In testing a conjecture about the following recursive relation
,
which claims a periodicity of some kind for the sequence of numbers, I wrote a python program which computes the sequences and prints them in a table.
1 # Consider the recursive relation x_{i+1} = p-1 - (p*i-1 mod x_i)
2 # with p prime and x_0 = 1. What is the shortest period of the
3 # sequence?
4
5 from __future__ import print_function
6 import numpy as np
7 from matplotlib import pyplot as plt
8
9 # The length of the sequences.
10 seq_length = 100
11
12 upperbound_primes = 30
13
14 # Computing a list of prime numbers up to n
15 def primes(n):
16 sieve = [True] * n
17 for i in xrange(3,int(n**0.5)+1,2):
18 if sieve[i]:
19 sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
20 return [2] + [i for i in xrange(3,n,2) if sieve[i]]
21
22 # The list of prime numbers up to upperbound_primes
23 p = primes(upperbound_primes)
24
25 # The amount of primes numbers
26 no_primes = len(p)
27
28 # Generate the sequence for the prime number p
29 def sequence(p):
30 x = np.empty(seq_length)
31 x[0] = 1
32 for i in range(1,seq_length):
33 x[i] = p - 1 - (p * (i-1) - 1) % x[i-1]
34 return x
35
36 # List with the sequences.
37 seq = [sequence(i) for i in p]
38 """
39 # Print the sequences in a table where the upper row
40 # indicates the prime numbers.
41 for i in range(seq_length):
42 if not i:
43 for n in p:
44 print('\t',n,end='')
45 print('')
46 print(i+1,'\t',end='')
47 for j in range(no_primes):
48 print(seq[j][i],end='\t')
49 print('\n',end='')
50 """
51 def autocor(x):
52 result = np.correlate(x,x,mode='full')
53 return result[result.size/2:]
54
55
56 fig = plt.figure('Finding period in the sequences')
57 k = 0
58 for s in seq:
59 k = k + 1
60 fig.add_subplot(no_primes,1,k)
61 plt.title("Prime number %d" % p[k-1])
62 plt.plot(autocor(s))
63 plt.show()
64
Now I want to investigate periodicities in these sequences that I computed. After looking around on the net I found myself two options it seems:
Preform autocorrelation on the data and look for the first peak. This should give an approximation of the period.
Preform a FFT on the data. This shows the frequency of the numbers. I do not see how this can give any useful information about the periodicity of a sequence of numbers.
The last lines show my attempt of using autocorrelation, inspired by the accepted answer of How can I use numpy.correlate to do autocorrelation?.
It gives the following plot
Clearly we see a descending sequence of numbers for all the primes.
When testing the same method on a sin function with the following simplyfied python-code snippet
1 # Testing the autocorrelation of numpy
2
3 import numpy as np
4 from matplotlib import pyplot as plt
5
6 num_samples = 1000
7 t = np.arange(num_samples)
8 dt = 0.1
9
10 def autocor(x):
11 result = np.correlate(x,x,mode='full')
12 return result[result.size/2:]
13
14 def f(x):
15 return [np.sin(i * 2 * np.pi * dt) for i in range(num_samples)]
16
17 plt.plot(autocor(f(t)))
18 plt.show()
I get a similar result, it giving the following plot for the sine function
How could I read off the periodicity in the sine-function case, for example?
Anyhow, I do not understand the mechanism of the autocorrelation leading to peaks that give information of the periodicity of a signal. Can someone elaborate on that? How do you properly use autocorrelation in this context?
Also what am I doing wrong in my implementation of the autocorrelation?
Suggestions on alternative methods of determining periodicity in a sequence of numbers are welcome.
There are quite a few questions here, so I'll start be describing how an autocorrelation produces the period from the case of "3", ie, your second sub-plot of the first image.
