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How to add on the left and on the right order like 1 to 8

Hi I have small problem I dont know how to add oder like from 1 to number 8 on the right and on the left of this program.Here is the list but How to add numbers on the left and on the righ. I did this with letters up and down
Here is my code
sachy = [[0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0, 1, 0], [0, 1, 0, 1, 0, 1, 0, 1], [0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0], [0, 2, 0, 2, 0, 2, 0, 2], [2, 0, 2, 0, 2, 0, 2, 0], [0, 2, 0, 2, 0, 2, 0, 2]]
poradi = ["a", "b", "c", "d", "e", "f", "g", "h"]
poradi_2 = [1, 2, 3, 4, 5, 6, 7, 8]
for prvek in poradi:
print(prvek, end=" ")
print()
print()
for seznam in sachy:
for prvek in seznam:
print(prvek, end=" ")
print(end="\n", )
print()
for prvek in poradi:
print(prvek, end=" ")
I try to write another list of order from 1 to 8 into the seznam but will always multiple by 8 becaouse of sachy that are 8x8.

You need to pair the row indices with the row itself, also use " ".join() for shorted code
print(" ", " ".join(poradi), "\n")
for idx, seznam in zip(poradi_2, sachy):
print(idx, " ".join(map(str, seznam)), idx)
print("\n ", " ".join(poradi), "\n\n")
a b c d e f g h
1 0 1 0 1 0 1 0 1 1
2 1 0 1 0 1 0 1 0 2
3 0 1 0 1 0 1 0 1 3
4 0 0 0 0 0 0 0 0 4
5 0 0 0 0 0 0 0 0 5
6 0 2 0 2 0 2 0 2 6
7 2 0 2 0 2 0 2 0 7
8 0 2 0 2 0 2 0 2 8
a b c d e f g h

Creating a 2D array from a one line txt file in Python

I'm attempting to read a one line text file into an array in python, but I am struggling with actually getting the file to transform into an 2D array. This is the text file:
6 4 0 0 1 0 0 0 2 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 3 0
The first number (6) represents the columns and the second number (4) represents the rows. Here is the code I have so far:
maze_1d_arr = open(sys.argv[1], 'r')
maze = []
maze_split = np.array([maze_1d_arr])
size_X = len(maze_split)
size_Y = len(maze_split[0])
maze_grid = [int(x) for x in maze_split[2:]]
maze = np.array(maze_grid).reshape(size_X, size_Y)
start = np.where(maze_split == 2)
end = np.where(maze_split == 3)
path = astar(maze, start, end)
print(path)
Sorry if this question has been asked before but I'm stumped at how to get it to work. Any help would be appreciated!

import numpy as np
x = np.array([6, 4, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0])
print(x[2:].reshape(x[[1,0]]))
[[0 0 1 0 0 0]
[2 0 1 0 1 1]
[0 0 1 0 0 0]
[0 0 0 0 3 0]]

Loading .dat file into one list and count entities in it

I have a .dat file with the following:
0 0 0 0 1 1
1 1 0 1 0 0
0 0 1 1 0 0
0 1 1 1 1 1
0 1 0 0 0 1
1 1 0 1 0 1
I'm trying to get this all into one list and then count the number of 1's and 0s.
I've got the code so far:
with open('image.dat', 'r') as a:
for line in a:
b = [line.strip()]
print(b)
c = b.count(0)
This just gives me:
['0 0 0 0 1 1']
['1 1 0 1 0 0']
['0 0 1 1 0 0']
['0 1 1 1 1 1']
['0 1 0 0 0 1']
['1 1 0 1 0 1']
0
I'm new to coding and I've tried everything.
Thanks for helping.

You can just count the number of times the string '0' (or '1') accures within the file:
with open("image.dat", "r") as a:
print(a.read().count('0'))

To load the data into one list, you can use list.extend method, for example:
data = []
with open('image.dat', 'r') as a:
for line in a:
data.extend(map(int, line.split()))
print(data)
Prints:
[0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1]
Then:
print('Number of 0:', data.count(0))
print('Number of 1:', data.count(1))
Prints:
Number of 0: 18
Number of 1: 18
EDIT: To load the data as list of lists:
lines = []
with open('image.dat', 'r') as a:
for line in a:
line = line.strip()
if not line:
continue
lines.append(list(map(int, line.split())))
print(lines)
Prints:
[[0, 0, 0, 0, 1, 1], [1, 1, 0, 1, 0, 0], [0, 0, 1, 1, 0, 0], [0, 1, 1, 1, 1, 1], [0, 1, 0, 0, 0, 1], [1, 1, 0, 1, 0, 1]]

Count string elements on a map based on where strings are and fill grid with the counts

