear predictor" in a probit model). I ignore the first month for each person for simplicity. For your information, this is simulating a probit DGP (but that is not necessary to know to solve the problem of computation speed).
# Outcome == 1 if and only if xb > -error
# The hard part: xb includes information about the previous month's outcome
start_time = Sys.time()
for(i in 1:nPerson){
# Determine the range of monthly ages to loop over for this person
AgeMonthMin = min(panel$AgeMonths[panel$id==i], na.rm=T)
AgeMonthMax = max(panel$AgeMonths[panel$id==i], na.rm=T)
# Loop over the monthly ages for this person and determine the outcome
for(t in (AgeMonthMin+1):AgeMonthMax){
# Indicator for whether Outcome was 1 last month
panel$Outcome1LastMonth[panel$id==i & panel$AgeMonths==t] = panel$Outcome[panel$id==i & panel$AgeMonths==t-1]
# xb = xbsmall + 4.47 if Outcome was 1 last month
# Otherwise, xb = xbsmall
panel$xb[panel$id==i & panel$AgeMonths==t] = with(panel[panel$id==i & panel$AgeMonths==t,], xbsmall + 4.47*Outcome1LastMonth)
# Outcome == 1 if xb > 0
panel$Outcome[panel$id==i & panel$AgeMonths==t] =
ifelse(panel$xb[panel$id==i & panel$AgeMonths==t] > - panel$error[panel$id==i & panel$AgeMonths==t], 1, 0)
}
}
end_time = Sys.time()
end_time - start_time
My thoughts for reducing computer time:
Something with cumsum()
Some wonderful panel data function that I do not know about
Find a way to make the t loop go through the same starting and ending points for each individual and then somehow use plyr::ddpl() or dplyr::gather_by()
Iterative solution: make an educated guess about the value of Outcome at each monthly age (say, the mode) and somehow adjust values that do not match the previous month. This would work better in my real application because xbsmall has a very clear trend in age.
Do the simulation only for smaller samples and then estimate the effect of sample size on the values I need (the distributions of regression coefficient estimates not calculated here)
One approach is to use a split-apply-combine method. I take out the for(t in (AgeMonthMin+1):AgeMonthMax) loop and put the contents in a function:
generate_outcome <- function(x) {
AgeMonthMin <- min(x$AgeMonths, na.rm = TRUE)
AgeMonthMax <- max(x$AgeMonths, na.rm = TRUE)
for (i in 2:(AgeMonthMax - AgeMonthMin + 1)){
x$xb[i] <- x$xbsmall[i] + 4.47 * x$Outcome[i - 1]
x$Outcome[i] <- ifelse(x$xb[i] > - x$error[i], 1, 0)
}
x
}
where x is a dataframe for one person. This allows us to simplify the panel$id==i & panel$AgeMonths==t construct. Now we can just do
out <- lapply(split(panel, panel$id), generate_outcome)
out <- do.call(rbind, out)
and all.equal(panel$Outcome, out$Outcome) returns TRUE. Computing 100 persons took 1.8 seconds using this method, compared to 1.5 minutes in the original code.
I just started using R for statistical purposes and I appreciate any kind of help.
As a first step, I ran a time series regression over my columns. Y values are dependent and the X is explanatory.
