IndexError: list assignment index out of range , Python - python

I am trying to achieve functionality. It's working should be this way:
It takes two lists.
Mark some indexes, preferably center few.
Both parents switches marked indexes.
Other indexes go sequentially to their parent element.
If the same element is already present in that parent, it maps and check where other parent same element was and goes there.
import random
def pm(indA, indB):
size = min(len(indA), len(indB))
c1, c2 = [0] * size, [0] * size
# Initialize the position of each indices in the individuals
for i in range(1,size):
c1[indA[i]] = i
c2[indB[i]] = i
crosspoint1 = random.randint(0, size)
crosspoint2 = random.randint(0, size - 1)
if crosspoint2 >= crosspoint1:
crosspoint2 += 1
else: # Swap the two cx points
crosspoint1, crosspointt2 = crosspoint2, crosspoint1
for i in range(crosspoint1, crosspoint2):
# Keep track of the selected values
temp1 = indA[i]
temp2 = indB[i]
# Swap the matched value
indA[i], indA[c1[temp2]] = temp2, temp1
indB[i], indB[c2[temp1]] = temp1, temp2
# Position bookkeeping
c1[temp1], c1[temp2] = c1[temp2], c1[temp1]
c2[temp1], c2[temp2] = c2[temp2], c2[temp1]
return indA, indB
a,b = pm([3, 4, 8, 2, 7, 1, 6, 5],[4, 2, 5, 1, 6, 8, 3, 7])
Error:
in pm
c1[indA[i]] = i
IndexError: list assignment index out of range

Not sure whether there are other errors in your code (I didn't run it), but here's the explanation for this one. In Python (as most of other languages), lists (sequences to be more precise) index is 0 based:
>>> l = [1, 2, 3, 4, 5, 6]
>>>
>>> for e in l:
... print(e, l.index(e))
...
1 0
2 1
3 2
4 3
5 4
6 5
>>>
>>> l[0]
1
>>> l[5]
6
>>> l[6]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: list index out of range
To summarize your problem:
Your indA and indB lists have each 6 elements ([1..6]), and their indexes: [0..5]
Your c1 and c2 lists also have 6 elements (indexes also [0..5])
But, your using values from #1. as indexes in lists from #2., and the value 6 is a problem, as there's no such index
To fix your problem, you should use valid index values. Either:
Have the proper values in indA and indB (this is the one I'd chose):
a, b = pmxCrossover([0, 3, 1, 2, 5, 4], [4, 0, 2, 3, 5, 1])
Subtract 1, wherever you encounter values from indA or indB used as indexes:
c1[indA[i] - 1] = i
As a general advice: whenever you encounter errors, add print statements before the faulty line (printing (partial) stuff from it), and that might give you clues that could lead to solving the problem yourself.
#EDIT0
Posting (a slightly modified version of) the original code, with the index conversion:
Before the algorithm: subtract 1 (from each element) to have valid indexes
After the algorithm: add 1 to come back to 1 based indexes
code00.py:
#!/usr/bin/env python3
import sys
import random
def pmx_crossover(ind_a, ind_b):
size = min(len(ind_a), len(ind_b))
c1, c2 = [0] * size, [0] * size
# Initialize the position of each indices in the individuals
for i in range(1, size):
c1[ind_a[i]] = i
c2[ind_b[i]] = i
# Choose crossover points
crosspoint1 = random.randint(0, size)
crosspoint2 = random.randint(0, size - 1)
if crosspoint2 >= crosspoint1:
crosspoint2 += 1
else: # Swap the two cx points
crosspoint1, crosspointt2 = crosspoint2, crosspoint1
# Apply crossover between cx points
for i in range(crosspoint1, crosspoint2):
# Keep track of the selected values
temp1 = ind_a[i]
temp2 = ind_b[i]
# Swap the matched value
ind_a[i], ind_a[c1[temp2]] = temp2, temp1
ind_b[i], ind_b[c2[temp1]] = temp1, temp2
# Position bookkeeping
c1[temp1], c1[temp2] = c1[temp2], c1[temp1]
c2[temp1], c2[temp2] = c2[temp2], c2[temp1]
return ind_a, ind_b
def main():
#initial_a, initial_b = [1, 2, 3, 4, 5, 6, 7, 8], [3, 7, 5, 1, 6, 8, 2, 4]
initial_a, initial_b = [1, 4, 2, 3, 6, 5], [5, 1, 3, 4, 6, 2]
index_offset = 1
temp_a = [i - index_offset for i in initial_a]
temp_b = [i - index_offset for i in initial_b]
a, b = pmx_crossover(temp_a, temp_b)
final_a = [i + index_offset for i in a]
final_b = [i + index_offset for i in b]
print("Initial: {0:}, {1:}".format(initial_a, initial_b))
print("Final: {0:}, {1:}".format(final_a, final_b))
if __name__ == "__main__":
print("Python {0:s} {1:d}bit on {2:s}\n".format(" ".join(item.strip() for item in sys.version.split("\n")), 64 if sys.maxsize > 0x100000000 else 32, sys.platform))
main()
print("\nDone.")
Output (one of the possibilities (due to random.randint)):
[cfati#CFATI-5510-0:e:\Work\Dev\StackOverflow\q058424002]> "e:\Work\Dev\VEnvs\py_064_03.07.03_test0\Scripts\python.exe" code00.py
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 22:22:05) [MSC v.1916 64 bit (AMD64)] 64bit on win32
Initial: [1, 4, 2, 3, 6, 5], [5, 1, 3, 4, 6, 2]
Final: [1, 3, 2, 4, 6, 5], [5, 1, 4, 3, 6, 2]
Done.

