My goal is to find much simpler code, which can generalize, that shows the relationships between responses to two survey questions. In the MWE, one question asked respondents to rank eight marketing selections from 1 to 8 and the other asked them to rank nine attribute selections from 1 to 9. Higher rankings indicate the respondent favored the selection more. Here is the data frame.
structure(list(Email = c("a", "b", "c", "d", "e", "f", "g", "h",
"i"), Ads = c(2, 1, 1, 1, 1, 2, 1, 1, 1), Alumni = c(3, 2, 2,
3, 2, 3, 2, 2, 2), Articles = c(6, 4, 3, 2, 3, 4, 3, 3, 3), Referrals = c(4,
3, 4, 8, 7, 8, 8, 6, 4), Speeches = c(7, 7, 6, 7, 4, 7, 4, 5,
5), Updates = c(8, 6, 6, 5, 5, 5, 5, 7, 6), Visits = c(5, 8,
7, 6, 6, 6, 6, 4, 8), `Business Savvy` = c(10, 6, 10, 10, 4,
4, 6, 8, 9), Communication = c(4, 3, 8, 3, 3, 9, 7, 6, 7), Experience = c(7,
7, 7, 9, 2, 8, 5, 9, 5), Innovation = c(2, 1, 4, 2, 1, 2, 2,
1, 1), Nearby = c(3, 2, 2, 1, 5, 3, 3, 2, 2), Personal = c(8,
10, 6, 8, 6, 10, 4, 3, 3), Rates = c(9, 5, 9, 6, 9, 7, 10, 5,
4), `Staffing Model` = c(6, 8, 5, 5, 7, 5, 8, 7, 8), `Total Cost` = c(5,
4, 3, 7, 8, 6, 9, 4, 6)), row.names = c(NA, -9L), class = c("tbl_df",
"tbl", "data.frame"))
If numeric rankings cannot be used for my solution to calculating relationships (correlations), please correct me.
Hoping they can be used, I arrived at the following plodding code, which I hope calculates the correlation matrix of each method selection against each attribute selection.
library(psych)
dataframe2 <- psych::corr.test(dataframe[ , c(2, 9:17)])[[1]][1:10] # the first method vs all attributes
dataframe3 <- psych::corr.test(dataframe[ , c(3, 9:17)])[[1]][1:10] # the 2nd method vs all attributes and so on
dataframe4 <- psych::corr.test(dataframe[ , c(4, 9:17)])[[1]][1:10]
dataframe5 <- psych::corr.test(dataframe[ , c(5, 9:17)])[[1]][1:10]
dataframe6 <- psych::corr.test(dataframe[ , c(6, 9:17)])[[1]][1:10]
dataframe7 <- psych::corr.test(dataframe[ , c(7, 9:17)])[[1]][1:10]
dataframe8 <- psych::corr.test(dataframe[ , c(8, 9:17)])[[1]][1:10]
# create a dataframe from the rbinded rows
bind <- data.frame(rbind(dataframe2, dataframe3, dataframe4, dataframe5, dataframe6, dataframe7, dataframe8))
Rename rows and columns:
colnames(bind) <- c("Sel", colnames(dataframe[9:17]))
rownames(bind) <- colnames(dataframe[2:8])
How can I accomplish the above more efficiently?
By the way, the bind data frame also allows one to produce a heat map with the DataExplorer package.
library(DataExplorer)
DataExplorer::plot_correlation(bind)
[Summary]
In the scope of our discussion, there are two ways to get the correlation data.
Use stats::cor, i.e., cor(subset(dataframe, select = -Email))
Use psych::corr.test, i.e., corr.test(subset(dataframe, select = -Email))[[1]]
Then you may subset the correlation matrix with the desired rows and columns.
In order to use DataExplorer::plot_correlation, you can simply do plot_correlation(dataframe, type = "c"). Note: the output heatmap will include correlations for all columns, so you can just ignore columns that are not of interests.
[Original Answer]
## Create data
dataframe <- structure(
list(
Email = c("a", "b", "c", "d", "e", "f", "g", "h", "i"),
Ads = c(2, 1, 1, 1, 1, 2, 1, 1, 1),
Alumni = c(3, 2, 2, 3, 2, 3, 2, 2, 2),
Articles = c(6, 4, 3, 2, 3, 4, 3, 3, 3),
Referrals = c(4, 3, 4, 8, 7, 8, 8, 6, 4),
Speeches = c(7, 7, 6, 7, 4, 7, 4, 5, 5),
Updates = c(8, 6, 6, 5, 5, 5, 5, 7, 6),
Visits = c(5, 8, 7, 6, 6, 6, 6, 4, 8),
`Business Savvy` = c(10, 6, 10, 10, 4, 4, 6, 8, 9),
Communication = c(4, 3, 8, 3, 3, 9, 7, 6, 7),
Experience = c(7, 7, 7, 9, 2, 8, 5, 9, 5),
Innovation = c(2, 1, 4, 2, 1, 2, 2, 1, 1),
Nearby = c(3, 2, 2, 1, 5, 3, 3, 2, 2),
Personal = c(8, 10, 6, 8, 6, 10, 4, 3, 3),
Rates = c(9, 5, 9, 6, 9, 7, 10, 5, 4),
`Staffing Model` = c(6, 8, 5, 5, 7, 5, 8, 7, 8),
`Total Cost` = c(5, 4, 3, 7, 8, 6, 9, 4, 6)
),
row.names = c(NA, -9L),
class = c("tbl_df", "tbl", "data.frame")
)
Following your example strictly, we can do the following:
## Calculate correlation
df2 <- subset(dataframe, select = -Email)
marketing_selections <- names(df2)[1:7]
attribute_selections <- names(df2)[8:16]
corr_matrix <- psych::corr.test(df2)[[1]]
bind <- subset(corr_matrix,
subset = rownames(corr_matrix) %in% marketing_selections,
select = attribute_selections)
DataExplorer::plot_correlation(bind)
WARNING
However, is this what you really want? psych::corr.test generates the correlation matrix, and DataExplorer::plot_correlation calculates the correlation again. It is like the correlation of the correlation.
