So say I have a 2D array such as:
[
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
And I want to set all the values 2 levels out around the 3 to a specific number like:
[
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 3, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
Note that the 3 could be in any position in the list, I'm using a random generator to get it. So how could I achieve this? Maybe using a for loop?
Carrying on from the comment - I find numpy super useful for slicing like this;
import numpy as np
arr = np.array([
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
])
xs, ys = np.where(arr == 3)
arr[xs[0] - 2: xs[0] + 3, ys[0] - 2: ys[0] + 3] = 1
arr[xs[0], ys[0]] = 3
Obviously possible in pure python/list form as well but you will be knee deep in double iteration probably
Here's a pure Python approach (albeit rather clumsy):
A = [
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
for j, a in enumerate(A):
try:
i = a.index(3)
P = A[j-1] if j > 0 else None
Q = A[j+1] if j < len(a) - 1 else None
for k in range(max(0, i-2), min(i+3, len(a))):
if P:
P[k] = 1
if Q:
Q[k] = 1
a[k] = 1
a[i] = 3
break
except ValueError:
pass
print(A)
pure python
A = [
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
location = [(i, j) for i, row in enumerate(A) for j, item in enumerate(row) if item == 3][0]
for i, row in enumerate(A):
for j, item in enumerate(row):
if (abs(i - location[0]) <= 2) and (abs(j - location[1]) <= 2) and not ((i, j) == location):
A[i][j] = 1
Related
TypeError: 'int' object is not subscriptable board[y][x]
Code:
board = []
bWidth = 10
bHeight = 20
for h in range(bHeight):
row = []
for w in range(bWidth):
row.append(0)
board.append(row)
for y in range(bHeight):
for x in range(bWidth):
if board[y][x] == 0:
Error:
File "C:\Users\TUF\PycharmProjects\Tetris\main.py", line 92, in drawWindow
if board[y][x] == 0:
TypeError: 'int' object is not subscriptable
it appears as though the code is missing something inside the second loop.
If i put a generic print statement there, it does get to the end and produces an array of zero's.
board = []
bWidth = 10
bHeight = 20
for h in range(bHeight):
row = []
for w in range(bWidth):
row.append(0)
board.append(row)
for y in range(bHeight):
for x in range(bWidth):
if board[y][x] == 0:
print('hello', x, y) # <--- this or pass
board
gives this:
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
I have array names y
and y.shape gives
y.shape
Out[6]: (9976, 158)
the value of y is
y
Out[3]:
array([[0, 0, 1, ..., 0, 0, 0],
[0, 0, 1, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
...,
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
[1, 0, 0, ..., 0, 0, 0]])
I want to get the Nth column of each array
I tried y[N] but that gave the Nth row
so
y[0]
Out[4]:
array([0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0,
0, 0, 0, 0])
how can I do that?
Use 2D slicing: y[:, n]
: -> all values in the first dimension (rows)
n -> just the nth column
I've been trying to change a single item in a 2-dimensional array in python using the syntax x[2][3]=1 but instead of just changing the item in the 2nd row 3rd column, it ends up changing the values of all of the 3rd column. My code is below:
population = [[0]*20]*5
population[2][3] = 1
for row in population:
print(row)
This outputs
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
but I only want
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
How would I index the item such that it only changes the 2nd row and 3rd column?
I'm using python 3.7.4 on repl.it
Link here: https://repl.it/#ajqe/2d-array-test
Use :
population = [[0]*20 for _ in range(5)]
to generate the lists instead. The method you are using is referencing the same object 5 times, instead of creating 5 separate lists. To check this you can use the is operator:
>>> population = [[0]*20]*5
>>> population[0] is population[1]
True
>>> population = [[0]*20 for _ in range(5)]
>>> population[0] is population[1]
False
Currently I am working on a project that involves creating an array with 10 binomial values 0 and 1 and a given success rate (= ci_rate[i]/1'000).
