I am working with triangular meshes in R. For those not familiar, the PLY format has two main components, a 3 by n matrix of vertex x,y,z coordinates, where n is the number of vertices, and a 3 by m matrix of faces where each number references one line from the vertex matrix, and so defining three corners of a triangular face. I am trying to find the mesh boundary edges, which are the "sides" of the triangles that are only referenced once in the faces matrix.
Therefore my question is, how do I find unique pairs of numbers across rows where there are three columns?
face 1 4 6 7
face 2 7 6 8
face 3 9 11 12
face 4 10 9 12
Here line (face) 1 has the edge 4-7 that only appears once, while 6-7 appears twice, as does 9-12.
unique() works across rows, but looks for unique rows, and expects the numbers to be in the same order. Any suggestions?
What you want to do is hash each pair, then make a table of the hashes. You also want (x,y)
to hash the same as (y,x).
R>data
V1 V2 V3 V4 V5
1 face 1 4 6 7
2 face 2 7 6 8
3 face 3 9 11 12
4 face 4 10 9 12
R>e1 <- pmin(data[3], data[4]) + pmax(data[3], data[4])/100
R>e2 <- pmin(data[3], data[5]) + pmax(data[3], data[5])/100
R>e3 <- pmin(data[4], data[5]) + pmax(data[4], data[5])/100
R>table(c(e1,e2,e3, recursive=TRUE))
4.06 4.07 6.07 6.08 7.08 9.1 9.11 9.12 10.12 11.12
1 1 2 1 1 1 1 2 1 1
Related
I'm currently reading "Practical Statistics for Data Scientists" and following along in R as they demonstrate some code. There is one chunk of code I'm particularly struggling to follow the logic of and was hoping someone could help. The code in question is creating a dataframe with 1000 rows where each observation is the mean of 5 randomly drawn income values from the dataframe loans_income. However, I'm getting confused about the logic of the code as it is fairly complicated with a tapply() function and nested rep() statements.
The code to create the dataframe in question is as follows:
samp_mean_5 <- data.frame(income = tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean),
type='mean_of_5')
In particular, I'm confused about the nested rep() statements and the 1000*5 portion of the sample() function. Any help understanding the logic of the code would be greatly appreciated!
For reference, the original dataset loans_income simply has a single column of 50,000 income values.
You have 50,000 loans_income in a single vector. Let's break your code down:
tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean)
I will replace 1000 with 10 and income with random numbers, so it's easier to explain. I also set set.seed(1) so the result can be reproduced.
sample(loans_income$income,1000*5)
We 50 random incomes from your vector without replacement. They are (temporarily) put into a vector of length 50, so the output looks like this:
> sample(runif(50000),10*5)
[1] 0.73283101 0.60329970 0.29871173 0.12637654 0.48434952 0.01058067 0.32337850
[8] 0.46873561 0.72334215 0.88515494 0.44036341 0.81386225 0.38118213 0.80978822
[15] 0.38291273 0.79795343 0.23622492 0.21318431 0.59325586 0.78340477 0.25623138
[22] 0.64621658 0.80041393 0.68511759 0.21880083 0.77455662 0.05307712 0.60320912
[29] 0.13191926 0.20816298 0.71600799 0.70328349 0.44408218 0.32696205 0.67845445
[36] 0.64438336 0.13241312 0.86589561 0.01109727 0.52627095 0.39207860 0.54643661
[43] 0.57137320 0.52743012 0.96631114 0.47151170 0.84099503 0.16511902 0.07546454
[50] 0.85970500
rep(1:1000,rep(5,1000))
Now we are creating an indexing vector of length 50:
> rep(1:10,rep(5,10))
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6
[29] 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 10 10
Those indices "group" the samples from step 1. So basically this vector tells R that the first 5 entries of your "sample vector" belong together (index 1), the next 5 entries belong together (index 2) and so on.
FUN = mean
Just apply the mean-function on the data.
tapply
So tapply takes the sampled data (sample-part) and groups them by the second argument (the rep()-part) and applies the mean-function on each group.
