Consider a 3 x 3 x 3 cube where each of the 27 elements is connected to other elements along faces. A cube-shaped element has 6 sides, thus a maximum of 6 connections is possible per element (for example, the center-most element in a 3 x 3 x 3 cube is bounded by 6 elements, and has 6 connections).
Then, let m1, m2, and m3 be the first, second, and third layers of the cube respectively. The name of each element is xyz, where x, y, z are the row number, column number, and layer number of the element. For example, the element 213 is in the second row, first column, and 3rd layer of the cube. This element is connected to 4 other elements: three are in its layer (113, 313, 223), and one is one layer above it (212).
x = 3 # nrow
y = 3 # ncol
z = 3 # nlay
# print each layer as a 2D matrix
for(k in 1:z){
m = paste0(rep(1:x, each=x), rep(1:y, times = y), k)
print(matrix(m, nrow=x, byrow=T))
}
[,1] [,2] [,3]
[1,] "111" "121" "131"
[2,] "211" "221" "231"
[3,] "311" "321" "331"
[,1] [,2] [,3]
[1,] "112" "122" "132"
[2,] "212" "222" "232"
[3,] "312" "322" "332"
[,1] [,2] [,3]
[1,] "113" "123" "133"
[2,] "213" "223" "233"
[3,] "313" "323" "333"
Is there an out-of-the-box function in igraph or a related package for creating either an adjacency matrix OR an edge list for a network like this? I need a solution that scales to any number of rows, columns, and layers. Python solutions are welcome.
I manually created the 2D adjacency matrix, where the rows and columns are given by c(m1, m2, m3) below:
m1 = paste0(rep(1:x, each=x), rep(1:y, times = y), 1)
m2 = paste0(rep(1:x, each=x), rep(1:y, times = y), 2)
m3 = paste0(rep(1:x, each=x), rep(1:y, times = y), 3)
c(m1, m2, m3)
[1] "111" "121" "131" "211" "221" "231" "311" "321" "331" "112" "122" "132" "212" "222" "232" "312" "322" "332"
[19] "113" "123" "133" "213" "223" "233" "313" "323" "333"
For this simple example the adjacency matrix is sparse, has 0's along the diagonal, and is symmetric. It looks like this:
And here's a dput() to C&P and validate with.
dput(temp)
structure(c(0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0), .Dim = c(27L,
27L), .Dimnames = list(c("111", "121", "131", "211", "221", "231",
"311", "321", "331", "112", "122", "132", "212", "222", "232",
"312", "322", "332", "113", "123", "133", "213", "223", "233",
"313", "323", "333"), c("111", "121", "131", "211", "221", "231",
"311", "321", "331", "112", "122", "132", "212", "222", "232",
"312", "322", "332", "113", "123", "133", "213", "223", "233",
"313", "323", "333")))
There's an edge when the Manhattan distance between the nodes is 1, so you can use dist() in R to create the adjacency matrix:
cube_mat = expand.grid(
x = 1:3,
y = 1:3,
z = 1:3
)
m_dist = as.matrix(dist(cube_mat[, 1:3], method = "manhattan", diag = TRUE))
# Zero out any distances != 1
m_dist[m_dist != 1] = 0
rownames(m_dist) = paste0(cube_mat$x, cube_mat$y, cube_mat$z)
colnames(m_dist) = paste0(cube_mat$x, cube_mat$y, cube_mat$z)
# Plot of the adjacency matrix (looks reversed because 111 is in the bottom left):
image(m_dist)
If you want to just use a package funtion from igraph:
#adj <- my.adjacency.matrix
as_edgelist(graph.adjacency(adj))
In general you can use the functions in the igraph package to go between edgelists, adjacency matrices, and also produce graphs using plot.igraph. Here's the default cube:
plot.igraph(graph.adjacency(adj))
Related
I am running KMeans on a singular column of a dataset containing the number of employees in companies. The graph of the data looks like this:
graph of dataset.
After looking at this and running KMeans on this particular dataset, it gives an output like this which is clearly not even close to correct:
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2,
2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
The code I've used for this is:
def perform_clustering_oneCol(data):
df2 = data.loc[:,"#Employees"]
print("hi: \n")
print(df2.head(15))
df2.dropna()
df2 = pd.get_dummies(df2)
arr = df2.to_numpy()
km = KMeans(n_clusters=3).fit(arr)
print(km.labels_.tolist())
I'm very new to this and would like to know how to get a more optimal result OR what other algorithms I can look at. Any help would be appreciated. Thanks.
