This is a typical use case for FEM/FVM equation systems, so is perhaps of broader interest. From a triangular mesh à la
I would like to create a scipy.sparse.csr_matrix. The matrix rows/columns represent values at the nodes of the mesh. The matrix has entries on the main diagonal and wherever two nodes are connected by an edge.
Here's an MWE that first builds a node->edge->cells relationship and then builds the matrix:
import numpy
import meshzoo
from scipy import sparse
nx = 1600
ny = 1000
verts, cells = meshzoo.rectangle(0.0, 1.61, 0.0, 1.0, nx, ny)
n = len(verts)
nds = cells.T
nodes_edge_cells = numpy.stack([nds[[1, 2]], nds[[2, 0]],nds[[0, 1]]], axis=1)
# assign values to each edge (per cell)
alpha = numpy.random.rand(3, len(cells))
vals = numpy.array([
[alpha**2, -alpha],
[-alpha, alpha**2],
])
# Build I, J, V entries for COO matrix
I = []
J = []
V = []
#
V.append(vals[0][0])
V.append(vals[0][1])
V.append(vals[1][0])
V.append(vals[1][1])
#
I.append(nodes_edge_cells[0])
I.append(nodes_edge_cells[0])
I.append(nodes_edge_cells[1])
I.append(nodes_edge_cells[1])
#
J.append(nodes_edge_cells[0])
J.append(nodes_edge_cells[1])
J.append(nodes_edge_cells[0])
J.append(nodes_edge_cells[1])
# Create suitable data for coo_matrix
I = numpy.concatenate(I).flat
J = numpy.concatenate(J).flat
V = numpy.concatenate(V).flat
matrix = sparse.coo_matrix((V, (I, J)), shape=(n, n))
matrix = matrix.tocsr()
With
python -m cProfile -o profile.prof main.py
snakeviz profile.prof
one can create and view a profile of the above:
The method tocsr() takes the lion share of the runtime here, but this is also true when building alpha is more complex. Consequently, I'm looking for ways to speed this up.
What I've already found:
Due to the structure of the data, the values on the diagonal of the matrix can be summed up in advance, i.e.,
V.append(vals[0, 0, 0] + vals[1, 1, 2])
I.append(nodes_edge_cells[0, 0]) # == nodes_edge_cells[1, 2]
J.append(nodes_edge_cells[0, 0]) # == nodes_edge_cells[1, 2]
This makes I, J, V shorter and thus speeds up tocsr.
Right now, edges are "per cell". I could identify equal edges with each other using numpy.unique, effectively saving about half of I, J, V. However, I found that this too takes some time. (Not surprising.)
One other thought that I had was that that I could replace the diagonal V, I, J by a simple numpy.add.at if there was a csr_matrix-like data structure where the main diagonal is kept separately. I know that this exists in some other software packages, but couldn't find it in scipy. Correct?
Perhaps there's a sensible way to construct CSR directly?
I would try creating the csr structure directly, especially if you are resorting to np.unique since this gives you sorted keys, which is half the job done.
I'm assuming you are at the point where you have i, j sorted lexicographically and overlapping v summed using np.add.at on the optional inverse output of np.unique.
Then v and j are already in csr format. All that's left to do is creating the indptr which you simply get by np.searchsorted(i, np.arange(M+1)) where M is the column length. You can pass these directly to the sparse.csr_matrix constructor.
Ok, let code speak:
import numpy as np
from scipy import sparse
from timeit import timeit
def tocsr(I, J, E, N):
n = len(I)
K = np.empty((n,), dtype=np.int64)
K.view(np.int32).reshape(n, 2).T[...] = J, I
S = np.argsort(K)
KS = K[S]
steps = np.flatnonzero(np.r_[1, np.diff(KS)])
ED = np.add.reduceat(E[S], steps)
JD, ID = KS[steps].view(np.int32).reshape(-1, 2).T
ID = np.searchsorted(ID, np.arange(N+1))
return sparse.csr_matrix((ED, np.array(JD, dtype=int), ID), (N, N))
def viacoo(I, J, E, N):
return sparse.coo_matrix((E, (I, J)), (N, N)).tocsr()
#testing and timing
# correctness
N = 1000
A = np.random.random((N, N)) < 0.001
I, J = np.where(A)
E = np.random.random((2, len(I)))
D = np.zeros((2,) + A.shape)
D[:, I, J] = E
D2 = tocsr(np.r_[I, I], np.r_[J, J], E.ravel(), N).A
print('correct:', np.allclose(D.sum(axis=0), D2))
# speed
N = 100000
K = 10
I, J = np.random.randint(0, N, (2, K*N))
E = np.random.random((2 * len(I),))
I, J, E = np.r_[I, I, J, J], np.r_[J, J, I, I], np.r_[E, E]
print('N:', N, ' -- nnz (with duplicates):', len(E))
print('direct: ', timeit('f(a,b,c,d)', number=10, globals={'f': tocsr, 'a': I, 'b': J, 'c': E, 'd': N}), 'secs for 10 iterations')
print('via coo:', timeit('f(a,b,c,d)', number=10, globals={'f': viacoo, 'a': I, 'b': J, 'c': E, 'd': N}), 'secs for 10 iterations')
Prints:
correct: True
N: 100000 -- nnz (with duplicates): 4000000
direct: 7.702431229001377 secs for 10 iterations
via coo: 41.813509466010146 secs for 10 iterations
Speedup: 5x
So, in the end this turned out to be the difference between COO's and CSR's sum_duplicates (just like #hpaulj suspected). Thanks to the efforts of everyone involved here (particularly #paul-panzer), a PR is underway to give tocsr a tremendous speedup.