c1 is out of range because in your for at the fourth index the value of indA[4] is 6.
And the range of c1 index it's 0-5 (it's lengh is 6).
With c1[indA[i]] = i
you try to do c1[6] = 4

Related

Index array based on value limits of another

Let's say I have an array (or even a list) that looks like:
tmp_data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
And then I have another ray that are distance values:
dist_data = [ 15.625 46.875 78.125 109.375 140.625 171.875 203.125 234.375 265.625 296.875]
Now, say I want to create a threshold of distance that I would like to perform an operation on from tmp_data. For this example, let's just take the max value. And let's set the threshold distance to 100. What I would like to do is take the n number of elements every 100 distance units and replace all elements in that with the maximum value in that small array. For example: I would want the final output to be
max_tmp_data_100 = [2,2,2,5,5,5,8,8,8,9]
This is because the first 3 elements in dist_data are below 100, so we take the first three elements of tmp_data (0,1,2), and get the maximum of this and replace all elements in there with that value, 2
Then, the next set of data that would be below the next 100 value would be
tmp_dist_array_100 = [109.375 140.625 171.875]
tmp_data_100 = [3,4,5]
max_tmp_data_100 = [5,5,5]
(append to [2,2,2])
I have come up with the following:
# Initialize
final_array = []
d_array = []
idx = 1
for i in range(0,10):
if dist_data[i] < idx * final_res:
d_array.append(tmp_data[i])
elif dist_data[i] > idx * final_res:
# Now get the values
max_val = np.amax(d_array)
new_array = np.ones(len(d_array)) * max_val
final_array.extend(new_array)
idx = idx + 1
But the outcome is
[2.0, 2.0, 2.0, 5.0, 5.0, 5.0, 5.0, 5.0]
When it should be [2,2,2,5,5,5,8,8,8,9]
With numpy:
import numpy as np
cdist_data = [15.625, 46.875, 78.125, 109.375, 140.625, 171.875, 203.125, 234.375,265.625, 296.875]
cut = 100
a = np.array(dist_data)
vals = np.searchsorted(a, np.r_[cut:a.max() + cut:cut]) - 1
print(vals[(a/cut).astype(int)])
It gives:
[2 2 2 5 5 5 9 9 9 9]
You can do with groupby
from itertools import groupby
dist_data = [ 15.625, 46.875 ,78.125 ,109.375 ,140.625 ,171.875 ,203.125 ,234.375, 265.625 ,296.875]
tmp_data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
result = []
index_list = [[dist_data.index(i) for i in l]
for k, l in groupby(dist_data, key=lambda x:x//100)]
for i in tmp_data:
for lst in index_list:
if i in lst:
result.append(max(lst))
print(result)
# [2, 2, 2, 5, 5, 5, 9, 9, 9, 9]
A per your requirements last 4 elements will comes under next threshold value, the max of last 4 element is 9.