I have 250 projects and 50 supervisors. Each of our 130 students rank their preferred choice of projects from 1-5 (1 being favourite) and don't score the other 245. I would like to assign each student to a supervisor but a supervisor can only have up to 12 students.
I'm trying to make some dummy data but struggling to make the cost matrix.
Defining a matrix with random integers of size (5 x 5).
import random
def createMatrix(n):
firstRow = random.sample(range(n),n)
permutes = random.sample(range(n),n)
return list(firstRow[i:]+firstRow[:i] for i in permutes)
N = 5
m = createMatrix(N)
for i in m:
print(i)
[1, 0, 3, 4, 2]
[3, 4, 2, 1, 0]
[4, 2, 1, 0, 3]
[2, 1, 0, 3, 4]
[0, 3, 4, 2, 1]
But this is a Latin square matrix and I would like to allow some rows to contain duplicate values representing that two students have ranked a given same project the same value. Any thoughts would be much appreciated.
You have 130 students and want some duplicated choices. So generate f.e. 110 choices (already minor chance of a duplicate - but with 5 out of 250 not much).
Then choose some of the generated choices as dupe-candidates and and add some of them again until you got back up to 130 choices:
from random import sample,shuffle
projects = 250
students = 130
dup_choices = 20 # means we generate 110 choices and at least 20 will be dupes
per_dup = 5 # add up to this amount of dupes for each dupe candidate
un_duped = students - dup_choices
# random samples
student_choices = [ sample(range(projects), k = 5) for _ in range(un_duped)]
# select dupe candidates
dups = sample(student_choices, k = max(0, dup_choices // per_dup) + 1)
# add enough of each duplicate to statisfy your numbers
for d in dups:
student_choices.extend( (d.copy() for _ in range(per_dup) ) )
# integer rounding + 1 => you will overshoot - so trim back to number of students
student_choices = student_choices[:students]
# mix the dupes into the data
# shuffle(student_choices)
This will generate 130 choices of wich at least 20 are some kind of duplicate of some other.
You create your cost matrix from this data - the position of the projects number inside the students choice is its priority (0 == 1st, 1 = 2nd, ...):
costs = [[0 if pr not in choice else choice.index(pr) + 1
for pr in range(projects)]
for choice in student_choices]
# print first 3 of choices / costs
for sc in student_choices[:3]:
print(sc)
for c in costs[:3]:
print (c)
Output:
[124, 174, 43, 181, 63]
[158, 110, 129, 120, 149]
[226, 238, 183, 249, 90]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4]
Working in an example I realized that there are at least two ways of computing morgan fingerprints for a molecule using rdkit. But using the exact same properties in both ways I get different vectors. Am I missing something?
First approach:
info = {}
mol = Chem.MolFromSmiles('C/C1=C\\C[C#H]([C+](C)C)CC/C(C)=C/CC1')
fp = AllChem.GetMorganFingerprintAsBitVect(mol, useChirality=True, radius=2, nBits = 124, bitInfo=info)
vector = np.array(fp)
vector
array([0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1])
Second approach:
morgan_fp_gen = rdFingerprintGenerator.GetMorganGenerator(includeChirality=True, radius=2, fpSize=124)
mol = Chem.MolFromSmiles('C/C1=C\\C[C#H]([C+](C)C)CC/C(C)=C/CC1')
fp = morgan_fp_gen.GetFingerprint(mol)
vector = np.array(fp)
vector
array([0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0,
1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0])
Which are clearly different, even using chirality in both cases.
Besides, there is a way to get the bitInfo from a bit vector using the second approach?
By default the Morgan Generator uses "count simulation": adding extra bits to a bit vector fingerprint in order to get bit-vector similarities. If you turn this off by passing useCountSimulation=False the fingerprints should be equivalent:
mol = Chem.MolFromSmiles('C/C1=C\\C[C#H]([C+](C)C)CC/C(C)=C/CC1')
fp1 = AllChem.GetMorganFingerprintAsBitVect(mol, useChirality=True, radius=2, nBits=124)
vec1 = np.array(fp1)
morgan_fp_gen = rdFingerprintGenerator.GetMorganGenerator(includeChirality=True, radius=2, fpSize=124, useCountSimulation=False)
fp2 = morgan_fp_gen.GetFingerprint(mol)
vec2 = np.array(fp2)
assert np.all(vec1 == vec2) == True
as for the bitInfo I am not sure this can be done with the second method, although someone may correct me
I have coded a sequence to sequence learning LSTM in keras myself using the knowledge gained from the web tutorials and my own intuitions. I converted my sample text to sequences and then padded using pad_sequence function in keras.