SciPy's tocsr does a lexsort on (I, J), so it helps organizing the indices in such a way that (I, J) will come out fairly sorted already.
For for nx=4, ny=2 in the above example, I and J are
[1 6 3 5 2 7 5 5 7 4 5 6 0 2 2 0 1 2 1 6 3 5 2 7 5 5 7 4 5 6 0 2 2 0 1 2 5 5 7 4 5 6 0 2 2 0 1 2 1 6 3 5 2 7 5 5 7 4 5 6 0 2 2 0 1 2 1 6 3 5 2 7]
[1 6 3 5 2 7 5 5 7 4 5 6 0 2 2 0 1 2 5 5 7 4 5 6 0 2 2 0 1 2 1 6 3 5 2 7 1 6 3 5 2 7 5 5 7 4 5 6 0 2 2 0 1 2 5 5 7 4 5 6 0 2 2 0 1 2 1 6 3 5 2 7]
First sorting each row of cells, then the rows by the first column like
cells = numpy.sort(cells, axis=1)
cells = cells[cells[:, 0].argsort()]
produces
[1 4 2 5 3 6 5 5 5 6 7 7 0 0 1 2 2 2 1 4 2 5 3 6 5 5 5 6 7 7 0 0 1 2 2 2 5 5 5 6 7 7 0 0 1 2 2 2 1 4 2 5 3 6 5 5 5 6 7 7 0 0 1 2 2 2 1 4 2 5 3 6]
[1 4 2 5 3 6 5 5 5 6 7 7 0 0 1 2 2 2 5 5 5 6 7 7 0 0 1 2 2 2 1 4 2 5 3 6 1 4 2 5 3 6 5 5 5 6 7 7 0 0 1 2 2 2 5 5 5 6 7 7 0 0 1 2 2 2 1 4 2 5 3 6]
For the number in the original post, sorting cuts down the runtime from about 40 seconds to 8 seconds.
Perhaps an even better ordering can be achieved if the nodes are numbered more appropriately in the first place. I'm thinking of Cuthill-McKee and friends.
Related
I am trying to edit a 5 * 5 square matrix in Python.And I initialize every element in this 5 * 5 matrix with the value 0. I initialize the matrix by using lists using this code:
h = []
for i in range(5):
h.append([0,0,0,0,0])
And now I want to change the matrix to something like this.
4 5 0 0 0
0 4 5 0 0
0 0 4 5 0
0 0 0 4 5
5 0 0 0 4
Here is the piece of code -
i = 0
a = 0
while i < 5:
h[i][a] = 4
h[i][a+1] = 5
a += 1
i += 1
where h[i][j] is the 2 D matrix. But the output is always is showing something like this -
4 4 4 4 4
4 4 4 4 4
4 4 4 4 4
4 4 4 4 4
4 4 4 4 4
Can you guys tell me what is wrong with it?
Do the update as follows using the modulo operator %:
for i in range(5):
h[i][i % 5] = 4
h[i][(i+1) % 5] = 5
The % 5 in the first line isn't strictly necessary but underlines the general principle for matrices of various dimensions. Or more generally, for random dimensions:
for i, row in enumerate(h):
n = len(row)
row[i % n] = 4
row[(i+1) % n] = 5
Question answered here: 2D list has weird behavor when trying to modify a single value
This should work:
#m = [[0]*5]*5 # Don't do this.
m = []
for i in range(5):
m.append([0]*5)
i = a = 0
while i < 5:
m[i][a] = 4
if a < 4:
m[i][a+1] = 5
a += 1
i += 1
Assume I have the following pandas data frame:
my_class value
0 1 1
1 1 2
2 1 3
3 2 4
4 2 5
5 2 6
6 2 7
7 2 8
8 2 9
9 3 10
10 3 11
11 3 12
I want to identify the indices of "my_class" where the class changes and remove n rows after and before this index. The output of this example (with n=2) should look like:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
My approach:
# where class changes happen
s = df['my_class'].ne(df['my_class'].shift(-1).fillna(df['my_class']))
# mask with `bfill` and `ffill`
df[~(s.where(s).bfill(limit=1).ffill(limit=2).eq(1))]
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
One of possible solutions is to:
Make use of the fact that the index contains consecutive integers.