If you are familiar with data.frames and the dplyr package, take a look at this (only the first 10 rows are displayed):
set.seed(1)
df <- data.frame(income=sample(runif(5000),10*5), index=rep(1:10,rep(5,10)))
income index
1 0.42585569 1
2 0.16931091 1
3 0.48127444 1
4 0.68357403 1
5 0.99374923 1
6 0.53227877 2
7 0.07109499 2
8 0.20754511 2
9 0.35839481 2
10 0.95615917 2
I attached the an index to the random numbers (your income). Now we calculate the mean per group:
df %>%
group_by(index) %>%
summarise(mean=mean(income))
which gives us
# A tibble: 10 x 2
index mean
<int> <dbl>
1 1 0.551
2 2 0.425
3 3 0.827
4 4 0.391
5 5 0.590
6 6 0.373
7 7 0.514
8 8 0.451
9 9 0.566
10 10 0.435
Compare it to
set.seed(1)
tapply(sample(runif(5000),10*5),
rep(1:10,rep(5,10)),
mean)
which yields basically the same result:
1 2 3 4 5 6 7 8 9
0.5507529 0.4250946 0.8273149 0.3905850 0.5902823 0.3730092 0.5143829 0.4512932 0.5658460
10
0.4352546
I have an adjancecy matrix stored in CSR format. Eg
xadj = 0 2 5 8 11 13 16 20 24 28 31 33 36 39 42 44
adjncy = 1 5 0 2 6 1 3 7 2 4 8 3 9 0 6 10 1 5 7 11 2 6 8 12 3 7 9 13 4 8 14 5 11 6 10 12 7 11 13 8 12 14 9 13
I am now paritioning said graph using METIS. This gives me the partition vector part of the graph. Basically a list that tells me in which partition each vertex is. Is there an efficient way to build the new adjacency matrix for this partitioning such that I can partition the new graph again? Eg a function rebuildAdjacency(xadj, adjncy, part). If possible reusing xadj and adjncy.
I'm assuming that what you mean by "rebuild" is removing the edges between vertices that have been assigned different partitions? If so, the (probably) best you can do is iterate your CSR list, generate a new CSR list, and skip all edges that are between partitions.
In pseudocode (actually, more or less Python):
new_xadj = []
new_adjcy = []
for row in range(0, n):
row_index = xadj[row]
next_row_index = xadj[row+1]
# New row index for the row we are currently building
new_xadj.append(len(new_adjcy))
for col in adjncy[row_index:next_row_index]:
if partition[row] != partition[col]:
pass # Not in the same partition
else:
# Put the row->col edge into the new CSR list
new_adjcy.append(col)
# Last entry in the row index field is the number of entries
new_xadj.append(len(new_adjcy))
I don't think that you can do this very efficiently re-using the old xadj and adjcy fields. However, if you are doing this recursively, you can save memory allocation / deallocation by having exacyly two copies of xadj and adjc, and alternating between them.
I have a dataset of Ages for the customer and I wanted to make a frequency distribution by 9 years of a gap of age.
Ages=c(83,51,66,61,82,65,54,56,92,60,65,87,68,64,51,
70,75,66,74,68,44,55,78,69,98,67,82,77,79,62,38,88,76,99,
84,47,60,42,66,74,91,71,83,80,68,65,51,56,73,55)
My desired outcome would be similar to below-shared table, variable names can be differed(as you wish)
Could I use binCounts code into it ? if yes could you help me out using the code as not sure of bx and idxs in this code?
binCounts(x, idxs = NULL, bx, right = FALSE) ??
Age Count
38-46 3
47-55 7
56-64 7
65-73 14
74-82 10
83-91 6
92-100 3
Much Appreciated!
I don't know about the binCounts or even the package it is in but i have a bare r function:
data.frame(table(cut(Ages,0:7*9+37)))
Var1 Freq
1 (37,46] 3
2 (46,55] 7
3 (55,64] 7
4 (64,73] 14
5 (73,82] 10
6 (82,91] 6
7 (91,100] 3
To exactly duplicate your results:
lowerlimit=c(37,46,55,64,73,82,91,101)
Labels=paste(head(lowerlimit,-1)+1,lowerlimit[-1],sep="-")#I add one to have 38 47 etc
group=cut(Ages,lowerlimit,Labels)#Determine which group the ages belong to
tab=table(group)#Form a frequency table
as.data.frame(tab)# transform the table into a dataframe
group Freq
1 38-46 3
2 47-55 7
3 56-64 7
4 65-73 14
5 74-82 10
6 83-91 6
7 92-100 3
All this can be combined as:
data.frame(table(cut(Ages,s<-0:7*9+37,paste(head(s+1,-1),s[-1],sep="-"))))
I will present my question in two ways. First, requesting a solution for a task; and second, as a description of my overall objective (in case I am overthinking this and there is an easier solution).
1) Task Solution
Data context: each row contains four price variables (columns) representing (a) the price at which the respondent feels the product is too cheap; (b) the price that is perceived as a bargain; (c) the price that is perceived as expensive; (d) the price that is too expensive to purchase.
## mock data set
a<-c(1,5,3,4,5)
b<-c(6,6,5,6,8)
c<-c(7,8,8,10,9)
d<-c(8,10,9,11,12)
df<-as.data.frame(cbind(a,b,c,d))
## result
# a b c d
#1 1 6 7 8
#2 5 6 8 10
#3 3 5 8 9
#4 4 6 10 11
#5 5 8 9 12
Task Objective: The goal is to create a single column in a new data frame that lists all of the unique values contained in a, b, c, and d.
price
#1 1
#2 3
#3 4
#4 5
#5 6
...