Add padding based on partial sum

I have four given variables:
group size
total of groups
partial sum
1-D tensor
and I want to add zeros when the sum within a group reached the partial sum. For example:
groupsize = 4
totalgroups = 3
partialsum = 15
d1tensor = torch.tensor([ 3, 12, 5, 5, 5, 4, 11])
The expected result is:
[ 3, 12, 0, 0, 5, 5, 5, 0, 4, 11, 0, 0]
I have no clue how can I achieve that in pure pytorch. In python it would be something like this:
target = [0]*(groupsize*totalgroups)
cursor = 0
current_count = 0
d1tensor = [ 3, 12, 5, 5, 5, 4, 11]
for idx, ele in enumerate(target):
subgroup_start = (idx//groupsize) *groupsize
subgroup_end = subgroup_start + groupsize
if sum(target[subgroup_start:subgroup_end]) < partialsum:
target[idx] = d1tensor[cursor]
cursor +=1
Can anyone help me with that? I have already googled it but couldn't find anything.
Some logic, Numpy and list comprehensions are sufficient here.
I will break it down step by step, you can make it slimmer and prettier afterwards:
import numpy as np
my_val = 15
block_size = 4
total_groups = 3
d1 = [3, 12, 5, 5, 5, 4, 11]
d2 = np.cumsum(d1)
d3 = d2 % my_val == 0 #find where sum of elements is 15 or multiple
split_points= [i+1 for i, x in enumerate(d3) if x] # find index where cumsum == my_val
#### Option 1
split_array = np.split(d1, split_points, axis=0)
padded_arrays = [np.pad(array, (0, block_size - len(array)), mode='constant') for array in split_array] #pad arrays
padded_d1 = np.concatenate(padded_arrays[:total_groups]) #put them together, discard extra group if present
#### Option 2
split_points = [el for el in split_points if el <len(d1)] #make sure we are not splitting on the last element of d1
split_array = np.split(d1, split_points, axis=0)
padded_arrays = [np.pad(array, (0, block_size - len(array)), mode='constant') for array in split_array] #pad arrays
padded_d1 = np.concatenate(padded_arrays)

Python Recursion in pascal's triangle [duplicate]