Find index values where class changes.
For each such index generate a sequence of indices from n-2
to n+1 and concatenate them.
Retrieve rows with indices not in this list.
The code to do it is:
ind = df[df['my_class'].diff().fillna(0, downcast='infer') == 1].index
df[~df.index.isin([item for sublist in
[ range(i-2, i+2) for i in ind ] for item in sublist])]
my_class = np.array([1] * 3 + [2] * 6 + [3] * 3)
cols = np.c_[my_class, np.arange(len(my_class)) + 1]
df = pd.DataFrame(cols, columns=['my_class', 'value'])
df['diff'] = df['my_class'].diff().fillna(0)
idx2drop = []
for i in df[df['diff'] == 1].index:
idx2drop += range(i - 2, i + 2)
print(df.drop(idx_drop)[['my_class', 'value']])
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
I am new to python and the last time I coded was in the mid-80's so I appreciate your patient help.
It seems .rolling(window) requires the window to be a fixed integer. I need a rolling window where the window or lookback period is dynamic and given by another column.
In the table below, I seek the Lookbacksum which is the rolling sum of Data as specified by the Lookback column.
d={'Data':[1,1,1,2,3,2,3,2,1,2],
'Lookback':[0,1,2,2,1,3,3,2,3,1],
'LookbackSum':[1,2,3,4,5,8,10,7,8,3]}
df=pd.DataFrame(data=d)
eg:
Data Lookback LookbackSum
0 1 0 1
1 1 1 2
2 1 2 3
3 2 2 4
4 3 1 5
5 2 3 8
6 3 3 10
7 2 2 7
8 1 3 8
9 2 1 3
You can create a custom function for use with df.apply, eg:
def lookback_window(row, values, lookback, method='sum', *args, **kwargs):
loc = values.index.get_loc(row.name)
lb = lookback.loc[row.name]
return getattr(values.iloc[loc - lb: loc + 1], method)(*args, **kwargs)
Then use it as:
df['new_col'] = df.apply(lookback_window, values=df['Data'], lookback=df['Lookback'], axis=1)
There may be some corner cases but as long as your indices align and are unique - it should fulfil what you're trying to do.
here is one with a list comprehension which stores the index and value of the column df['Lookback'] and the gets the slice by reversing the values and slicing according to the column value:
df['LookbackSum'] = [sum(df.loc[:e,'Data'][::-1].to_numpy()[:i+1])
for e,i in enumerate(df['Lookback'])]
print(df)
Data Lookback LookbackSum
0 1 0 1
1 1 1 2
2 1 2 3
3 2 2 4
4 3 1 5
5 2 3 8
6 3 3 10
7 2 2 7
8 1 3 8
9 2 1 3
An exercise in pain, if you want to try an almost fully vectorized approach. Sidenote: I don't think it's worth it here. At all.
Inspired by Divakar's answer here
Given:
import numpy as np
import pandas as pd
d={'Data':[1,1,1,2,3,2,3,2,1,2],
'Lookback':[0,1,2,2,1,3,3,2,3,1],
'LookbackSum':[1,2,3,4,5,8,10,7,8,3]}
df=pd.DataFrame(data=d)
Using the function from Divakar's answer, but slightly modified
from skimage.util.shape import view_as_windows as viewW
def strided_indexing_roll(a, r, fill_value=np.nan):
# Concatenate with sliced to cover all rolls
p = np.full((a.shape[0],a.shape[1]-1),fill_value)
a_ext = np.concatenate((p,a,p),axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = a.shape[1]
return viewW(a_ext,(1,n))[np.arange(len(r)), -r + (n-1),0]
Now, we just need to prepare a 2d array for the data and independently shift the rows according to our desired lookback values.
arr = df['Data'].to_numpy().reshape(1, -1).repeat(len(df), axis=0)
shifter = np.arange(len(df) - 1, -1, -1) #+ d['Lookback'] - 1
temp = strided_indexing_roll(arr, shifter, fill_value=0)
out = strided_indexing_roll(temp, (len(df) - 1 - df['Lookback'])*-1, 0).sum(-1)
Output:
array([ 1, 2, 3, 4, 5, 8, 10, 7, 8, 3], dtype=int64)
We can then just assign it back to the dataframe as needed and check.