For prime number 3, the sequence is (after a less consistent start) 1,2,1,2,1,2,1,2,.... To calculate the autocorrelation, the array is basically translated relative to itself, all the elements that align are multiplied, and all of these results are added. So it looks something like this, for a few test cases, where A is the autocorrelation:
0 1 2 3 4 5 6 7 ... 43 44 45 46 47 48 49 # indices 0
1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 # values 0
1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 # values 1
0 1 2 3 4 5 6 7 ... 43 44 45 46 47 48 49 # indices 1
1 4 1 4 1 4 1 4 4 1 4 1 4 1 4 # products
# above result A[0] = 5*25 5=1+4 25=# of pairs # A[0] = 125
0 1 2 3 4 5 6 7 ... 43 44 45 46 47 48 49 # indices 0
1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 # values 0
1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 # values 1
0 1 2 3 4 5 6 7 ... 43 44 45 46 47 48 49 # indices 1
2 2 2 2 2 2 2 2 2 2 2 2 2 2 # products
# above result A[1] = 4*24 4=2+2 24=# of pairs # A[1] = 96
0 1 2 3 4 5 6 7 ... 43 44 45 46 47 48 49 # indices 0
1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 # values 0
1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 # values 1
0 1 2 3 4 5 6 7 ... 43 44 45 46 47 48 49 # indices 1
1 4 1 4 1 4 1 4 4 1 4 1 4 # products
# above result A[2] = 5*23 5=4+1 23=# of pairs # A[2] = 115
There are three take-home messages from the above: 1. the autocorrelation, A, has larger value when like elements are lined up and multiplied, here at every other step. 2. The index of the autocorrelation corresponds to the relative shift. 3. When doing the autocorrelation over the full arrays, as shown here, there's always a downward ramp since the number of points added together to produce the value are reduced at each successive shift.
So here you can see why there's a periodic 20% bump in your graph from "Prime number 3": because the terms that are summed are 1+4 when they are aligned, vs 2+2 when they aren't, ie, 5 vs 4. It's this bump that you're looking for in reading off the period. That is, here is shows that the period is 2, since that is the index of your first bump. (Also, note btw, in the above I only do the calculation as number of pairs to see how this known periodicity leads to the result you see in the autocorrelation, that is, one doesn't in general want to think of number of pairs.)
In these calculations, the values of the bump relative to the base will be increased if you first subtract the mean before doing the autocorrelation. The ramp can be removed if you do the calculation using arrays with trimmed ends, so there's always the same overlap; this usually makes sense since usually one is looking for a periodicity of much shorter wavelength than the full sample (because it takes many oscillations to define a good period of oscillation).
For the autocorrelation of the sine wave, the basic answer is that the period is shown as the first bump. I redid the plot except with the time axis applied. It's always clearest in these things to use a real time axis, so I changed your code a bit to include that. (Also, I replaced the list comprehension with a proper vectorized numpy expression for calculating the sin wave, but that's not important here. And I also explicitly defined the frequency in f(x), just to make it more clear what's going on -- as an implicitly frequency of 1 in confusing.)
The main point is that since the autocorrelation is calculated by shifting along the axis one point at a time, the axis of the autocorrelation is just the time axis. So I plot that as the axis, and then can read the period off of that. Here I zoomed in to see it clearly (and the code is below):
# Testing the autocorrelation of numpy
import numpy as np
from matplotlib import pyplot as plt
num_samples = 1000
dt = 0.1
t = dt*np.arange(num_samples)
def autocor(x):
result = np.correlate(x,x,mode='full')
return result[result.size/2:]
def f(freq):
return np.sin(2*np.pi*freq*t)
plt.plot(t, autocor(f(.3)))
plt.xlabel("time (sec)")
plt.show()
That is, in the above, I set the frequency to 0.3, and the graph shows the period as about 3.3, which is what's expected.
All of this said, in my experience, the autocorrelation generally works well for physical signals but not so reliably for algorithmic signals. It's fairly easy to throw off, for example, if a periodic signal skips a step, which can happen with an algorithm, but is less likely with a vibrating object. You'd think that it should be trivial to calculate that period of an algorithmic signal, but a bit of searching around will show that it's not, and it's even difficult to define what's meant by period. For example, the series:
1 2 1 2 1 2 0 1 2 1 2 1 2
won't work well with the autocorrelation test.
Update.
#tom10 gave a thorough survey of autocorrelation and explained why the first bump in the autocorrelation could give the period of the periodic signal.
I tried both approaches, FFT as well as autocorrelation. Their results agree, although I would prefer FFT over autocorrelation since it gives you the period more directly.