Due to the fact that the rate is different for each of the 10 years, I run a loop 10 times that is creating 20'000 binomial values each time (for 20'000 scenarios).
The success rate for the binomial values is very small, but is an absorbing state for the following years. Simplified for only 10 scenarios and 10 years I would like to output the following:
[1,0,0,0,0,0,0,0,0,0]
[1,0,0,0,0,0,0,1,0,0]
[1,0,0,1,0,0,0,1,0,0]
[1,0,0,1,0,0,0,1,0,0]
[1,0,0,1,0,0,0,1,0,0]
[1,0,0,1,0,0,0,1,0,0]
[1,0,0,1,0,1,0,1,0,0]
[1,0,0,1,0,1,0,1,0,0]
[1,0,0,1,0,1,0,1,0,0]
[1,0,0,1,0,1,0,1,0,0]
Currently I am solving the problem in this way:
for j in range(20000):
tem = np.zeros(len(ci_rate))
for i in range(len(ci_rate)):
if i == 0:
tem[0] = (np.random.binomial(1, p = ci_rate[i] / 1000))
else:
tem[i]= int(np.where(tem[i-1]==1, 1, np.random.binomial(1, p = ci_rate[i] / 1000)))
ci_sim.append(tem)
Is anyone creative enough to solve this more time efficient?
This solution first ignores the persistence rule and enforces it afterwards using maximum.accumulate.
ci_rate = np.random.uniform(0, 0.1, 10)
res = np.maximum.accumulate(np.random.random((20000, ci_rate.size))<ci_rate, axis=1).view(np.int8)
res[:20]
#
# array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 0, 1, 1, 1, 1],
# [0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]], dtype=int8)
My attempt would be:
import numpy as np
ci_rate = np.random.normal(size=20)
ci_rate = (ci_rate - min(ci_rate)) /(max(ci_rate) - min(ci_rate)) - 0.7
ci_rate[ci_rate < 0] = 0
r = []
for i in range(100):
t = np.random.binomial(1, ci_rate)
r += [t.tolist()]
ci_rate = [1 if j == 1 else i for i, j in zip(ci_rate, t)]
#output
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0],
I am suggesting geometric distribution, since it looks like you are trying to see the number of trials for first success.
I am comparing the usefulness of using the geomentric distribution in terms of computation time
EDIT:
%%timeit
ci_rate = np.random.uniform(0, 0.1, nb_years)
successful_trail = np.random.geometric(p=ci_rate)
ci_sim=np.zeros((nb_scenarios,nb_years))
for i in range(nb_years):
ci_sim[i,successful_trail[i]:]=1
## 10000 loops, best of 3: 41.4 µs per loop
%%timeit
ci_rate = np.random.uniform(0, 0.1, nb_years)
res = np.maximum.accumulate(np.random.random((nb_scenarios, ci_rate.size))<ci_rate, axis=1).view(np.int8)
## 100 loops, best of 3: 2.97 ms per loop
This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 4 years ago.
def create_octahedron(size):
x = []
y = []
z = []
if size % 2 == 0 or size <= 1:
return x
for i in range(size):
x.append(0)
for i in range(size):
y.append(x)
for i in range(size):
z.append(y)
for i in range(size):
for u in range(size):
for v in range(size):
if i == len(z)//2:
if u == len(y)//2:
if v == len(x)//2:
z[3][3][3] = 1
print(z)
create_octahedron(7)
[[[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]], [[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]], [[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]], [[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]], [[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]], [[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]], [[0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0]]]
this is the output i keep getting but the output I'm expecting is to only have a 1 at the middle of the entire equation not at. i am much less interested in how to fix this as i already know how. What i want to know is why this is giving this output.
Because you append the same list. In z, each row points the same Y, and in Y each row points to the same X. If you try z[0][0][0] = 2, you could see that every row's first element changes to 2.
To avoid this, create a new x/y list before append.