I have a list called my_map that contains two different kinds of string values '.' and '&'. Now, for each value [x][y] that is a '.' I want to count the number of times an '&' was found in any of the eight directions next to the '.'
I created a grid to store the counts but I am just not able to formulate my conditions correctly. I can not use numpy arrays.
Note: 'S' and 'E' are treated like '.'
my_map = ['................' '....&...........' '..........E.....'
'&&..&...........' '....&&&.........' '......&&&&..&&..'
'................' '.......&........' '.....&.&........'
'....S...........' '.......&.&&.....']
def create_grid(my_map):
grid = [[0]*(len(my_map[0])) for x in range(len(my_map))]
return grid
grid = create_grid(my_map)
for x, y in [(x,y) for x in range(len(my_map)) for y in range(len(my_map[0]))]:
#any '&' north ?
if my_map[x][y+1]== '&' and my_map[x][y]=='.':
grid[x][y]+= 1
#any '&' west ?
if my_map[x-1][y]== '&' and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' south ?
if my_map[x][y-1]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' east ?
if my_map[x+1][y]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' north-east ?
if my_map[x+1][y+1] == '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' south-west ?
if my_map[x-1][y-1] == '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' south-east ?
if my_map[x+1][y-1]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' north-west?
if my_map[x-1][y+1]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#desired output for first 3 rows
grid = [[0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0],[2,2,1,1,2,1,0,0,0,0,0,0,0,0,0,0]]
At the moment, I get an 'IndexError: string index out of range'. I dont know how to limit the range so it will still be correct.The only thing I managed so far was a grid displaying 1s for all '.' and 0s for all '&'.

I don't think the nested conditionals are appropriate here; each outer conditional must be true for the inner ones to be evaluated. They should be independent of each other and sequential.
It's also a lot of work and error-prone to enumerate every conditional by hand. For each cell, there are up to 8 directions in which a neighbor might live, and we do the exact same check on each direction. A loop is the appropriate construct for doing this; each loop iteration checks one neighboring cell, determining whether it's in bounds and of the appropriate character.
Furthermore, since your grid has few &, it makes sense to only perform neighbor checks for & characters. For each one, increment counts for neighboring .s. Do the opposite if the grid is predominantly & characters.
my_map = [
'................',
'....&...........',
'..........E.....',
'&&..&...........',
'....&&&.........',
'......&&&&..&&..',
'................',
'.......&........',
'.....&.&........',
'....S...........',
'.......&.&&.....'
]
grid = [[0] * len(x) for x in my_map]
directions = [
[-1, 0], [1, 0], [0, 1], [0, -1],
[-1, -1], [1, 1], [1, -1], [-1, 1]
]
for row in range(len(my_map)):
for col in range(len(my_map[row])):
if my_map[row][col] == "&":
for x, y in directions:
y += row
x += col
if y < len(my_map) and y >= 0 and \
x < len(my_map[y]) and x >= 0 and \
my_map[y][x] != "&":
grid[y][x] += 1
for row in grid:
print(row)
Output:
[0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[2, 2, 1, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 1, 2, 0, 4, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0]
[2, 2, 1, 2, 0, 0, 0, 4, 3, 2, 1, 1, 2, 2, 1, 0]
[0, 0, 0, 1, 2, 4, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0]
[0, 0, 0, 0, 0, 1, 3, 4, 4, 2, 1, 1, 2, 2, 1, 0]
[0, 0, 0, 0, 1, 1, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 1, 0, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 1, 1, 3, 2, 3, 2, 2, 1, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 0, 0]
And a version that overlays counts with the original map Minesweeper-style:
0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 1 & 1 0 0 0 0 0 0 0 0 0 0
2 2 1 2 2 2 0 0 0 0 0 0 0 0 0 0
& & 1 2 & 4 2 1 0 0 0 0 0 0 0 0
2 2 1 2 & & & 4 3 2 1 1 2 2 1 0
0 0 0 1 2 4 & & & & 1 1 & & 1 0
0 0 0 0 0 1 3 4 4 2 1 1 2 2 1 0
0 0 0 0 1 1 3 & 2 0 0 0 0 0 0 0
0 0 0 0 1 & 3 & 2 0 0 0 0 0 0 0
0 0 0 0 1 1 3 2 3 2 2 1 0 0 0 0
0 0 0 0 0 0 1 & 2 & & 1 0 0 0 0
Try it!

Pseudo-random generation of python list containing binary values with run-length and frequency constraints

I want to pseudo-randomly create a list with 48 entries -- 24 zeros and 24 ones -- where the same value never occurs three times in a row. I have the following code:
import random
l = list()
for i in range(48):
if len(l) < 2:
l.append(random.choice([0,1]))
else:
if l[i-1] == l[i-2]:
if l[i-1] == 0:
l.append(1)
else:
l.append(0)
else:
l.append(random.choice([0,1]))
But sometimes the count of 0s and 1s is uneven.