When using autocorrelation, we simply determine the coordinate of the first peak. A manual inspection of the autocorrelation graph will reveal if you have the 'right' peak, since you can notice the period (although for primes above 7 this becomes less clear). I'm sure you could also work out a simple algorithm which calculates the 'right' peak. Perhaps someone could elaborate on some simple algortihm which does the job?
See, for instance, the following plot of the sequences next to their autocorrelation.
Code:
1 # Plotting sequences satisfying, x_{i+1} = p-1 - (p*i-1 mod x_i)
2 # with p prime and x_0 = 1, next to their autocorrelation.
3
4 from __future__ import print_function
5 import numpy as np
6 from matplotlib import pyplot as plt
7
8 # The length of the sequences.
9 seq_length = 10000
10
11 upperbound_primes = 12
12
13 # Computing a list of prime numbers up to n
14 def primes(n):
15 sieve = [True] * n
16 for i in xrange(3,int(n**0.5)+1,2):
17 if sieve[i]:
18 sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
19 return [2] + [i for i in xrange(3,n,2) if sieve[i]]
20
21 # The list of prime numbers up to upperbound_primes
22 p = primes(upperbound_primes)
23
24 # The amount of primes numbers
25 no_primes = len(p)
26
27 # Generate the sequence for the prime number p
28 def sequence(p):
29 x = np.empty(seq_length)
30 x[0] = 1
31 for i in range(1,seq_length):
32 x[i] = p - 1 - (p * (i-1) - 1) % x[i-1]
33 return x
34
35 # List with the sequences.
36 seq = [sequence(i) for i in p]
37
38 # Autocorrelation function.
39 def autocor(x):
40 result = np.correlate(x,x,mode='full')
41 return result[result.size/2:]
42
43 fig = plt.figure("The sequences next to their autocorrelation")
44 plt.suptitle("The sequences next to their autocorrelation")
45
46 # Proper spacing between subplots.
47 fig.subplots_adjust(hspace=1.2)
48
49 # Set up pyplot to use TeX.
50 plt.rc('text',usetex=True)
51 plt.rc('font',family='serif')
52
53 # Maximize plot window by command.
54 mng = plt.get_current_fig_manager()
55 mng.resize(*mng.window.maxsize())
56
57 k = 0
58 for s in seq:
59 k = k + 1
60 fig.add_subplot(no_primes,2,2*(k-1)+1)
61 plt.title("Sequence of the prime %d" % p[k-1])
62 plt.plot(s)
63 plt.xlabel(r"Index $i$")
64 plt.ylabel(r"Sequence number $x_i$")
65 plt.xlim(0,100)
66
67 # Constrain the number of ticks on the y-axis, for clarity.
68 plt.locator_params(axis='y',nbins=4)
69
70 fig.add_subplot(no_primes,2,2*k)
71 plt.title(r"Autocorrelation of the sequence $^{%d}x$" % p[k-1])
72 plt.plot(autocor(s))
73 plt.xlabel(r"Index $i$")
74 plt.xticks
75 plt.ylabel("Autocorrelation")
76
77 # Proper scaling of the y-axis.
78 ymin = autocor(s)[1]-int(autocor(s)[1]/10)
79 ymax = autocor(s)[1]+int(autocor(s)[1]/10)
80 plt.ylim(ymin,ymax)
81 plt.xlim(0,500)
82
83 plt.locator_params(axis='y',nbins=4)
84
85 # Use scientific notation when 0< t < 1 or t > 10
86 plt.ticklabel_format(style='sci',axis='y',scilimits=(0,1))
87
88 plt.show()
When using FFT, we Fourier transform our sequence and look for the first peak. The coordinate of this first peak, gives the frequency which represents our signal the coarsest. This will give our period since the coarsest frequency is the frequency by which our sequence (ideally) oscillates.
See the following plot of the sequences next to their Fourier transforms.
Code:
1 # Plotting sequences satisfying, x_{i+1} = p-1 - (p*i-1 mod x_i)
2 # with p prime and x_0 = 1, next to their Fourier transforms.
3
4 from __future__ import print_function
5 import numpy as np
6 from matplotlib import pyplot as plt
7
8 # The length of the sequences.
