Household Size 0 1 2 3 4 5+
Bedrooms Bedrooms Bedrooms Bedrooms Bedrooms Bedrooms
1 253 4486 2033 930 105 8
2 10 666 3703 947 85 7
3 4 68 1972 1621 52 5
4 1 12 680 1835 164 11
5+ 0 6 147 1230 721 122
I have the above dataframe where 'Bedrooms' is a label on the columns.
I'm trying to change this into a data table I can then use within rmarkdown to add into a flexdashboard. When I use the below code:
DT::datatable(df, rownames = FALSE, extensions = 'FixedColumns', escape=TRUE,options= list(bPaginate = FALSE))
I get the output:
Household Size 0 1 2 3 4 5+
1 253 4486 2033 930 105 8
2 10 666 3703 947 85 7
3 4 68 1972 1621 52 5
4 1 12 680 1835 164 11
5+ 0 6 147 1230 721 122
I have a few problems with this:
The lables that say 'Bedrooms' don't show, so there's no way of knowing what these numbers in the columns actually mean. I'd like to include the labels or have a Row on top of the column names that says "Number of Bedrooms" that covers all of the rows?
The column Household Size and 5+ have a wider width than the rest of the columns, I want these to either be the same or Household Size to be slightly bigger than the rest
I think it's worth noting that the row 5+ and the column 5+ are both a new row/column that count any value above 5.
Also, this is just an extra but I'd like to colour the bottom left cells red and the top right cells green, is this possible?
I've figured out how to keep 'Bedrooms' in the column titles. It's possible to set the column names within DT::datatable using the code below;
DT::datatable(HS_BED_ALL, rownames = FALSE, colnames=c('Household Size','0 Bedrooms','1 Bedroom','2 Bedrooms','3 Bedrooms','4 Bedrooms','5+ Bedrooms'), extensions = 'FixedColumns', escape=TRUE, options= list(bPaginate = FALSE, dom = 't',buttons = c('excel')))%>%formatStyle(1:7,fontSize = '14px')
Which gives the desired output.
Related
This question already has answers here:
Replacing values from a column using a condition in R
(2 answers)
Closed 4 years ago.
Here is my code
nutrients<- read.csv("nutrients.csv", head = TRUE, sep = ",")
> plot(nutrients)
> head(nutrients)
crop Nutrient.dens N..tons.acre. P2O5 K2O sum.nut
1 broccoli 340.0 210 245 100 555
2 carrot 458.0 70 250 50 370
3 cauliflower 315.0 25 35 80 140
4 letuce 318.5 165 150 90 405
5 onion 109.0 120 30 150 300
6 tomato 186.0 175 85 275 535
> df_nutrients<- as.data.frame(nutrients)
> df_nutrients<- df_nutrients[1,1=="broc"]
I am sure this is easy, and Ive tried searching anything i can find to get the answer but i cannot find it. I just need to change that one variable to "broc". is there a specific function i need or something?
If crop is a character type, then a simple subset should work
nutrients$crop[nutrients$crop == "broccoli"] <- "broc"
If crop is a factor, then use this:
levels(nutrients$crop)[levels(nutrients$crop) == "broccoli"] <- "proc"
I asked the question earlier here Using variations of `apply` in R. Now I have an extension to that question. In my 40 variables, some variables are categorical. I need the number of observations for each unique quality. I would like to use some form of apply because I have been using sapply and tapply on various parts of this code, but it is not required. Here is a bit of the data:
Age Wt Ht Type Color Width
79 134 66 C red small
67 199 64 C green small
39 135 78 T yellow small
92 149 61 C yellow medium
33 138 75 T green medium
68 139 71 C yellow medium
95 198 62 T red large
65 132 65 T blue large
56 138 81 C green large
71 193 78 T blue large
What the last two columns should look like is
C T
red 1 1
green 2 1
blue 0 2
yellow 2 1
small 2 1
medium 2 1
large 1 3
Also, I know I could use 'table', but how do I send multiple variables one at a time against Type in order to get it to look something like this? Using table as opposed to apply is fine with me.
Thanks!
We can use table after unlisting the 'Color' and 'Width' columns and replicating the 'Type'.
Un1 <- unlist(df1[5:6])
Un2 <- df1$Type[row(df1[5:6])]
If we need a customer order, convert to factor and specify the levels in the same order.
table(factor(Un1, levels = c("red", "green", "blue", "yellow", "small",
"medium", "large")), Un2)
# Un2
# C T
# red 1 1
# green 2 1
# blue 0 2
# yellow 2 1
# small 2 1
# medium 2 1
# large 1 3
Or if the order is based on the order of appearance of unique elements in each of the columns
table(factor(Un1, levels = unique(Un1)), Un2)
Say, I have a data.frame() like this
>head(Acquisition)
original_date first_payment_date LTV DTI FICO
1 01/2007 03/2007 56 37 734
2 02/2007 04/2007 80 11 762
3 12/2006 02/2007 80 28 656
4 12/2006 03/2007 70 50 700
I want to discretize the Acquisition$LTV and Acquisition$DTI by the step size 0.05 and Acquisition$FICO by the step size 10.