df['out'] = out
#output:
Data Lookback LookbackSum out
0 1 0 1 1
1 1 1 2 2
2 1 2 3 3
3 2 2 4 4
4 3 1 5 5
5 2 3 8 8
6 3 3 10 10
7 2 2 7 7
8 1 3 8 8
9 2 1 3 3
I have two dataframes net and M.
net =
i j d
0 5 3 3
1 2 0 2
2 3 2 1
3 4 5 2
4 0 1 3
5 0 3 4
M =
0 1 2 3 4 5
0 0 3 2 4 1 5
1 3 0 2 0 3 3
2 2 2 0 1 1 4
3 4 0 1 0 3 3
4 1 3 1 3 0 2
5 5 3 4 3 2 0
I want to find in M the same values of net['d'], choose randomly a cell in M and create a new dataframe containing the coordinate of that cell. For instance
net['d'][0] = 3
so in M I find:
M[0][1]
M[1][0]
M[1][4]
M[1][5]
...
Finally net1 would be something like that
net1 =
i1 j1 d1
0 1 5 3
1 5 4 2
2 2 3 1
3 1 2 2
4 1 5 3
5 3 0 4
This what I am doing:
I1 = []
J1 = []
for i in net.index:
tmp = net['d'][i]
ds = np.where( M == tmp)
size = len(ds[0])
ind = randint(size) ## find two random locations with distance ds
h = ds[0][ind]
w = ds[1][ind]
I1.append(h)
J1.append(w)
net1 = pd.DataFrame()
net1['i1'] = I1
net1['j1'] = J1
net1['d1'] = net['d']
I am wondering which is the best way to avoid that loop
You can stack the columns of M and then just sample it with replacement
net = pd.DataFrame({'i':[5,2,3,4,0,0],
'j':[3,0,2,5,1,3],
'd':[3,2,1,2,3,4]})
M = pd.DataFrame({0:[0,3,2,4,1,5],
1:[3,0,2,0,3,3],
2:[2,2,0,1,1,4],
3:[4,0,1,0,3,3],
4:[1,3,1,3,0,2],
5:[5,3,4,3,2,0]})
def random_net(net, M):
# make long table and randomize order of rows and rename columns
net1 = M.stack().reset_index()
net1.columns =['i1', 'j1', 'd1']
# get size of each group for random mapping
net1_id_length = net1.groupby('d1').size()
# add id column to uniquely identify row in net
net_copy = net.copy()
# first map gets size of each group and second gets random integer
net_copy['id'] = net_copy['d'].map(net1_id_length).map(np.random.randint)
net1['id'] = net1.groupby('d1').cumcount()
# make for easy lookup
net_copy = net_copy.set_index(['d', 'id'])
net1 = net1.set_index(['d1', 'id'])
# choose from net1 only those from original net
return net1.reindex(net_copy.index).reset_index('d').reset_index(drop=True).rename(columns={'d':'d1'})
random_net(net, M)
output
d1 i1 j1
0 3 5 1
1 2 0 2
2 1 3 2
3 2 1 2
4 3 3 5
5 4 0 3
Timings on 6 million rows
n = 1000000
net = pd.DataFrame({'i':[5,2,3,4,0,0] * n,
'j':[3,0,2,5,1,3] * n,
'd':[3,2,1,2,3,4] * n})
M = pd.DataFrame({0:[0,3,2,4,1,5],
1:[3,0,2,0,3,3],
2:[2,2,0,1,1,4],
3:[4,0,1,0,3,3],
4:[1,3,1,3,0,2],
5:[5,3,4,3,2,0]})
%timeit random_net(net, M)
1 loop, best of 3: 13.7 s per loop
I am attempting to print my python program that implements floyds algorithm.
n=5
for k in range(n):
for j in range(n):
for i in range(n):
if A[i][k]+A[k][j]<A[i][j]:
A[i][j]=A[i][k]+A[k][j]
I am trying to print the solution in the same format as below: (not including the first column and row)
0 1 2 3 4
-----------
0|0 1 4 500 3
1|1 0 2 500 4
2|4 2 0 1 5
3|500 500 1 0 3
4|3 4 5 3 0
500 indicates infinity
Any ideas? I am hoping indices will do the trick.
Also does anyone know the order of magnitude of this algorithm?
You could use this.
>>> A = [[0, 1, 4, 500, 3], [1, 0, 2, 500, 4],[4,2,0,1,5],[500,500,1,0,3],[3,4,5,3,0]]
>>> for elem in A:
print "\t".join(['Inf' if val == 500 else str(val) for val in elem])
0 1 4 Inf 3
1 0 2 Inf 4
4 2 0 1 5
Inf Inf 1 0 3
3 4 5 3 0