9 seq_length = 10000
10
11 upperbound_primes = 12
12
13 # Computing a list of prime numbers up to n
14 def primes(n):
15 sieve = [True] * n
16 for i in xrange(3,int(n**0.5)+1,2):
17 if sieve[i]:
18 sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
19 return [2] + [i for i in xrange(3,n,2) if sieve[i]]
20
21 # The list of prime numbers up to upperbound_primes
22 p = primes(upperbound_primes)
23
24 # The amount of primes numbers
25 no_primes = len(p)
26
27 # Generate the sequence for the prime number p
28 def sequence(p):
29 x = np.empty(seq_length)
30 x[0] = 1
31 for i in range(1,seq_length):
32 x[i] = p - 1 - (p * (i-1) - 1) % x[i-1]
33 return x
34
35 # List with the sequences.
36 seq = [sequence(i) for i in p]
37
38 fig = plt.figure("The sequences next to their FFT")
39 plt.suptitle("The sequences next to their FFT")
40
41 # Proper spacing between subplots.
42 fig.subplots_adjust(hspace=1.2)
43
44 # Set up pyplot to use TeX.
45 plt.rc('text',usetex=True)
46 plt.rc('font',family='serif')
47
48
49 # Maximize plot window by command.
50 mng = plt.get_current_fig_manager()
51 mng.resize(*mng.window.maxsize())
52
53 k = 0
54 for s in seq:
55 f = np.fft.rfft(s)
56 f[0] = 0
57 freq = np.fft.rfftfreq(seq_length)
58 k = k + 1
59 fig.add_subplot(no_primes,2,2*(k-1)+1)
60 plt.title("Sequence of the prime %d" % p[k-1])
61 plt.plot(s)
62 plt.xlabel(r"Index $i$")
63 plt.ylabel(r"Sequence number $x_i$")
64 plt.xlim(0,100)
65
66 # Constrain the number of ticks on the y-axis, for clarity.
67 plt.locator_params(nbins=4)
68
69 fig.add_subplot(no_primes,2,2*k)
70 plt.title(r"FFT of the sequence $^{%d}x$" % p[k-1])
71 plt.plot(freq,abs(f))
72 plt.xlabel("Frequency")
73 plt.ylabel("Amplitude")
74 plt.locator_params(nbins=4)
75
76 # Use scientific notation when 0 < t < 0 or t > 10
77 plt.ticklabel_format(style='sci',axis='y',scilimits=(0,1))
78
79 plt.show()
To see why the FFT method is more convenient then autocorrelation notice that we have a clear algorithm of determining the period: find the first peak of the Fourier transform. For a sufficient number of samples this always works.
See the following table, attained by the FFT method, which agrees with the autocorrelation method.
prime frequency period
2 0.00 1000.00
3 0.50 2.00
5 0.08 12.00
7 0.02 59.88
11 0.00 1000.00
The following code implements the algorithm, printing a table specifying the frequency and period of the sequences per prime number.
1 # Print a table of periods, determined by the FFT method,
2 # of sequences satisfying,
3 # x_{i+1} = p-1 - (p*i-1 mod x_i) with p prime and x_0 = 1.
4
5 from __future__ import print_function
6 import numpy as np
7 from matplotlib import pyplot as plt
8
9 # The length of the sequences.
10 seq_length = 10000
11
12 upperbound_primes = 12
13
14 # Computing a list of prime numbers up to n
15 def primes(n):
16 sieve = [True] * n
17 for i in xrange(3,int(n**0.5)+1,2):
18 if sieve[i]:
19 sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
20 return [2] + [i for i in xrange(3,n,2) if sieve[i]]
21
22 # The list of prime numbers up to upperbound_primes
23 p = primes(upperbound_primes)
24
25 # The amount of primes numbers
26 no_primes = len(p)
27
28 # Generate the sequence for the prime number p
29 def sequence(p):
30 x = np.empty(seq_length)
31 x[0] = 1
32 for i in range(1,seq_length):
33 x[i] = p - 1 - (p * (i-1) - 1) % x[i-1]
34 return x
35
36 # List with the sequences.
37 seq = [sequence(i) for i in p]
38
39 # Function that finds the first peak.
40 # Assumption: seq_length >> 10 so the Fourier transformed
41 # signal is sufficiently smooth.
