Largest strongly connected components of a directed graph - python

I am working on an Networkx .MultiDiGraph() object built from a total of 82927 directed email data. At current stage, I am trying to get the largest strongly connected components from the .MultiDiGraph() object and its corresponding subgraph.
The text data can be accessed here.
Here's my working code:
import networkx as nx
import pandas as pd
import matplotlib.pyplot as plt
email_df = pd.read_csv('email_network.txt', delimiter = '->')
edge_groups = email_df.groupby(["#Sender", "Recipient"], as_index=False).count().rename(columns={"time":"weight"})
email = nx.from_pandas_dataframe(edge_groups, '#Sender', 'Recipient', edge_attr = 'weight')
G = nx.MultiDiGraph()
G.add_edges_from(email.edges(data=True))
# G is a .MultiDiGraph object
# using .strongly_connected_components() to get the part of G that has the most nodes
# using list comprehension
number_of_nodes = [len(n) for n in sorted(nx.strongly_connected_components(G))]
number_of_nodes
# 'number_of_nodes' return a list of [1, 1, 1,...,1] of length 167 (which is the exact number of nodes in the network)
# using the recommended method in networkx documentation
largest = max(nx.strongly_connected_components(G), key=len)
largest
# 'largest' returns {92}, not sure what this means...
As I noted in the above code block, the list comprehension method returns a list of [1, 1, 1,..., 1] of length 167 (which is the total number of nodes in my data), while the max(nx.strongly_connected_components(G), key=len) returned {92}, I am not sure what this means.
It looks like there's something wrong with my code and I might have missed several key steps in processing the data. Could anyone care to take a look at and enlighten me on this?
Thank you.
Note: Revised code (kudos to Eric and Joel)
import networkx as nx
import pandas as pd
import matplotlib.pyplot as plt
email_df = pd.read_csv('email_network.txt', delimiter = ' ')
edge_groups = email_df.groupby(["#Sender", "Recipient"], as_index=False).count().rename(columns={"time":"weight"})
# per #Joel's comment, adding 'create_using = nx.DiGraph()'
email = nx.from_pandas_dataframe(edge_groups, '#Sender', 'Recipient', edge_attr = 'weight', create_using = nx.DiGraph())
# adding this 'directed' edge list to .MultiDiGraph() object
G = nx.MultiDiGraph()
G.add_edges_from(email.edges(data=True))
We now examine the largest strongly connected component (in terms of the number of nodes) in this network.
In [1]: largest = max(nx.strongly_connected_components(G), key=len)
In [2]: len(largest)
Out [2]: 126
The largest strongly connected component consists of 126 nodes.
[Updates]
Upon further trial and error, I found that one needs to use create_using = .MultiDiGraph() (instead of .DiGraph()) when loading data onto networkx, otherwise, even if you get correct number of nodes for your MultiDiGraph and its weakly/strongly connected subgraphs, you might still get the number of edges wrong! This will reflect in you .strongly_connected_subgraphs() outputs.
For my case here, I will recommend others to use this one-liner
import networkx as nx
import pandas as pd
import matplotlib.pyplot as plt
G = nx.read_edgelist(path="email_network.txt", data=[('time', int)], create_using=nx.MultiDiGraph(), nodetype=str)
And we can implement .strongly_connected_components(G) and strongly_connected_subgraphs to verify.
If you use the networkx output G from the first code block, max(nx.strongly_connected_components(G), key=len) will give an output with 126 nodes and 52xx something edges, but if you apply the one-liner I listed above, you will get:
In [1]: largest = max(nx.strongly_connected_components(G), key=len)
In [2]: G_sc = max(nx.strongly_connected_subgraphs(G), key=len)
In [3]: nx.number_of_nodes(G_sc)
Out [3]: 126
In [4]: nx.number_of_nodes(G_sc)
Out [4]: 82130
You will get the same number of nodes with both methods but different number of edges owing to different counting mechanisms associated with different networkx graph classes.