I have found the answer just use cut function is okay.
dis.LTV=cut(Acquisition$LTV,(max(Acquisition$LTV)-min(Acquisition$LTV))/0.05)
I work with neuralnet package to predict values of stocks (diploma thesis). The example data are below
predict<-runif(23,min=0,max=1)
day<-c(369:391)
ChoosedN<-c(2,5,5,5,5,5,4,3,5,5,5,2,1,1,5,5,4,3,2,3,4,3,2)
Profit<-runif(23,min=-2,max=5)
df<-data.frame(predict,day,ChoosedN,Profit)
colnames(df)<-c('predict','day','ChoosedN','Profit')
But I haven't always same period for investments (ChoodedN). For backtest the neural site I have to skip the days when I am still in position even if the neural site says 'buy it' (i.e.predict > 0.5). The frame looks like this
predict day ChoosedN Profit
1 0.6762981061 369 2 -1.6288823350
2 0.0195611224 370 5 1.5682195597
3 0.2442795106 371 5 0.6195915225
4 0.9587601107 372 5 -1.9701975542
5 0.7415729680 373 5 3.7826137026
6 0.4814927997 374 5 4.1228808255
7 0.1340754859 375 4 3.7818792837
8 0.6316874851 376 3 0.7670884461
9 0.1107241728 377 5 -1.3367400097
10 0.5850426450 378 5 2.2848396166
11 0.2809308425 379 5 2.5234691438
12 0.2835292015 380 2 -0.3291319925
13 0.3328713216 381 1 4.7425349397
14 0.4766904986 382 1 -0.4062103292
15 0.5005860797 383 5 4.8612083721
16 0.2734292494 384 5 -0.2320077328
17 0.1488479455 385 4 2.6195679584
18 0.9446908936 386 3 0.4889716264
19 0.8222738281 387 2 0.7362413658
20 0.7570014759 388 3 4.6661250258
21 0.9988698252 389 4 2.6340743946
22 0.8384663551 390 3 1.0428046484
23 0.1938821415 391 2 0.8855748393
And I need to create new data.frame this way.For example:If predict (in first row) > 0.5,delete second and third row (because ChoosedN in first row is 2 so next two after first row has to be delete, because there we were still in position). And continue on fourth the same way (if predict (fourth row) > 0.5, delete next five rows and so. And of course, if predict <=0.5 delete this row too.
Any straightforward way how to do it with some loop?
Thanks
I would create a new dataframe, then bind the rows you want using rbind inside of a for loop
newDF <- data.frame() # New, Empty Dataframe
i = 1 # Loop index Variable
while (i < nrow(df)) {
if (df$predict[i] > 0.5) { # If predict > 0.5,
newDF <- rbind(newDF, df[i,]) # Bind the row
i = i + df$ChoosedN[i] # Adjust for ChoosedN rows
}
i = i + 1 # Move to the next row
}
I have a data frame having 20 columns. I need to filter / remove noise from one column. After filtering using convolve function I get a new vector of values. Many values in the original column become NA due to filtering process. The problem is that I need the whole table (for later analysis) with only those rows where the filtered column has values but I can't bind the filtered column to original table as the number of rows for both are different. Let me illustrate using the 'age' column in 'Orange' data set in R:
> head(Orange)
Tree age circumference
1 1 118 30
2 1 484 58
3 1 664 87
4 1 1004 115
5 1 1231 120
6 1 1372 142
Convolve filter used
smooth <- function (x, D, delta){
z <- exp(-abs(-D:D/delta))
r <- convolve (x, z, type='filter')/convolve(rep(1, length(x)),z,type='filter')
r <- head(tail(r, -D), -D)
r
}
Filtering the 'age' column
age2 <- smooth(Orange$age, 5,10)
data.frame(age2)
The number of rows for age column and age2 column are 35 and 15 respectively. The original dataset has 2 more columns and I like to work with them also. Now, I only need 15 rows of each column corresponding to the 15 rows of age2 column. The filter here removed first and last ten values from age column. How can I apply the filter in a way that I get truncated dataset with all columns and filtered rows?
You would need to figure out how the variables line up. If you can add NA's to age2 and then do Orange$age2 <- age2 followed by na.omit(Orange) you should have what you want. Or, equivalently, perhaps this is what you are looking for?
df <- tail(head(Orange, -10), -10) # chop off the first and last 10 observations
df$age2 <- age2
df
Tree age circumference age2
11 2 1004 156 915.1678
12 2 1231 172 876.1048
13 2 1372 203 841.3156
14 2 1582 203 911.0914
15 3 118 30 948.2045
16 3 484 51 1008.0198
17 3 664 75 955.0961
18 3 1004 108 915.1678
19 3 1231 115 876.1048
20 3 1372 139 841.3156
21 3 1582 140 911.0914
22 4 118 32 948.2045
23 4 484 62 1008.0198
24 4 664 112 955.0961
25 4 1004 167 915.1678
Edit: If you know the first and last x observations will be removed then the following works:
x <- 2
df <- tail(head(Orange, -x), -x) # chop off the first and last x observations
df$age2 <- age2