42 def firstpeak(x):
43 for i in range(10,len(x)-1):
44 if x[i+1] < x[i]:
45 return i
46 return len(x)-1
47
48 k = 0
49 for s in seq:
50 f = np.fft.rfft(s)
51 freq = np.fft.rfftfreq(seq_length)
52 k = k + 1
53 if k == 1:
54 print("prime \t frequency \t period")
55 print(p[k-1],'\t %.2f' % float(freq[firstpeak(abs(f))]), \
56 '\t\t %.2f' % float(1/freq[firstpeak(abs(f))]))
I used 10000 samples (seq_length) in all the above code. As we increase the number of samples the periods are seen to converge to a certain integral value (Using the FFT method).
The FFT method seems to me like an ideal tool to determine periods in algorithmic signals, one only being limited by how high a sample number your equipment can handle.
Related
#Print the user board
s = ''
for i in range(n):
for j in range(n):
z = i * n + j
s += ' '
if z < 10:
s += ' '
s += str(z)
s += '\n'
print(dedent(s))
queens = list(map(int, input("Queens: ").split()))
I keep getting an error from my testcase environment of a last blank line before proceeding to the queens input below. What can I approach to fix
I have tried cleandoc from textwrap and while it works, it disrupts every required spacing and new distinctive lines for z from the "s" string which is a perfect 8 by 8 from 0 to 63.
Creating the right way instead of removing things after is easier:
from itertools import count
import toolz
n = 8
start_value = 0
counter = count(start_value)
rows = [" ".join(f"{num:2}" for num in toolz.take(n, counter)) for _ in range(n)]
print("\n".join(rows))
The itertools.count creates an iterable counter that gives the next value every time you want. Here the take function will get 8 items each time.
If you do not want the initial space, you'd better use join, which only adds the separator between consecutive items:
s = '\n'.join(' '.join('{:2d}'.format(i * n + j)
for j in range(n))
for i in range(n))
print(s)
gives as expected:
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 50 51 52 53 54 55
56 57 58 59 60 61 62 63
Being struggle with this for a while but how am I able to skip for loops in python by X and continue for Y then repeat.
For example:
# This loop should loop until 99, but after it reaches a number that is a multiple of 10 e.g. 10,20,30 etc it should continue for 5 iterations, then skip to next multiple of 10.
# E.g 0,1,2,3,4,5,10,11,12,13,14,15,20,21,22,23,24,25,30...etc
for page_num in range(100):
# Use page_num however
Modify the loop to use a step of 10, and add a sub-loop to have iterations that break off after the 5th element.
for j in range(0,100,10):
for page_num in range(j, j+6):
# use page_num
Use continue to skip the rest of the loop if the digit in units place is greater than 5.
skip_after = 5
for page_num in range(100):
if page_num % 10 > skip_after: continue
# ... Do the rest of your loop
print(page_num)
page_num % 10 uses the modulo % operator to give the remainder from dividing page_num by 10 (which is the digit in the units place).
Output (joined into a single line for readability):
0 1 2 3 4 5 10 11 12 13 14 15 20 21 22 23 24 25 30
Using itertools.filterfalse
from itertools import filterfalse
for x in (filterfalse(lambda x: (x % 10) > 5, range(0, 100))):
print(x, end=' ')
Output:
0 1 2 3 4 5 10 11 12 13 14 15 20 21 22 23 24 25 30 31 32 33 34 35 40 41 42 43 44 45 50 51 52 53 54 55 60 61 62 63 64 65 70 71 72 73 74 75 80 81 82 83 84 85 90 91 92 93 94 95
I would recommend list comprehension. Im not positive if this is the complete answer but I would do it as:
def find_divisors_2(page_num):
"""
You can insert an algorithmic phrase after if and it will produce the answer.
"""
find_divisors_2 = [x for x in range(100) if (Insert your algorithm)]
return find_divisors_2
You can directly modify the value of page_num inside the for loop with a certain condition:
for page_num in range(100):
if str(page_num)[-1]=="6":
page_num += 4
print(page_num)
I have a Pandas DataFrame with the following hypothetical data:
ID Time X-coord Y-coord
0 1 5 68 5
1 2 8 72 78
2 3 1 15 23
3 4 4 81 59
4 5 9 78 99
5 6 12 55 12
6 7 5 85 14
7 8 7 58 17
8 9 13 91 47
9 10 10 29 87
For each row (or ID), I want to find the ID with the closest proximity in time and space (X & Y) within this dataframe. Bonus: Time should have priority over XY.