The underlying cause of your error is that nx.from_pandas_dataframe defaults to creating an undirected graph. So email is an undirected graph. When you then create the directed graph, each edge appears in only one direction.
To fix it use nx.from_pandas_dataframe with the argument create_using = DiGraph
older comments related to the output you were getting
All your strongly connected components have a single node.
When you do max(nx.strongly_connected_components(G), key=len) it finds the set of nodes which has the longest length and returns it. In your case, they all have length 1, so it returns one of them (I believe whichever networkx happened to put into nx.strongly_connected_components(G) first). But it's returning the set, not the length. So {92} is the set of nodes it is returning.
It happens that {92} was chosen to be the "longest" length 1 component in nx.strongly_connected_components(G) by the tiebreaker.
Example:
max([{1}, {3}, {5}], key = len)
> {1}

[1, 1, 1,...,1] of length 167 (which is the exact number of nodes in the network)
This means that there's basically no strongly connected component in your graph (except for lone vertices, that is).
If you sort those components by length, you get a randon component of one single vertex since the components all have the same length (1). In your example, {92}, which could have been any other vertex.
The import looks correct and there's really no strongly connected component, it means that nobody ever replied to any email.
To check if the problem doesn't come from pandas, MultiDiGraph or your import, I wrote:
G = nx.DiGraph()
with open('email_network.txt') as f:
for line in f:
n1, n2, time = line.split()
if n1.isdigit():
G.add_edge(int(n1),int(n2))
It didn't change the result.
Just adding an edge with G.add_edge(2,1) creates a large strongly connected component, though:
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 126, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 115, 117, 118, 119, 120, 121, 122, 123, 124, 128, 129, 134, 149, 151}