As a learning experience for Python, I am trying to code my own version of Pascal's triangle. It took me a few hours (as I am just starting), but I came out with this code:
pascals_triangle = []
def blank_list_gen(x):
while len(pascals_triangle) < x:
pascals_triangle.append([0])
def pascals_tri_gen(rows):
blank_list_gen(rows)
for element in range(rows):
count = 1
while count < rows - element:
pascals_triangle[count + element].append(0)
count += 1
for row in pascals_triangle:
row.insert(0, 1)
row.append(1)
pascals_triangle.insert(0, [1, 1])
pascals_triangle.insert(0, [1])
pascals_tri_gen(6)
for row in pascals_triangle:
print(row)
which returns
[1]
[1, 1]
[1, 0, 1]
[1, 0, 0, 1]
[1, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 0, 0, 1]
However, I have no idea where to go from here. I have been banging my head against the wall for hours. I want to emphasize that I do NOT want you to do it for me; just push me in the right direction. As a list, my code returns
[[1], [1, 1], [1, 0, 1], [1, 0, 0, 1], [1, 0, 0, 0, 1], [1, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1]]
Thanks.
EDIT: I took some good advice, and I completely rewrote my code, but I am now running into another problem. Here is my code.
import math
pascals_tri_formula = []
def combination(n, r):
return int((math.factorial(n)) / ((math.factorial(r)) * math.factorial(n - r)))
def for_test(x, y):
for y in range(x):
return combination(x, y)
def pascals_triangle(rows):
count = 0
while count <= rows:
for element in range(count + 1):
[pascals_tri_formula.append(combination(count, element))]
count += 1
pascals_triangle(3)
print(pascals_tri_formula)
However, I am finding that the output is a bit undesirable:
[1, 1, 1, 1, 2, 1, 1, 3, 3, 1]
How can I fix this?
OK code review:
import math
# pascals_tri_formula = [] # don't collect in a global variable.
def combination(n, r): # correct calculation of combinations, n choose k
return int((math.factorial(n)) / ((math.factorial(r)) * math.factorial(n - r)))
def for_test(x, y): # don't see where this is being used...
for y in range(x):
return combination(x, y)
def pascals_triangle(rows):
result = [] # need something to collect our results in
# count = 0 # avoidable! better to use a for loop,
# while count <= rows: # can avoid initializing and incrementing
for count in range(rows): # start at 0, up to but not including rows number.
# this is really where you went wrong:
row = [] # need a row element to collect the row in
for element in range(count + 1):
# putting this in a list doesn't do anything.
# [pascals_tri_formula.append(combination(count, element))]
row.append(combination(count, element))
result.append(row)
# count += 1 # avoidable
return result
# now we can print a result:
for row in pascals_triangle(3):
print(row)
prints:
[1]
[1, 1]
[1, 2, 1]
Explanation of Pascal's triangle:
This is the formula for "n choose k" (i.e. how many different ways (disregarding order), from an ordered list of n items, can we choose k items):
from math import factorial
def combination(n, k):
"""n choose k, returns int"""
return int((factorial(n)) / ((factorial(k)) * factorial(n - k)))
A commenter asked if this is related to itertools.combinations - indeed it is. "n choose k" can be calculated by taking the length of a list of elements from combinations:
from itertools import combinations
def pascals_triangle_cell(n, k):
"""n choose k, returns int"""
result = len(list(combinations(range(n), k)))
# our result is equal to that returned by the other combination calculation:
assert result == combination(n, k)
return result
Let's see this demonstrated:
from pprint import pprint
ptc = pascals_triangle_cell
>>> pprint([[ptc(0, 0),],
[ptc(1, 0), ptc(1, 1)],
[ptc(2, 0), ptc(2, 1), ptc(2, 2)],
[ptc(3, 0), ptc(3, 1), ptc(3, 2), ptc(3, 3)],
[ptc(4, 0), ptc(4, 1), ptc(4, 2), ptc(4, 3), ptc(4, 4)]],
width = 20)
[[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1]]
We can avoid repeating ourselves with a nested list comprehension:
def pascals_triangle(rows):
return [[ptc(row, k) for k in range(row + 1)] for row in range(rows)]
>>> pprint(pascals_triangle(15))
[[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1],
[1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1],
[1, 7, 21, 35, 35, 21, 7, 1],
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1],
[1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1],
[1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1],
[1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1],
[1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1],