Getting uniformity without using rejection is tricky.
The rejection approach is straightforward, something like
def brute(n):
seq = [0]*n+[1]*n
while True:
random.shuffle(seq)
if not any(len(set(seq[i:i+3])) == 1 for i in range(len(seq)-2)):
break
return seq
which will be very slow at large n but is reliable.
There's probably a slick way to take a non-rejection sample where it's almost trivial, but I couldn't see it and instead I fell back on methods which work generally. You can make sure that you're uniformly sampling the space if at each branch point, you weight the options by the number of successful sequences you generate if you take that choice.
So, we use dynamic programming to make a utility which counts the number of possible sequences, and extend to the general case where we have (#zeroes, #ones) bits left, and then use this to provide the weights for our draws. (We could actually refactor this into one function but I think they're clearer if they're separate, even if it introduces some duplication.)
from functools import lru_cache
import random
def take_one(bits_left, last_bits, choice):
# Convenience function to subtract a bit from the bits_left
# bit count and shift the last bits seen.
bits_left = list(bits_left)
bits_left[choice] -= 1
return tuple(bits_left), (last_bits + (choice,))[-2:]
#lru_cache(None)
def count_seq(bits_left, last_bits=()):
if bits_left == (0, 0):
return 1 # hooray, we made a valid sequence!
if min(bits_left) < 0:
return 0 # silly input
if 0 in bits_left and max(bits_left) > 2:
return 0 # short-circuit if we know it won't work
tot = 0
for choice in [0, 1]:
if list(last_bits).count(choice) == 2:
continue # can't have 3 consec.
new_bits_left, new_last_bits = take_one(bits_left, last_bits, choice)
tot += count_seq(new_bits_left, new_last_bits)
return tot
def draw_bits(n):
bits_left = [n, n]
bits_drawn = []
for bit in range(2*n):
weights = []
for choice in [0, 1]:
if bits_drawn[-2:].count(choice) == 2:
weights.append(0) # forbid this case
continue
new_bits_left, new_last_bits = take_one(bits_left, tuple(bits_drawn[-2:]), choice)
weights.append(count_seq(new_bits_left, new_last_bits))
bit_drawn = random.choices([0, 1], weights=weights)[0]
bits_left[bit_drawn] -= 1
bits_drawn.append(bit_drawn)
return bits_drawn
First, we can see how many such valid sequences there are:
In [1130]: [count_seq((i,i)) for i in range(12)]
Out[1130]: [1, 2, 6, 14, 34, 84, 208, 518, 1296, 3254, 8196, 20700]
which is A177790 at the OEIS, named
Number of paths from (0,0) to (n,n) avoiding 3 or more consecutive east steps and 3 or more consecutive north steps.
which if you think about it is exactly what we have, treating a 0 as an east step and a 1 as a north step.
Our random draws look good:
In [1145]: draw_bits(4)
Out[1145]: [0, 1, 1, 0, 1, 0, 0, 1]
In [1146]: draw_bits(10)
Out[1146]: [0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0]
and are quite uniform:
In [1151]: Counter(tuple(draw_bits(4)) for i in range(10**6))
Out[1151]:
Counter({(0, 0, 1, 0, 1, 0, 1, 1): 29219,
(1, 0, 1, 0, 0, 1, 0, 1): 29287,
(1, 1, 0, 0, 1, 0, 1, 0): 29311,
(1, 0, 1, 0, 1, 0, 1, 0): 29371,
(1, 0, 1, 0, 1, 1, 0, 0): 29279,
(0, 1, 0, 1, 0, 0, 1, 1): 29232,
(0, 1, 0, 1, 1, 0, 1, 0): 29824,
(0, 1, 1, 0, 0, 1, 1, 0): 29165,
(0, 1, 1, 0, 1, 0, 0, 1): 29467,
(1, 1, 0, 0, 1, 1, 0, 0): 29454,
(1, 0, 1, 1, 0, 0, 1, 0): 29338,
(0, 0, 1, 1, 0, 0, 1, 1): 29486,
(0, 1, 1, 0, 1, 1, 0, 0): 29592,
(0, 0, 1, 1, 0, 1, 0, 1): 29716,
(1, 1, 0, 1, 0, 0, 1, 0): 29500,
(1, 0, 0, 1, 0, 1, 0, 1): 29396,
(1, 0, 1, 0, 0, 1, 1, 0): 29390,
(0, 1, 1, 0, 0, 1, 0, 1): 29394,
(0, 1, 1, 0, 1, 0, 1, 0): 29213,
(0, 1, 0, 0, 1, 0, 1, 1): 29139,
(0, 1, 0, 1, 0, 1, 1, 0): 29413,
(1, 0, 0, 1, 0, 1, 1, 0): 29502,
(0, 1, 0, 1, 0, 1, 0, 1): 29750,
(0, 1, 0, 0, 1, 1, 0, 1): 29097,
(0, 0, 1, 1, 0, 1, 1, 0): 29377,
(1, 1, 0, 0, 1, 0, 0, 1): 29480,
(1, 1, 0, 1, 0, 1, 0, 0): 29533,
(1, 0, 0, 1, 0, 0, 1, 1): 29500,
(0, 1, 0, 1, 1, 0, 0, 1): 29528,
(1, 0, 1, 0, 1, 0, 0, 1): 29511,
(1, 0, 0, 1, 1, 0, 0, 1): 29599,
(1, 0, 1, 1, 0, 1, 0, 0): 29167,
(1, 0, 0, 1, 1, 0, 1, 0): 29594,
(0, 0, 1, 0, 1, 1, 0, 1): 29176})
Coverage is also correct, in that we can recover the A177790 counts by randomly sampling (and with some luck):
In [1164]: [len(set(tuple(draw_bits(i)) for _ in range(20000))) for i in range(9)]
Out[1164]: [1, 2, 6, 14, 34, 84, 208, 518, 1296]