Related

Convert multidimensional array of indices clusters to a 1D categorical array

I have a function which returns a multidimensional array of k clusters. My algorith works for the most part, but I need it to return a categorical array instead of a multidimensional array. Here is my code:
import numpy as np
import pandas as pd
import random
from bokeh.sampledata.iris import flowers
from typing import List, Tuple
def get_closest(data_point: np.ndarray, centroids: np.ndarray):
"""
Takes a data_point and a nd.array of multiple centroids and returns the index of the centroid closest to data_point
by computing the euclidean distance for each centroid and picking the closest.
"""
N = centroids.shape[0]
dist = np.empty(N)
for i, c in enumerate(centroids):
dist[i] = np.linalg.norm(c - data_point)
index_min = np.argmin(dist)
return index_min
# Use these centroids in the first iteration of you algorithm if "Random Centroids" is set to False in the Dashboard
DEFAULT_CENTROIDS = np.array([[5.664705882352942, 3.0352941176470587, 3.3352941176470585, 1.0176470588235293],
[5.446153846153847, 3.2538461538461543, 2.9538461538461536, 0.8846153846153846],
[5.906666666666667, 2.933333333333333, 4.1000000000000005, 1.3866666666666667],
[5.992307692307692, 3.0230769230769234, 4.076923076923077, 1.3461538461538463],
[5.747619047619048, 3.0714285714285716, 3.6238095238095243, 1.1380952380952383],
[6.161538461538462, 3.030769230769231, 4.484615384615385, 1.5307692307692309],
[6.294117647058823, 2.9764705882352938, 4.494117647058823, 1.4],
[5.853846153846154, 3.215384615384615, 3.730769230769231, 1.2076923076923078],
[5.52857142857143, 3.142857142857143, 3.107142857142857, 1.007142857142857],
[5.828571428571429, 2.9357142857142855, 3.664285714285714, 1.1]])
def k_means(data_np: np.ndarray, k:int=3, n_iter:int=500, random_initialization=False) -> Tuple[np.ndarray, int]:
"""
:param data: your data, a numpy array with shape (n_entries, n_features)
:param k: The number of clusters to compute
:param n_iter: The maximal numnber of iterations
:param random_initialization: If False, DEFAULT_CENTROIDS are used as the centroids of the first iteration.
:return: A tuple (cluster_indices: A numpy array of cluster_indices,
n_iterations: the number of iterations it took until the algorithm terminated)
"""
# Initialize the algorithm by assigning random cluster labels to each entry in your dataset
k=k+1
centroids = data_np[random.sample(range(len(data_np)), k)]
labels = np.array([np.argmin([(el - c) ** 2 for c in centroids]) for el in data_np])
clustering = []
for k in range(k):
clustering.append(data_np[labels == k])
# Implement K-Means with a while loop, which terminates either if the centroids don't move anymore, or
# if the number of iterations exceeds n_iter
counter = 0
while counter < n_iter:
# Compute the new centroids, if random_initialization is false use DEFAULT_CENTROIDS in the first iteration
# if you use DEFAULT_CENTROIDS, make sure to only pick the k first entries from them.
if random_initialization is False and counter == 0:
centroids = DEFAULT_CENTROIDS[random.sample(range(len(DEFAULT_CENTROIDS)), k)]
# Update the cluster labels using get_closest
labels = np.array([get_closest(el, centroids) for el in data_np])
clustering = []
for i in range(k):
clustering.append(np.where(labels == i)[0])
counter += 1
new_centroids = np.zeros_like(centroids)
for i in range(k):
if len(clustering[i]) > 0:
new_centroids[i] = data_np[clustering[i]].mean(axis=0)
else:
new_centroids[i] = centroids[i]
# if the centroids didn't move, exit the while loop
if clustering is not None and (centroids == new_centroids).sum() == 0:
break
else:
centroids = new_centroids
pass
# return the final cluster labels and the number of iterations it took
return clustering, counter
# read and store the dataset
data: pd.DataFrame = flowers.copy(deep=True)
data = data.drop(['species'], axis=1)
data_np = np.asarray(data)
clustering, counter = k_means(data_np,4,500,False)
So clustering looks like so
clustering
[array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33,
34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 57,
98], dtype=int64),
array([60, 93], dtype=int64),
array([ 50, 51, 52, 53, 54, 55, 56, 58, 61, 62, 63, 65, 66,
67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79,
80, 81, 82, 83, 86, 87, 89, 90, 91, 92, 94, 95, 96,
97, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110,
111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123,
124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136,
137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149],
dtype=int64),
array([59, 64, 84, 85, 88], dtype=int64)]
However, what I'm looking for is an array like
clustering
array([1, 3, 2, ..., 4, 1, 4], dtype=int64)]
Also, the while loop is always terminating after 1 iteration which shouldn't be the case.
counter
1
EDIT1:
The code continues as follows.
def callback(attr, old, new):
# recompute the clustering and update the colors of the data points based on the result
k = slider_k.valued_throttled
init = select_init.value
clustering_new, counter_new = k_means(data_np,k,500,init)
pass
# Create the dashboard
# 1. A Select widget to choose between random initialization or using the DEFAULT_CENTROIDS on top
select_init = Select(title='Random Centroids',value='False',options=['True','False'])
# 2. A Slider to choose a k between 2 and 10 (k being the number of clusters)
slider_k = Slider(start=2,end=10,value=3,step=1,title='k')
# 4. Connect both widgets to the callback
select_init.on_change('value',callback)
slider_k.on_change('value_throttled',callback)
# 3. A ColumnDataSource to hold the data and the color of each point you need
source = ColumnDataSource(dict(petal_length=data['petal_length'],sepal_length=data['sepal_length'],petal_width=data['petal_width'],clustering=clustering))
# 4. Two plots displaying the dataset based on the following table, have a look at the images
# in the handout if this confuses you.
#
# Axis/Plot Plot1 Plot2
# X Petal length Petal width
# Y Sepal length Petal length
#
# Use a categorical color mapping, such as Spectral10, have a look at this section of the bokeh docs:
# https://docs.bokeh.org/en/latest/docs/user_guide/categorical.html#filling
plot1 = figure(plot_width=100,plot_height=100,title='Scatterplot of flowers distribution by petal length and sepal length')
plot1.yaxis.axis_label = 'Sepal length'
plot1.xaxis.axis_label = 'Petal length'
scatter1 = plot1.scatter(x='petal_length',y='sepal_length',source=source,fill_color=factor_cmap('clustering', palette=Spectral10, factors=clustering))
plot2 = figure(plot_width=100,plot_height=100,title='Scatterplot of flowers distribution by petal width and petal length')
plot2.yaxis.axis_label = 'Petal length'
plot2.xaxis.axis_label = 'Petal width'
scatter2 = plot2.scatter(x='petal_width',y='petal_length',source=source,fill_color=factor_cmap('clustering', palette=Spectral10, factors=clustering))
# 5. A Div displaying the currently number of iterations it took the algorithm to update the plot.
div = Div(text='Number of iterations: ')
Thus the end result should look like so
I'm not sure I understand what you need.
If clustering contains a list of arrays where each array represent a cluster and the ith array contains the indices of the samples that belong to the ith cluster and what you need is to convert this to a single vector of size number_of_samples that represent the cluster each sample belongs to you can do it like this:
def to_classes(clustering):
# Get number of samples (you can pass it directly to the function)
num_samples = sum(x.shape[0] for x in clustering)
indices = np.empty((num_samples,)) # An empty array with correct size
for ith, cluster in enumerate(clustering):
# use cluster indices to assign to correct the cluster index
indices[cluster] = ith
return indices
The loops exists after a single iteration because the break condition is wrong, I think what you want is actually
# note the !=
if clustering is not None and (centroids != new_centroids).sum() == 0:
break