[1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]
Recursively defined:
We can define this recursively (a less efficient, but perhaps more mathematically elegant definition) using the relationships illustrated by the triangle:
def choose(n, k): # note no dependencies on any of the prior code
if k in (0, n):
return 1
return choose(n-1, k-1) + choose(n-1, k)
And for fun, you can see each row take progressively longer to execute, because each row has to recompute nearly each element from the prior row twice each time:
for row in range(40):
for k in range(row + 1):
# flush is a Python 3 only argument, you can leave it out,
# but it lets us see each element print as it finishes calculating
print(choose(row, k), end=' ', flush=True)
print()
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 ...
Ctrl-C to quit when you get tired of watching it, it gets very slow very fast...
I know you want to implement yourself, but the best way for me to explain is to walk through an implementation. Here's how I would do it, and this implementation relies on my fairly complete knowledge of how Python's functions work, so you probably won't want to use this code yourself, but it may get you pointed in the right direction.
def pascals_triangle(n_rows):
results = [] # a container to collect the rows
for _ in range(n_rows):
row = [1] # a starter 1 in the row
if results: # then we're in the second row or beyond
last_row = results[-1] # reference the previous row
# this is the complicated part, it relies on the fact that zip
# stops at the shortest iterable, so for the second row, we have
# nothing in this list comprension, but the third row sums 1 and 1
# and the fourth row sums in pairs. It's a sliding window.
row.extend([sum(pair) for pair in zip(last_row, last_row[1:])])
# finally append the final 1 to the outside
row.append(1)
results.append(row) # add the row to the results.
return results
usage:
>>> for i in pascals_triangle(6):
... print(i)
...
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
Without using zip, but using generator:
def gen(n,r=[]):
for x in range(n):
l = len(r)
r = [1 if i == 0 or i == l else r[i-1]+r[i] for i in range(l+1)]
yield r
example:
print(list(gen(15)))
output:
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1], [1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56, 28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1], [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]
DISPLAY AS TRIANGLE
To draw it in beautiful triangle(works only for n < 7, beyond that it gets distroted. ref draw_beautiful for n>7)
for n < 7
def draw(n):
for p in gen(n):
print(' '.join(map(str,p)).center(n*2)+'\n')
eg:
draw(10)
output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
for any size
since we need to know the max width, we can't make use of generator
def draw_beautiful(n):
ps = list(gen(n))
max = len(' '.join(map(str,ps[-1])))
for p in ps:
print(' '.join(map(str,p)).center(max)+'\n')
example (2) :
works for any number:
draw_beautiful(100)
Here is my attempt:
def generate_pascal_triangle(rows):
if rows == 1: return [[1]]
triangle = [[1], [1, 1]] # pre-populate with the first two rows
row = [1, 1] # Starts with the second row and calculate the next
for i in range(2, rows):
row = [1] + [sum(column) for column in zip(row[1:], row)] + [1]
triangle.append(row)
return triangle
for row in generate_pascal_triangle(6):
print row
Discussion
The first two rows of the triangle is hard-coded
The zip() call basically pairs two adjacent numbers together
We still have to add 1 to the beginning and another 1 to the end because the zip() call only generates the middle of the next row
# combining the insights from Aaron Hall and Hai Vu,
# we get:
def pastri(n):
rows = [[1]]
for _ in range(1, n+1):
rows.append([1] +
[sum(pair) for pair in zip(rows[-1], rows[-1][1:])] +
[1])
return rows
# thanks! learnt that "shape shifting" data,
# can yield/generate elegant solutions.
def pascal(n):
if n==0:
return [1]
else:
N = pascal(n-1)
return [1] + [N[i] + N[i+1] for i in range(n-1)] + [1]
def pascal_triangle(n):
for i in range(n):
print pascal(i)
Beginner Python student here. Here's my attempt at it, a very literal approach, using two For loops:
pascal = [[1]]
num = int(input("Number of iterations: "))
print(pascal[0]) # the very first row
for i in range(1,num+1):
pascal.append([1]) # start off with 1
for j in range(len(pascal[i-1])-1):
# the number of times we need to run this loop is (# of elements in the row above)-1
pascal[i].append(pascal[i-1][j]+pascal[i-1][j+1])