# example
Y1 <- runif(100, 5.0, 17.5)
Y2 <- runif(100, 4.0, 27.5)
Y3 <- runif(100, 3.0, 14.5)
Y4 <- runif(100, 2.0, 12.5)
Y5 <- runif(100, 5.0, 17.5)
X <- runif(100, 5.0, 7.5)
df1 <- data.frame(X, Y1, Y2, Y3, Y4, Y5)
# calculating log returns to provide data for the first regression
n <- nrow(df1)
X_logret <- log(X[2:n])-log(X[1:(n-1)])
Y1_logret <- log(Y1[2:n])-log(Y1[1:(n-1)])
Y2_logret <- log(Y2[2:n])-log(Y2[1:(n-1)])
Y3_logret <- log(Y3[2:n])-log(Y3[1:(n-1)])
Y4_logret <- log(Y4[2:n])-log(Y4[1:(n-1)])
Y5_logret <- log(Y5[2:n])-log(Y5[1:(n-1)])
# bringing the calculated log returns together in one data frame
df2 <- data.frame(X_logret, Y1_logret, Y2_logret, Y3_logret, Y4_logret, Y5_logret)
# running the time series regression
Regression <- lm(as.matrix(df2[c('Y1_logret', 'Y2_logret', 'Y3_logret', 'Y4_logret', 'Y5_logret')]) ~ df2$X)
# extracting the coefficients for further calculation
Regression$coefficients[2,(1:5)]
As a second step I want to run a regression row by row, which is day by day, since the data contains daily observed values. I also have a column "DATE" but I didn't know how to bring it in here in the example. The format of the DATE column is POSIXct, maybe someone has an idea how to refer to a certain period in it on which the regression should be done.
In the row by row regression I would like to use the 5 calculated coefficients (from the first regression) as an explanatory variable. The 5 Y_logret values, I would like to use as dependent variable.
Y_logret(1 to 5) = Beta * Regression$coefficients[2,(1:5)] + error value. The intercept is not needed, so I would set it to zero by adding +0 in the lm function.
My goal is to run this regression over a period of time, for example over 20 days. Day by day, this would provide a total of 20 Beta estimates (for one regression per day), but I would also need all errors for further calculation. So I have to extract 5 errors per day, that is a total of 20*5 error values.
This is just an example, in the original dataset I have 20 of the Y values and over 4000 rows. I would like to run the regression over certain intervals with 900-1000 day. Since I am completely new to R, I have no idea how to proceed. Especially how to code this in a few lines.
I really appreciate any kind of help.
I received some good help getting my data formatted properly produce a multinomial logistic model with mlogit here (Formatting data for mlogit)
However, I'm trying now to analyze the effects of covariates in my model. I find the help file in mlogit.effects() to be not very informative. One of the problems is that the model appears to produce a lot of rows of NAs (see below, index(mod1) ).
Can anyone clarify why my data is producing those NAs?
Can anyone help me get mlogit.effects to work with the data below?
I would consider shifting the analysis to multinom(). However, I can't figure out how to format the data to fit the formula for use multinom(). My data is a series of rankings of seven different items (Accessible, Information, Trade offs, Debate, Social and Responsive) Would I just model whatever they picked as their first rank and ignore what they chose in other ranks? I can get that information.
Reproducible code is below:
#Loadpackages
library(RCurl)
library(mlogit)
library(tidyr)
library(dplyr)
#URL where data is stored
dat.url <- 'https://raw.githubusercontent.com/sjkiss/Survey/master/mlogit.out.csv'
#Get data
dat <- read.csv(dat.url)
#Complete cases only as it seems mlogit cannot handle missing values or tied data which in this case you might get because of median imputation
dat <- dat[complete.cases(dat),]
#Change the choice index variable (X) to have no interruptions, as a result of removing some incomplete cases
dat$X <- seq(1,nrow(dat),1)
#Tidy data to get it into long format
dat.out <- dat %>%
gather(Open, Rank, -c(1,9:12)) %>%
arrange(X, Open, Rank)
#Create mlogit object
mlogit.out <- mlogit.data(dat.out, shape='long',alt.var='Open',choice='Rank', ranked=TRUE,chid.var='X')
#Fit Model
mod1 <- mlogit(Rank~1|gender+age+economic+Job,data=mlogit.out)
Here is my attempt to set up a data frame similar to the one portrayed in the help file. It doesnt work. I confess although I know the apply family pretty well, tapply is murky to me.