from keras.preprocessing.text import Tokenizer,base_filter
from keras.preprocessing.sequence import pad_sequences
def shift(seq, n):
n = n % len(seq)
return seq[n:] + seq[:n]
txt="abcdefghijklmn"*100
tk = Tokenizer(nb_words=2000, filters=base_filter(), lower=True, split=" ")
tk.fit_on_texts(txt)
x = tk.texts_to_sequences(txt)
#shifing to left
y = shift(x,1)
#padding sequence
max_len = 100
max_features=len(tk.word_counts)
X = pad_sequences(x, maxlen=max_len)
Y = pad_sequences(y, maxlen=max_len)
After a carefully inspection I found my padded sequence looks like this
>>> X[0:6]
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7]], dtype=int32)
>>> X
array([[ 0, 0, 0, ..., 0, 0, 1],
[ 0, 0, 0, ..., 0, 0, 3],
[ 0, 0, 0, ..., 0, 0, 2],
...,
[ 0, 0, 0, ..., 0, 0, 13],
[ 0, 0, 0, ..., 0, 0, 12],
[ 0, 0, 0, ..., 0, 0, 14]], dtype=int32)
Is the padded sequence suppose to look like this? Except the last column in the array the rest are all zeros. I think I made some mistake in padding the text to sequence and if so can you tell me where I made the error?
If you want to tokenize by char, you can do it manually, it's not too complex:
First build a vocabulary for your characters:
txt="abcdefghijklmn"*100
vocab_char = {k: (v+1) for k, v in zip(set(txt), range(len(set(txt))))}
vocab_char['<PAD>'] = 0
This will associate a distinct number for every character in your txt. The character with index 0 should be preserved for the padding.
Having the reverse vocabulary will be usefull to decode the output.
rvocab = {v: k for k, v in vocab.items()}
Once you have this, you can first split your text into sequences, say you want to have sequences of length seq_len = 13 :
[[vocab_char[char] for char in txt[i:(i+seq_len)]] for i in range(0,len(txt),seq_len)]
your output will look like :
[[9, 12, 6, 10, 8, 7, 2, 1, 5, 13, 11, 4, 3],
[14, 9, 12, 6, 10, 8, 7, 2, 1, 5, 13, 11, 4],
...,
[2, 1, 5, 13, 11, 4, 3, 14, 9, 12, 6, 10, 8],
[7, 2, 1, 5, 13, 11, 4, 3, 14]]
Note that the last sequence doesn't have the same length, you can discard it or pad your sequence to max_len = 13, it will add 0's to it.
You can build your targets Y the same way, by shifting everything by 1. :-)
I hope this helps.
The problem is in this line:
tk = Tokenizer(nb_words=2000, filters=base_filter(), lower=True, split=" ")
When you set such split (by " "), due to nature of your data, you'll get each sequence consisting of a single word. That's why your padded sequences have only one non-zero element. To change that try:
txt="a b c d e f g h i j k l m n "*100
The argument padding controls padding either before or after each sequence. Use like this:
X = pad_sequences(x, maxlen=max_len, padding='post')
Y = pad_sequences(y, maxlen=max_len, padding='post')
I am trying to add lists l as values for different dict d keys. For the array a, having [6,12,18,24,30] I am trying to get the dict d to contain the following key-value pairs:
d[6] = [0, 0, 0.....0]
d[12] = [6, 0, 0, ..0]
d[18] = [6, 12, 0, ...0]
d[24] = [6, 12, 18, 0, ..0]
Where there are 59 elements in each of the lists above.
I use the code below to do this, but my output for the key 24 is:
{24: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 18, 12, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24, 18, 12, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]}
I am trying to understand where I am going wrong.
d = {}
l = []
a =numpy.array([6,12,18,24,30])
for x, value in numpy.ndenumerate(a):
months_to_maturity = value
for i in range(6, 354, 6):
if i <= months_to_maturity:
l.append(months_to_maturity - i)
else:
l.append(0)
d[months_to_maturity] = l
You're always appending to the same list. Thus all the dictionary values end up pointing to the same list. You want to append to a different list each time:
d = {}
a = numpy.array([6, 12, 18, 24, 30])
for months_to_maturity in a:
l = []
for i in range(6, 354, 6):
if i <= months_to_maturity:
l.append(months_to_maturity - i)
else:
l.append(0)
d[months_to_maturity] = l