I am learning to use various forecasting packages available in R, and came across bsts(). The data I deal with is a time series of demands.
data=c(27, 2, 7, 7, 9, 4, 3, 3, 3, 9, 6, 2, 6, 2, 3, 8, 6, 1, 3, 8, 4, 5, 8, 5, 4, 4, 6, 1, 6, 5, 1, 3, 0, 2, 6, 7, 1, 2, 6, 2, 8, 6, 1, 1, 3, 2, 1, 3, 1, 6, 3, 4, 3, 7, 3, 4, 1, 7, 5, 6, 3, 4, 3, 9, 2, 1, 7, 2, 2, 9, 4, 5, 3, 4, 2, 4, 4, 8, 6, 3, 9, 2, 9, 4, 1, 3, 8, 1, 7, 7, 6, 0, 1, 4, 8, 9, 2, 5)
ts.main=ts(data, start=c(1910,1), frequency=12)
ss <- AddLocalLinearTrend(list(), y=ts.main)
ss <- AddSeasonal(ss, y=as.numeric(ts.temp), nseasons=12)
model <- bsts(as.numeric(ts.temp),
state.specification = ss,
niter = 1000)
pred <- predict(model, horizon = 12)
Is there way I can restrict pred$mean from becoming negative?
Since your data are a time series of counts, you need to take that into account rather than assume Gaussian errors; for some discussion on this and elaboration of some approaches, see for example Brandt et al 2000 and Brandt and Williams 2001. Luckily, the bsts package has a built-in functionality for this, the family option (see pages 24 to 26 of the documentation).
So, you can just do this
model <- bsts(as.numeric(ts.main),
state.specification = ss,
family = 'poisson',
niter = 1000)
so that the bsts() function correctly considers the data as counts, which will solve your issue, since the draws from the posterior predictive distribution will then be non-negative by definition.
df <- data.frame(
num =
c(5, 7, 3,
4, 2, 6,
5, 3, 6,
5, 6, 0,
7, 4, 0,
7, 7, 0,
6, 6, 0,
4, 6, 1,
6, 4, 0,
7, 7, 0,
2, 4, 0,
5, 7, 4,
7, 5, 0,
4, 5, 0,
6, 6, 3
),
x1 = factor(rep(c("xx", "pp", "tru"), 15)),
x2 = factor(rep(c("A", "B", "C"), 15)),
x3 = factor(rep(1:15, rep(3, 15))))
I would like to calculate significance for:
x1
x2
x3
interaction x1/x2
interaction x1/x3
interaction x2/x3
interaction x1/x2/x3
I think I have to do a linear model lm so I have tried
lm(df[,"num"] ~ df[,"x1"] * df[,"x2"] * df[,"x3"])
I am not sure if this is correct.
The rule of thumb is to fit a linear model then perform an ANOVA:
fit <- lm(num ~ x1 * x2 * x3, data = df)
anova(fit)
However, your provided toy example is really a bad one, so nothing interesting will be seen.
You have x1 and x2 identical (so they have perfect nesting). In this regard, you will get lots of NA coefficients;
You have no replication. For each factor combination you only have one observation, so you will end up with an exact fit with zero residuals.
I have a vector of calls made on each days of a certain month.
callsperDayforMonth <- c(3, 1, 2, 1, 1, 3, 9, 1, 4, 2, 6, 4, 9, 13, 15, 2, 5, 5, 2, 7, 3, 0, 1, 2, 7, 1, 8, 6, 9, 4)
I also have a vector of factors which spans the range of the "callsperDayforMonth" vector.
"0-2" "3-5" "6-8" "9-11" "12-14" "16+"
I need to create a histogram, with the factors on the horizontal axis.
How can this be done.
The hist command has an argument breaks that can be a vector of the breakpoints to be used. That should do what you want.
Or you could use table and cut to do the counts yourself and create a barplot from the result.
For example:
library(ggplot2)
cuts <- cut(callsperDayforMonth,
breaks = c(-Inf,2, 5, 8, 11, 14, 16, Inf),
labels = c("0-2", "3-5", "6-8", "9-11", "12-14", "15-16", "16+"))
df <- data.frame(cuts, callsperDayforMonth)
ggplot(df, aes(x=cuts)) + geom_bar(stat = "count")