Here's a reasonably efficient solution that gives you fairly random output that obeys the constraints, although it doesn't cover the full solution space.
We can ensure that the number of zeroes and ones are equal by ensuring that the number of single zeros equals the number of single ones, and the number of pairs of zeros equals the number of pairs of ones. In a perfectly random output list we'd expect the number of singles to be roughly double the number of pairs. This algorithm makes that exact: each list has 12 singles of each type, and 6 pairs.
Those run lengths are stored in a list named runlengths. On each round, we shuffle that list to get the sequence of run lengths for the zeros, and shuffle it again to get the sequence of run lengths for the ones. We then fill the output list by alternating between runs of zeroes and ones.
To check that the lists are correct we use the sum function. If there are equal numbers of zeroes and ones the sum of a list is 24.
from random import seed, shuffle
seed(42)
runlengths = [1] * 12 + [2] * 6
bits = [[0], [1]]
for i in range(10):
shuffle(runlengths)
a = runlengths[:]
shuffle(runlengths)
b = runlengths[:]
shuffle(bits)
out = []
for u, v in zip(a, b):
out.extend(bits[0] * u)
out.extend(bits[1] * v)
print(i, ':', *out, ':', sum(out))
output
0 : 0 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 1 1 0 1 : 24
1 : 0 1 0 0 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 1 0 0 1 0 1 0 1 1 0 0 1 0 1 1 0 1 : 24
2 : 0 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 0 1 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 : 24
3 : 0 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 : 24
4 : 1 1 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 0 : 24
5 : 1 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 0 : 24
6 : 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 : 24
7 : 0 1 0 0 1 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 0 1 : 24
8 : 0 0 1 1 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 1 0 1 1 0 1 0 1 1 0 1 1 0 0 1 0 1 : 24
9 : 1 0 1 1 0 1 1 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 1 0 : 24

Here is a simple code that obeys your constraints:
import random
def run():
counts = [24, 24]
last = [random.choice([0, 1]), random.choice([0, 1])]
counts[last[0]] -= 1
counts[last[1]] -= 1
while sum(counts) > 0:
can_pick_ones = sum(last[-2:]) < 2 and counts[1] > 0.33 * (48 - len(last))
can_pick_zeros = sum(last[-2:]) > 0 and counts[0] > 0.33 * (48 - len(last))
if can_pick_ones and can_pick_zeros:
value = random.choice([0, 1])
elif can_pick_ones:
value = 1
elif can_pick_zeros:
value = 0
counts[value] -= 1
last.append(value)
return last
for i in range(4):
r = run()
print(sum(r), r)
Output
24 [1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0]
24 [0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1]
24 [0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1]
24 [1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0]
Rationale
At each step of the while loop you can either choose 1, choose 0 or both. You can choose:
1 if the last two elements are not one and the counts of 1 is larger than 1/3 the amount of remaining slots: sum(last[-2:]) < 2 and counts[1] > 0.33 * (48 - len(last))
0 if the last two elements are not 0 and the counts of 0 is larger than 1/3 the amount of remaining slots: sum(last[-2:]) > 0 and counts[0] > 0.33 * (48 - len(last))
Both if you can choose 1 or 0
The sum of the two last elements can be 0, 1 or 2, if equals 0 it means that the last two elements were 0 so you can only pick 0 if sum(last[-2:]) > 0. If equals 2 it means that the last two elements where 1, so you can only pick 1 if sum(last[-2:]) < 2. Finally you need to check that the amount of elements of both 1 and 0 are at least a third of the remaining positions to assign, otherwise you are going to be forced to create a run of three consecutive equal elements.