# add two adjacent numbers of the row above together
pascal[i].append(1) # and cap it with 1
print(pascal[i])
Here is an elegant and efficient recursive solution. I'm using the very handy toolz library.
from toolz import memoize, sliding_window
#memoize
def pascals_triangle(n):
"""Returns the n'th row of Pascal's triangle."""
if n == 0:
return [1]
prev_row = pascals_triangle(n-1)
return [1, *map(sum, sliding_window(2, prev_row)), 1]
pascals_triangle(300) takes about 15 ms on a macbook pro (2.9 GHz Intel Core i5). Note that you can't go much higher without increasing the default recursion depth limit.
I am cheating from the popular fibonacci sequence solution. To me, the implementation of Pascal's triangle would have the same concept of fibonacci's. In fibonacci we use a single number at a time and add it up to the previous one. In pascal's triangle use a row at a time and add it up to the previous one.
Here is a complete code example:
>>> def pascal(n):
... r1, r2 = [1], [1, 1]
... degree = 1
... while degree <= n:
... print(r1)
... r1, r2 = r2, [1] + [sum(pair) for pair in zip(r2, r2[1:]) ] + [1]
... degree += 1
Test
>>> pascal(3)
[1]
[1, 1]
[1, 2, 1]
>>> pascal(4)
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
>>> pascal(6)
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
Note: to have the result as a generator, change print(r1) to yield r1.
# call the function ! Indent properly , everything should be inside the function
def triangle():
matrix=[[0 for i in range(0,20)]for e in range(0,10)] # This method assigns 0's to all Rows and Columns , the range is mentioned
div=20/2 # it give us the most middle columns
matrix[0][div]=1 # assigning 1 to the middle of first row
for i in range(1,len(matrix)-1): # it goes column by column
for j in range(1,20-1): # this loop goes row by row
matrix[i][j]=matrix[i-1][j-1]+matrix[i-1][j+1] # this is the formula , first element of the matrix gets , addition of i index (which is 0 at first ) with third value on the the related row
# replacing 0s with spaces :)
for i in range(0,len(matrix)):
for j in range(0,20):
if matrix[i][j]==0: # Replacing 0's with spaces
matrix[i][j]=" "
for i in range(0,len(matrix)-1): # using spaces , the triangle will printed beautifully
for j in range(0,20):
print 1*" ",matrix[i][j],1*" ", # giving some spaces in two sides of the printing numbers
triangle() # calling the function
would print something like this
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Here is a simple way of implementing the pascal triangle:
def pascal_triangle(n):
myList = []
trow = [1]
y = [0]
for x in range(max(n,0)):
myList.append(trow)
trow=[l+r for l,r in zip(trow+y, y+trow)]
for item in myList:
print(item)
pascal_triangle(5)
Python zip() function returns the zip object, which is the iterator of tuples where the first item in each passed iterator is paired together, and then the second item in each passed iterator are paired together. Python zip is the container that holds real data inside.
Python zip() function takes iterables (can be zero or more), makes an iterator that aggregates items based on the iterables passed, and returns the iterator of tuples.
I did this when i was working with my son on intro python piece. It started off as rather simple piece, when we targeted -
1
1 2
1 2 3
1 2 3 4
However, as soon as we hit the actual algorithm, complexity overshot our expectations. Anyway, we did build this -
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Used some recursion -
def genRow(row:list) :
# print(f"generatig new row below {row}")
# printRow(row)
l = len(row) #2
newRow : list = []
i = 0
# go through the incoming list
while i <= l:
# print(f"working with i = {i}")
# append an element in the new list
newRow.append(1)
# set first element of the new row to 1
if i ==0:
newRow[i] = 1
# print(f"1:: newRow = {newRow}")
# if the element is in the middle somewhere, add the surroundng two elements in
# previous row to get the new element
# e.g. row 3[2] = row2[1] + row2[2]
elif i <= l-1:
# print(f"2:: newRow = {newRow}")
newRow[i] = row[i-1] + row[i]
else:
# print(f"3 :: newRow = {newRow}")
newRow[i] = 1
i+=1
# print(newRow)
return newRow
def printRow(mx : int, row:list):
n = len(row)
spaces = ' ' *((mx - n)*2)
print(spaces,end=' ')
for i in row:
print(str(i) + ' ',end = ' ')
print(' ')
r = [1,1]
mx = 7
printRow(mx,[1])
printRow(mx,r)
for a in range(1,mx-1):
# print(f"working for Row = {a}")
if len(r) <= 2:
a1 = genRow(r)
r=a1
else:
a2 = genRow(a1)
a1 = a2
printRow(mx,a1)
Hopefully it helps.

