Create a list of multiples of a number - python

Problem:
List of Multiples
Create a Python 3 function that takes two numbers (value, length) as arguments and returns a list of multiples of value until the size of the list reaches length.
Examples
list_of_multiples(value=7, length=5) ➞ [7, 14, 21, 28, 35]
list_of_multiples(value=12, length=10) ➞ [12, 24, 36, 48, 60, 72, 84, 96, 108, 120]
list_of_multiples(value=17, length=6) ➞ [17, 34, 51, 68, 85, 102]
def multiples (value,length):
"""
Value is number to be multiply
length is maximum number of iteration up to
which multiple required.
"""
for i in range(length):
out=i
return i


Most Pythonic Way
def multiples(value, length):
return [*range(value, length*value+1, value)]
print(multiples(7, 5))
# [7, 14, 21, 28, 35]
print(multiples(12, 10))
# [12, 24, 36, 48, 60, 72, 84, 96, 108, 120]
print(multiples(17, 6))
# [17, 34, 51, 68, 85, 102]


Pythonic way:
def multiples(value, length):
return [value * i for i in range(1, length + 1)]
print(multiples(7, 5))
# [7, 14, 21, 28, 35]
print(multiples(12, 10))
# [12, 24, 36, 48, 60, 72, 84, 96, 108, 120]
print(multiples(17, 6))
# [17, 34, 51, 68, 85, 102]


def multiples(value, length):
list_multiples = []
i = 0
while i < length:
list_multiples.append(value*(i+1))
i+=1
return list_multiples


The easy / not in-line way would be :
def multiples(value, length):
l = []
for i in range(1, length+1):
l.append(value*i)
return l


The best answer for small values of length (< 100) is given by Nite Block.
However, in case length becomes bigger, using numpy is significantly faster than python loops:
numpy.arange(1, length+1) * value
With a length of 1000, python loops take almost 4 times longer than numpy. See code below:
import timeit
testcode_numpy = '''
import numpy
def multiples_numpy(value, length):
return numpy.arange(1, length+1) * value
multiples_numpy(5, 1000)
'''
testcode = '''
def multiples(value, length):
return [*range(value, length*value+1, value)]
multiples(5, 1000)
'''
print(timeit.timeit(testcode_numpy))
print(timeit.timeit(testcode))
# Result:
# without numpy: 9.7 s
# with numpy: 2.4 s

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'))

Creating a limit to find the sum of array values to a set number Numpy Python

I want to make a function where the sum of the Arrays and Arrays2 array is equivalent to val. The function should modify the Arrays and Arrays2 values so that the last index will output the sum of all values in the array to be val. How will be able to get the Expected Output?
import numpy as np
Arrays = np.array([50, 30, 25, 87, 44, 68, 45])
Arrays2 = np.array([320])
val = 300
Expected output:
[50, 30, 25, 87, 44, 64]
[300]

something like this?
import numpy as np
Arrays = np.array([50, 30, 25, 87, 44, 68, 45])
Arrays2 = np.array([320])
val = 300
def thisRareFunction(arr):
outArrays = []
acum = 0
for x in arr:
acum += x
if acum <=val:
outArrays.append(x)
else:
outArrays.append(x -(acum-val))
break
return outArrays
print(thisRareFunction(Arrays))
print(thisRareFunction(Arrays2))

Python - cut only the descending part of the dataset

I have a timeseries with various downcasts. My question is how do I slice a pandas dataframe (or in this case the array, just to keep it simple) to get the data and its indexes of the descending bits of the timeseries?
import matplotlib.pyplot as plt
import numpy as np
b = np.asarray([ 1.3068586 , 1.59882279, 2.11291473, 2.64699527,
3.23948166, 3.81979878, 4.37630243, 4.97740025,
5.59247254, 6.18671493, 6.77414586, 7.43078595,
8.02243495, 8.59612224, 9.22302662, 9.83263379,
10.43125902, 11.0956864 , 11.61107838, 12.09616684,
12.63973254, 12.49437955, 11.6433792 , 10.61083269,
9.50534291, 8.47418827, 7.40571742, 6.56611512,
5.66963658, 4.89748187, 4.10543794, 3.44828054,
2.76866318, 2.24306623, 1.68034463, 1.26568186,
1.44548443, 2.01225076, 2.60715524, 3.21968562,
3.8622007 , 4.57035958, 5.14021305, 5.77879484,
6.42776897, 7.09397923, 7.71722028, 8.30860725,
8.96652218, 9.66157193, 10.23469208, 10.79889453,
10.5788411 , 9.38270646, 7.82070643, 6.74893389,
5.68200335, 4.73429009, 3.78358222, 3.05924946,
2.30428171, 1.78052369, 1.27897065, 1.16840532,
1.59452726, 2.13085096, 2.70989933, 3.3396291 ,
3.97318058, 4.62429262, 5.23997774, 5.91232803,
6.5906609 , 7.21099657, 7.82936331, 8.49636247,
9.15634983, 9.76450244, 10.39680729, 11.04659976,
11.69287237, 12.35692643, 12.99957563, 13.66228386,
14.31806385, 14.91871927, 15.57212978, 16.22288287,
16.84697357, 17.50502002, 18.15907842, 18.83068151,
19.50945548, 20.18020639, 20.84441358, 21.52792846,
22.17933087, 22.84614545, 23.51212887, 24.18308399,
24.8552263 , 25.51709528, 26.18724379, 26.84531493,
27.50690265, 28.16610365, 28.83394822, 29.49621179,
30.15118676, 30.8019521 , 31.46714114, 32.1213546 ,
32.79366952, 33.45233007, 34.12158193, 34.77502197,
35.4532211 , 36.11018053, 36.76540453, 37.41746323])
plt.plot(-b)
plt.show()

You can just change the negative diffs to NaN and then plot:
bb = pd.Series(-b)
bb[bb.diff().ge(0)] = np.nan
bb.plot()
To get the indexes of descending values, use:
bb.index[bb.diff().lt(0)]
Int64Index([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20, 37, 38, 39, 40, 41, 42,
43, 44, 45, 46, 47, 48, 49, 50, 51, 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, 102, 103, 104, 105, 106, 107,
108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119],
dtype='int64')

create a second dataframe where you move everyting from one index then you do it by substracting them term to term. you should get what you want (getting only the ones with negative diff)
here:
df = DataFrame(b)
df = concat([df.shift(1),df],axis = 1)
df.columns = ['t-1','t']
df.reset_index()
df = df.drop(df.index[0])
df['diff'] = df['t']-df['t-1']
res = df[df['diff']<0]

There is also an easy numpy-only solution (the question is tagged pandas but the code uses only numpy) using np.where. You want the points where the graph is descending which means the data is ascending.
# the indices where the data is ascending.
ix, = np.where(np.diff(b) > 0)
# the values
c = b[ix]
Note that this will give you the first value in each ascending pair of consecutive values, while the pandas-based solution gives the second one. To get the same indices just add 1 to ix.
s = pd.Series(b)
assert np.all(s[s.diff() > 0].index == ix + 1)
assert np.all(s[s.diff() > 0] == b[ix + 1])

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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