How to calculate third central moment?

Description: I have a sample: sample = [100, 86, 51, 100, 95, 100, 12, 61, 0, 0, 12, 86, 0, 52, 62, 76, 91, 91, 62, 91, 65, 91, 9, 83, 67, 58, 56]. I need to calculate third central moment of this sample.
My approach:
I'm making a table with top row being unique values from the sample and bottom row - frequency of each value from the top row:
table = dict(Counter(sample))
Then I'm calculating empirical k-th central moment with this formula:
def empirical_central_moment(table: dict, k):
mean = sum([value * frequency for value, frequency in table.items()]) / sum(list(table.values()))
N = sum(list(table.values()))
return sum([(value - mean)**k * frequency / N for value, frequency in table.items()])
Program:
from collections import Counter
def empirical_central_moment(table: dict, k):
mean = sum([value * frequency for value, frequency in table.items()]) / sum(list(table.values()))
N = sum(list(table.values()))
return sum([(value - mean)**k * frequency / N for value, frequency in table.items()])
sample = [100, 86, 51, 100, 95, 100, 12, 61, 0, 0, 12, 86, 0, 52, 62, 76, 91, 91, 62, 91, 65, 91, 9, 83, 67, 58, 56]
table = dict(Counter(sample))
print(empirical_central_moment(table, 3))
Problem: Instead of desired -545.33983 ... I'm getting -26721.65147589292 and I just can't wrap my head around as to why I'm gettting wrong. Will appreciate any help, thanks in advance.
Your answer is correct. Not sure what other answer you might be looking for. In general, and unless the purpose of this code is to exercise programming the logic of it, you don't need to reinvent the wheel and you'll be much faster and safer by doing something as simple as:
from scipy.stats import moment
sample = [100, 86, 51, 100, 95, 100, 12, 61, 0, 0, 12, 86, 0, 52, 62, 76, 91, 91, 62, 91, 65, 91, 9, 83, 67, 58, 56]
print(scipy.stats.moment(sample, moment=3, axis=0, nan_policy='propagate'))

Why doesn't numpy create an array when executing a list method

Playing around with numpy:
import numpy as np
l = [39, 54, 72, 46, 89, 53, 96, 64, 2, 75]
nl = np.array(l.append(3))
>> array(None, dtype=object)
Now, if I call on l, I'll get the list: [39, 54, 72, 46, 89, 53, 96, 64, 2, 75, 3]
My question is, why doesn't numpy create that list as an array?
If I do something like this:
nl = np.array(l.extend([45])) I get the same thing.
But, if I try to concatenate without a method: nl = np.array(l+[45]) it works.
What is causing this behaviour?
The append function will always return None. You must do this in two different lines of code:
import numpy as np
l = [39, 54, 72, 46, 89, 53, 96, 64, 2, 75]
l.append(3)
nl = np.array(l)
append and extend are in-place methods and return None.
print(l.append(3)) # None
print(l.extend([3])) # None