with(mlogit.out, data.frame(economic=tapply(economic, index(mod1)$alt, mean)))
Compare from the help:
data("Fishing", package = "mlogit")
Fish <- mlogit.data(Fishing, varying = c(2:9), shape = "wide", choice = "mode")
m <- mlogit(mode ~ price | income | catch, data = Fish)
# compute a data.frame containing the mean value of the covariates in
# the sample data in the help file for effects
z <- with(Fish, data.frame(price = tapply(price, index(m)$alt, mean),
catch = tapply(catch, index(m)$alt, mean),
income = mean(income)))
# compute the marginal effects (the second one is an elasticity
effects(m, covariate = "income", data = z)
I'll try Option 3 and switch to multinom(). This code will model the log-odds of ranking an item as 1st, compared to a reference item (e.g., "Debate" in the code below). With K = 7 items, if we call the reference item ItemK, then we're modeling
log[ Pr(Itemk is 1st) / Pr(ItemK is 1st) ] = αk + xTβk
for k = 1,...,K-1, where Itemk is one of the other (i.e. non-reference) items. The choice of reference level will affect the coefficients and their interpretation, but it will not affect the predicted probabilities. (Same story for reference levels for the categorical predictor variables.)
I'll also mention that I'm handling missing data a bit differently here than in your original code. Since my model only needs to know which item gets ranked 1st, I only need to throw out records where that info is missing. (E.g., in the original dataset record #43 has "Information" ranked 1st, so we can use this record even though 3 other items are NA.)
# Get data
dat.url <- 'https://raw.githubusercontent.com/sjkiss/Survey/master/mlogit.out.csv'
dat <- read.csv(dat.url)
# dataframe showing which item is ranked #1
ranks <- (dat[,2:8] == 1)
# for each combination of predictor variable values, count
# how many times each item was ranked #1
dat2 <- aggregate(ranks, by=dat[,9:12], sum, na.rm=TRUE)
# remove cases that didn't rank anything as #1 (due to NAs in original data)
dat3 <- dat2[rowSums(dat2[,5:11])>0,]
# (optional) set the reference levels for the categorical predictors
dat3$gender <- relevel(dat3$gender, ref="Female")
dat3$Job <- relevel(dat3$Job, ref="Government backbencher")
# response matrix in format needed for multinom()
response <- as.matrix(dat3[,5:11])
# (optional) set the reference level for the response by changing
# the column order
ref <- "Debate"
ref.index <- match(ref, colnames(response))
response <- response[,c(ref.index,(1:ncol(response))[-ref.index])]
# fit model (note that age & economic are continuous, while gender &
# Job are categorical)
library(nnet)
fit1 <- multinom(response ~ economic + gender + age + Job, data=dat3)
# print some results
summary(fit1)
coef(fit1)
cbind(dat3[,1:4], round(fitted(fit1),3)) # predicted probabilities
I didn't do any diagnostics, so I make no claim that the model used here provides a good fit.
You are working with Ranked Data, not just Multinomial Choice Data. The structure for the Ranked data in mlogit is that first set of records for a person are all options, then the second is all options except the one ranked first, and so on. But the index assumes equal number of options each time. So a bunch of NAs. We just need to get rid of them.
> with(mlogit.out, data.frame(economic=tapply(economic, index(mod1)$alt[complete.cases(index(mod1)$alt)], mean)))
economic
Accessible 5.13
Debate 4.97
Information 5.08
Officials 4.92
Responsive 5.09
Social 4.91
Trade.Offs 4.91
I am learning r and trying to understand one concept in building a model:
The data:
Time Counts
1 0 126.6
2 1 101.8
3 2 71.6
etc...
The model:
Time2 <- Time^2
quadratic.model <-lm(Counts ~ Time + Time2)
The prediction:
timevalues <- seq(0, 30, 0.1)
predictedcounts <- predict(quadratic.model,list(Time=timevalues, Time2=timevalues^2))
I don't understand this part of the above function.
list(Time=timevalues, Time2=timevalues^2)
What exactly is the list doing? Is there a more intuitive way to accomplish the same thing?
The list is specifying what values of Time and Time2 should be used for prediction. If you had different time values (say from a cross validation set) called TimeValuesB, then by setting list(Time = TimeValuesB, Time2 = TimeValuesB^2) you could obtain the predicted output for these new data values.
However, if you just want to obtain the predictions from the original data you can omit the list. So in your case
predictedcounts <- predict(quadratic.model)
should work just fine.