Adding to a specific variable in a list

I'm trying to add two lists. If the last variable is greater than 10, it needs to carry over to the previous variable in the list. For example :
1 / 2 / 3 (List 1)
7 / 8 / 9 (List 2)
Should equal
9 / 1 / 2 not 8/10/12
So far, I have
list1 = [1, 2, 3]
list2 = [7, 8, 9]
SumOfLists = [x+y for x,y in zip(list1, list2)]
That adds the lists together, but I'm not sure how to make the number carry over.
You can try this code.
list1 = [1, 2, 3]
list2 = [7, 8, 9]
def add_list(a,b):
carry = 0
res_list = []
for i,j in zip(a[::-1],b[::-1]): # Iterate through the lists in reverse
val = (i+j+carry)%10 # Store the sum in val
carry = (i+j+carry)//10 # Store the carry
res_list.append(val) # Append to the returning list
return res_list[::-1] # Return the list
print add_list(list1,list2)
Wil print
[9, 1, 2]
Algorithm
Loop through each of the values in reverse. Add each corresponding values. If the values are above 10 then find the exceeding value and put it to carry. Finally return the reverse of the list.
list1 = [1, 2, 3]
list2 = [7, 8, 9]
cur = 0 # num to carry over
result = []
for x,y in zip(reversed(list2),reversed(list1)):
if x + y + cur > 10: # if sum greater than 10, remember to add 1 on
t = x+y + cur # the next loop
d = str(t)[1] # get the rightmost digit
result.append(int(d))
cur = 1
else: # nothing to curry over, but still add cur,
# it may be 1
result.append(x+y+cur)
cur = 0
print(list(reversed(result)) )
[9, 1, 2]
just subtract 10 if it's more then 10 and add 1 to it's previous element. Do this proccess for all element in sum list
if SumOfLists[2] >= 10:
SumOfLists[2] -= 10
SumOfLists[1] += 1
And at last check
if SumOfLists[0] >= 10:
for i in range(len(SumOfLists)-1,0,-1):
SumOfLists[i] = SumOfLists[i-1]
SumOfLists[0] = 1

Python sort array by another positions array

Assume I have two arrays, the first one containing int data, the second one containing positions
a = [11, 22, 44, 55]
b = [0, 1, 10, 11]
i.e. I want a[i] to be be moved to position b[i] for all i. If I haven't specified a position, then insert a -1
i.e
sorted_a = [11, 22,-1,-1,-1,-1,-1,-1,-1,-1, 44, 55]
^ ^ ^ ^
0 1 10 11
Another example:
a = [int1, int2, int3]
b = [5, 3, 1]
sorted_a = [-1, int3, -1, int2, -1, int1]
Here's what I've tried:
def sort_array_by_second(a, b):
sorted = []
for e1 in a:
sorted.appendAt(b[e1])
return sorted
Which I've obviously messed up.
Something like this:
res = [-1]*(max(b)+1) # create a list of required size with only -1's
for i, v in zip(b, a):
res[i] = v
The idea behind the algorithm:
Create the resulting list with a size capable of holding up to the largest index in b
Populate this list with -1
Iterate through b elements
Set elements in res[b[i]] with its proper value a[i]
This will leave the resulting list with -1 in every position other than the indexes contained in b, which will have their corresponding value of a.
I would use a custom key function as an argument to sort. This will sort the values according to the corresponding value in the other list:
to_be_sorted = ['int1', 'int2', 'int3', 'int4', 'int5']
sort_keys = [4, 5, 1, 2, 3]
sort_key_dict = dict(zip(to_be_sorted, sort_keys))
to_be_sorted.sort(key = lambda x: sort_key_dict[x])
This has the benefit of not counting on the values in sort_keys to be valid integer indexes, which is not a very stable thing to bank on.
>>> a = ["int1", "int2", "int3", "int4", "int5"]
>>> b = [4, 5, 1, 2, 3]
>>> sorted(a, key=lambda x, it=iter(sorted(b)): b.index(next(it)))
['int4', 'int5', 'int1', 'int2', 'int3']
Paulo Bu answer is the best pythonic way. If you want to stick with a function like yours:
def sort_array_by_second(a, b):
sorted = []
for n in b:
sorted.append(a[n-1])
return sorted
will do the trick.
Sorts A by the values of B:
A = ['int1', 'int2', 'int3', 'int4', 'int5']
B = [4, 5, 1, 2, 3]
from operator import itemgetter
C = [a for a, b in sorted(zip(A, B), key = itemgetter(1))]
print C
Output
['int3', 'int4', 'int5', 'int1', 'int2']
a = [11, 22, 44, 55] # values
b = [0, 1, 10, 11] # indexes to sort by
sorted_a = [-1] * (max(b) + 1)
for index, value in zip(b, a):
sorted_a[index] = value
print(sorted_a)
# -> [11, 22, -1, -1, -1, -1, -1, -1, -1, -1, 44, 55]

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