how to include first element of a list before computing the difference

I'm creating a histogram. I currently have this block of code:
g = [479, 481, 503, 525, 554, 586, 614, 669, 683]
and then i've written this for the x and y axis:
x =[28, 27, 26, 25, 24, 23, 22, 21, 20]
y = diff(g)
This is what it computes y as:
array([ 2, 22, 22, 29, 32, 28, 55, 14])
However, I realized that my histogram doesn't include 479 (first element in g) before it starts computing the difference from there onwards, which is what I was hoping to do. My desired output is
array([ 479, 2, 22, 22, 29, 32, 28, 55, 14])
Is there a way that I can do this? I don't want to manually append it as I need to automate it for various files.
There are two main ways of prepending elements to a diff: before or after the fact. If you want to prepend a zero before, you can use the the prepend argument available as of v1.16.0:
y = np.diff(g, prepend=0)
This is equivalent to manually inserting a zero into your array (in case your version of numpy is older):
y = np.diff(np.insert(g, 0, 0))
You can do something very similar after the diff, by inserting g[0] into the beginning:
y = np.insert(np.diff(g), 0, g[0])
However, all the options shown here are inefficient because they copy all your data (g or the diff). A space-efficient solution would allocate an output buffer, and compute the difference manually:
y = np.empty_like(g)
y[1:] = g[1:] - g[:-1]
y[0] = g[0]

Numpy Array Slicing

I have a 1D numpy array, and some offset/length values. I would like to extract from this array all entries which fall within offset, offset+length, which are then used to build up a new 'reduced' array from the original one, that only consists of those values picked by the offset/length pairs.
For a single offset/length pair this is trivial with standard array slicing [offset:offset+length]. But how can I do this efficiently (i.e. without any loops) for many offset/length values?
Thanks,
Mark
>>> import numpy as np
>>> a = np.arange(100)
>>> ind = np.concatenate((np.arange(5),np.arange(10,15),np.arange(20,30,2),np.array([8])))
>>> a[[ind]]
array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 22, 24, 26, 28, 8])
There is the naive method; just doing the slices:
>>> import numpy as np
>>> a = np.arange(100)
>>>
>>> offset_length = [(3,10),(50,3),(60,20),(95,1)]
>>>
>>> np.concatenate([a[offset:offset+length] for offset,length in offset_length])
array([ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 50, 51, 52, 60, 61, 62, 63,
64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 95])
The following might be faster, but you would have to test/benchmark.
It works by constructing a list of the desired indices, which is valid method of indexing a numpy array.
>>> indices = [offset + i for offset,length in offset_length for i in xrange(length)]
>>> a[indices]
array([ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 50, 51, 52, 60, 61, 62, 63,
64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 95])
It's not clear if this would actually be faster than the naive method but it might be if you have a lot of very short intervals. But I don't know.
(This last method is basically the same as #fraxel's solution, just using a different method of making the index list.)
Performance testing
I've tested a few different cases: a few short intervals, a few long intervals, lots of short intervals. I used the following script:
import timeit
setup = 'import numpy as np; a = np.arange(1000); offset_length = %s'
for title, ol in [('few short', '[(3,10),(50,3),(60,10),(95,1)]'),
('few long', '[(3,100),(200,200),(600,300)]'),
('many short', '[(2*x,1) for x in range(400)]')]:
print '**',title,'**'
print 'dbaupp 1st:', timeit.timeit('np.concatenate([a[offset:offset+length] for offset,length in offset_length])', setup % ol, number=10000)
print 'dbaupp 2nd:', timeit.timeit('a[[offset + i for offset,length in offset_length for i in xrange(length)]]', setup % ol, number=10000)
print ' fraxel:', timeit.timeit('a[np.concatenate([np.arange(offset,offset+length) for offset,length in offset_length])]', setup % ol, number=10000)
This outputs:
** few short **
dbaupp 1st: 0.0474979877472
dbaupp 2nd: 0.190793991089
fraxel: 0.128381967545
** few long **
dbaupp 1st: 0.0416231155396
dbaupp 2nd: 1.58000087738
fraxel: 0.228138923645
** many short **
dbaupp 1st: 3.97210478783
dbaupp 2nd: 2.73584890366
fraxel: 7.34302687645
This suggests that my first method is the fastest when you have a few intervals (and it is significantly faster), and my second is the fastest when you have lots of intervals.

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