#12 12
My initial thought was to use rbind() and unique()...
price<-rbind(df$a,df$b,df$c,df$d)
price<-unique(price)
...expecting that a, b, c and d would stack vertically.
[Pseudo illustration]
a[1]
a[2]
a[...]
a[n]
b[1]
b[2]
b[...]
b[n]
etc.
Instead, the "columns" are treated as rows and stacked horizontally.
V1 V2 V3 V4 V5
1 1 5 3 4 5
2 6 6 5 6 8
3 7 8 8 10 9
4 8 10 9 11 12
How may I stack a, b, c and d such that price consists of only one column ("V1") that contains all twenty responses? (The unique part I can handle separately afterwards).
2) Overall Objective: The Bigger Picture
Ultimately, I want to create a cumulative share of population for each price (too cheap, bargain, expensive, too expensive) at each price point (defined by the unique values described above). For example, what percentage of respondents felt $1 was too cheap, what percentage felt $3 or less was too cheap, etc.
The cumulative shares for bargain and expensive are later inverted to become not.bargain and not.expensive and the four vectors reside in a data frame like this:
buckets too.cheap not.bargain not.expensive too.expensive
1 0.01 to 0.50 0.000000000 1 1 0
2 0.51 to 1.00 0.000000000 1 1 0
3 1.01 to 1.50 0.000000000 1 1 0
4 1.51 to 2.00 0.000000000 1 1 0
5 2.01 to 2.50 0.001041667 1 1 0
6 2.51 to 3.00 0.001041667 1 1 0
...
from which I may plot something that looks like this:
Above, I accomplished my plotting objective using defined price buckets ($0.50 ranges) and the hist() function.
However, the intersections of these lines have meanings and I want to calculate the exact price at which any of the lines cross. This is difficult when the x-axis is defined by price range buckets instead of a specific value; hence the desire to switch to exact values and the need to generate the unique price variable.
[Postscript: This analysis is based on Peter Van Westendorp's Price Sensitivity Meter (https://en.wikipedia.org/wiki/Van_Westendorp%27s_Price_Sensitivity_Meter) which has known practical limitations but is relevant in the context of my research which will explore consumer perceptions of value under different treatments rather than defining an actual real-world price. I mention this for two reasons 1) to provide greater insight into my objective in case another approach comes to mind, and 2) to keep the thread focused on the mechanics rather than whether or not the Price Sensitivity Meter should be used.]
We can unlist the data.frame to a vector and get the sorted unique elements
sort(unique(unlist(df)))
When we do an rbind, it creates a matrix and unique of matrix calls the unique.matrix
methods('unique')
#[1] unique.array unique.bibentry* unique.data.frame unique.data.table* unique.default unique.IDate* unique.ITime*
#[8] unique.matrix unique.numeric_version unique.POSIXlt unique.warnings
which loops through the rows as the default MARGIN is 1 and then looks for unique elements. Instead, if we use the 'price', either as.vector or c(price) converts into vector
sort(unique(c(price)))
#[1] 1 3 4 5 6 7 8 9 10 11 12
If we use unique.default
sort(unique.default(price))
#[1] 1 3 4 5 6 7 8 9 10 11 12
I have a big dataset of about 35000 cases X 32 variables
one of those variables is Description in which a description of status is given. for example: patient suffered ischemic stroke.
Now I would like to make a dataframe in which I place all cases in which the word "stroke", "STROKE" or "Stroke" is found in the variable Description.
Could anyone suggest a efficient way to do this. Because now I just added all by hand in a very inefficient way:
df1<-rbind(df[1,],df[2,],df[3,]
It works but it's unbelievably inelegant and prone to mistakes.
Here I create some example data to work with.
a <- c(1:10)
b <- c(11:20)
description <- c("Stroke","ALS","Parkinsons","STROKE","STROKE","stroke","Alzheimers","Stroke","ALS","Parkinsons")
df<-data.frame(a,b,description)
df
a b description
1 1 11 Stroke
2 2 12 ALS
3 3 13 Parkinsons
4 4 14 STROKE
5 5 15 STROKE
6 6 16 stroke
7 7 17 Alzheimers
8 8 18 Stroke
9 9 19 ALS
10 10 20 Parkinsons
With this code you can remove every case (row) that is not associated with "Stroke", "STROKE" or "stroke":
df1<-df[!(df$description!="STROKE" & df$description!="Stroke" & df$description!="stroke"),]
df1
a b description
1 1 11 Stroke
4 4 14 STROKE
5 5 15 STROKE
6 6 16 stroke
8 8 18 Stroke
Hope this was what you were looking for.