Ideally, in the end I would like to have a new column called "Closest_ID" containing the most proximal ID within the dataframe.
I'm having trouble coming up with a function for this.
I would really appreciate any help or hint that points me in the right direction!
Thanks a lot!
Let's denote df as our dataframe. Then you can do something like:
from sklearn.metrics import pairwise_distances
space_vals = df[['X-coord', 'Y-coord']]
time_vals =df['Time']
space_distance = pairwise_distance(space_vals)
time_distance = pairwise_distance(time_vals)
space_distance[space_distance == 0] = 1e9 # arbitrary large number
time_distance[time_distance == 0] = 1e9 # again
closest_space_id = np.argmin(space_distance, axis=0)
closest_time_id = np.argmin(time_distance, axis=0)
Then, you can store the last 2 results in 2 columns, or somehow decide which one is closer.
Note: this code hasn't been checked, and it might have a few bugs...
def tablesOneToTen(): # a function that will print out multiplication tables from 1-10
x = 1
y = 1
while x <= 10 and y <= 12:
f = x * y
print(f)
y = y + 1
x = x + 1
tablesOneToTen()
I am trying to make a function that will give me values from the multiplication table from 1-10.
Should I add if and elif statements in addition to nested while loops to make this code work?
For these sort of iteration tasks you're better off using the for loop since you already know the boundaries you're working with, also Python makes creating for loops especially easy.
With while loops you have to check that you are in range using conditionals while also explicitly incrementing your counters making mistakes all the more likely.
Since you know you need multiplication tables for values of x and y ranging from 1-10 you can, to get you familiar with loops, create two for loops:
def tablesOneToTen(): # a function that will print out multiplication tables from 1-10
# This will iterate with values for x in the range [1-10]
for x in range(1, 11):
# Print the value of x for reference
print("Table for {} * (1 - 10)".format(x))
# iterate for values of y in a range [1-10]
for y in range(1, 11):
# Print the result of the multiplication
print(x * y, end=" ")
# Print a new Line.
print()
Running this will give you the tables you need:
Table for 1 * (1 - 10)
1 2 3 4 5 6 7 8 9 10
Table for 2 * (1 - 10)
2 4 6 8 10 12 14 16 18 20
Table for 3 * (1 - 10)
3 6 9 12 15 18 21 24 27 30
With a while loop, the logic is similar but of course just more verbose than it need to since you must initialize, evaluate the condition and increment.
As a testament to its uglyness, the while loop would look something like this:
def tablesOneToTen():
# initialize x counter
x = 1
# first condition
while x <= 10:
# print reference message
print("Table for {} * [1-10]".format(x))
# initialize y counter
y = 1
# second condition
while y <=10:
# print values
print(x*y, end=" ")
# increment y
y += 1
# print a new line
print(" ")
# increment x
x += 1
Using Python 3
for i in range(1, 10+1):
for j in range(i, (i*10)+1):
if (j % i == 0):
print(j, end="\t")
print()
or:
for i in range(1, 10+1):
for j in range(i, (i*10)+1, i):
print(j, end="\t")
print()
Output:
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
6 12 18 24 30 36 42 48 54 60
7 14 21 28 35 42 49 56 63 70
8 16 24 32 40 48 56 64 72 80
9 18 27 36 45 54 63 72 81 90
10 20 30 40 50 60 70 80 90 100
Hope it would help you to get 1 to 10 table.
a = [1,2,3,4,5,6,7,8,9,10]
for i in a:
print(*("{:3}" .format (i*col) for col in a))
print()
Here is my code.
#Prime numbers between 0-100
for i in range(2,100):
flg=0
for j in range(2,int(i/2)):
if i%j==0:
flg=1
break
if flg!=1:
print(i)
And the output is
2
3
4 <-
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
I have no idea why there is this 4.
Pardon me if i made some noobish mistake, as i can't seem to figure it out.
The reason is that range is not inclusive, i.e.
>>> range(2,2)
[]
So when you access 4, you don't check for divisors. Change for example to range(2,int(i/2)+1)
To speed up your calculation, you can use math.sqrt instead of /2 operation, for example as:
import math
and then
for j in range(2, int(math.sqrt(i)+1)):