I have a matrix of 10 classes (2089 rows and 112 colunms).
0 1 2 3 4 5 6 7 8 9
482 60 404 134 60 339 376 66 63 105
I want to split the matrix randomly in three sets of proportion: 60, 20 and 20 % respectively with keeping the proportion classes in each set as in the original matrix.
I cheked Stratified random sampling from data frame but it's not the same question.
The first column of the matrix contains the classes indexes from 0 to 9. I want to do the split in 60, 20 and 20 % according to this column. For example the 9th class contains 63 observations (3%). The three parts must contains 3% of this class.
Related
I am attempting to work with a large dataset in R where I need to create a column that compares the value in an existing column to all values that follow it (ex: row 1 needs to compare rows 1-10,000, row 2 needs to compare rows 2-10,000, row 3 needs to compare rows 3-10,000, etc.), but cannot figure out how to write the range.
I currently have a column of raw numeric values and a column of row values generated by:
samples$row = seq.int(nrow(samples))
I have attempted to generate the column with the following command:
samples$processed = min(samples$raw[samples$row:10000])
but get the error "numerical expression has 10000 elements: only the first used" and the generated column only has the value for row 1 repeated for each of the 10,000 rows.
How do I need to write this command so that the lower bound of the range is the row currently being calculated instead of 1?
Any help would be appreciated, as I have minimal programming experience.
If all you need is the min of the specific row and all following rows, then
rev(cummin(rev(samples$val)))
# [1] 24 24 24 24 24 24 24 24 24 24 24 24 165 165 165 165 410 410 410 882
If you have some other function that doesn't have a cumulative variant (and your use of min is just a placeholder), then one of:
mapply(function(a, b) min(samples$val[a:b]), seq.int(nrow(samples)), nrow(samples))
# [1] 24 24 24 24 24 24 24 24 24 24 24 24 165 165 165 165 410 410 410 882
sapply(seq.int(nrow(samples)), function(a) min(samples$val[a:nrow(samples)]))
The only reason to use mapply over sapply is if, for some reason, you want window-like operations instead of always going to the bottom of the frame. (Though if you wanted windows, I'd suggest either the zoo or slider packages.)
Data
set.seed(42)
samples <- data.frame(val = sample(1000, size=20))
samples
# val
# 1 561
# 2 997
# 3 321
# 4 153
# 5 74
# 6 228
# 7 146
# 8 634
# 9 49
# 10 128
# 11 303
# 12 24
# 13 839
# 14 356
# 15 601
# 16 165
# 17 622
# 18 532
# 19 410
# 20 882
I'm struggeling to get a good performing script for this problem: I have a table with a score, x, y. I want to sort the table by score and than build groups based on the x value. Each group should have an equal sum (not counts) of x. x is a metric number in the dataset and resembles the historic turnover of a customer.
score x y
0.436024136 3 435
0.282303336 46 56
0.532358015 24 34
0.644236597 0 2
0.99623626 0 4
0.557673456 56 46
0.08898779 0 7
0.702941303 453 2
0.415717835 23 1
0.017497461 234 3
0.426239166 23 59
0.638896238 234 86
0.629610596 26 68
0.073107526 0 35
0.85741877 0 977
0.468612039 0 324
0.740704267 23 56
0.720147257 0 68
0.965212467 23 0
a good way to do so is adding a group variable to the data.frame with cumsum! Now you can easily sum the groups with e. g. subset.
data.frame$group <-cumsum(as.numeric(data.frame$x)) %/% (ceiling(sum(data.frame$x) / 3)) + 1
remarks:
in big data.frames cumsum(as.numeric()) works reliably
%/% is a division where you get an integer back
the '+1' just let your groups start with 1 instead of 0
thank you #Ronak Shah!
Assume I have this matrix
set.seed(123)
x <- matrix(rnorm(410),205,2)
x[8,] <- c(0.13152348, -0.05235148) #similar to x[5,]
x[16,] <- c(1.21846582, 1.695452178) #similar to x[11,]
The values are very similar to the rows specified above, and in the context of the whole data, they are semi-duplicates. What could I do to find and remove them? My original data is an array that contains many such matrices, but the position of the semi duplicates is the same across all matrices.
I know of agrep but the function operates on vectors as far as I understand.
You will need to set a threshold, but you can just compute the distance between each row using dist and find the points that are sufficiently close together. Of course, Each point is near itself, so you need to ignore the diagonal of the distance matrix.
DM = as.matrix(dist(x))
diag(DM) = 1 ## ignore diagonal
which(DM < 0.025, arr.ind=TRUE)
row col
8 8 5
5 5 8
16 16 11
11 11 16
48 48 20
20 20 48
168 168 71
91 91 73
73 73 91
71 71 168
This finds the "close" points that you created and a few others that got generated at random.
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