I'm trying to change elements of the diagonal(1) of a matrix but I can't do it because of the error "assignment destination is read-only.
x=np.loadtxt('matrice.txt')
print(x.diagonal(1)) #2
x.diagonal(1)[0]=3
ValueError: assignment destination is read-only
In numpy docs for diagonal it is said
Starting in NumPy 1.9 it returns a read-only view on the original
array. Attempting to write to the resulting array will produce an
error.
In some future release, it will return a read/write view and writing
to the returned array will alter your original array. The returned
array will have the same type as the input array.
If you don’t write to the array returned by this function, then you
can just ignore all of the above.
If you depend on the current behavior, then we suggest copying the
returned array explicitly, i.e., use np.diagonal(a).copy() instead of
just np.diagonal(a). This will work with both past and future versions
of NumPy.
So you should use np.diagonal(a).copy() to get modifiable array.
If you need to edit a diagonal of your matrix you can use this answer by Hans Then.
def kth_diag_indices(a, k):
rows, cols = np.diag_indices_from(a)
if k < 0:
return rows[-k:], cols[:k]
elif k > 0:
return rows[:-k], cols[k:]
else:
return rows, cols
x[kth_diag_indices(x,1)] = 2 # to make them all equal to 2
Related
I want to make my algorithm more efficient via deleting the items it already sorted, but i don't know how I can do it efficiently. The only way I found was to rewrite the whole list.
l = [] #Here you put your list
sl = [] # this is to store the list when it is sorted
a = 0 # variable to store which numbers he already looked for
while True: # loop
if len(sl) == len(l): #if their size is matching it will stop
print(sl) # print the sorted list
break
a = a + 1
if a in l: # check if it is in list
sl.append(a) # add to sorted list
#here i want it to be deleted from the list.
The variable a is a little awkward. It starts at 0 and increments 1 by 1 until it matches elements from the list l
Imagine if l = [1000000, 1200000, -34]. Then your algorithm will first run for 1000000 iterations without doing anything, just incrementing a from 0 to 1000000. Then it will append 1000000 to sl. Then it will run again 200000 iterations without doing anything, just incrementing a from 1000000 to 1200000.
And then it will keep incrementing a looking for the number -34, which is below zero...
I understand the idea behind your variable a is to select the elements from l in order, starting from the smallest element. There is a function that does that: it's called min(). Try using that function to select the smallest element from l, and append that element to sl. Then delete this element from l; otherwise, the next call to min() will select the same element again instead of selecting the next smallest element.
Note that min() has a disadvantage: it returns the value of the smallest element, but not its position in the list. So it's not completely obvious how to delete the element from l after you've found it with min(). An alternative is to write your own function that returns both the element, and its position. You can do that with one loop: in the following piece of code, i refers to a position in the list (0 is the position of the first element, 1 the position of the second, etc) and a refers to the value of that element. I left blanks and you have to figure out how to select the position and value of the smallest element in the list.
....
for i, a in enumerate(l):
if ...:
...
...
If you managed to do all this, congratulations! You have implemented "selection sort". It's a well-known sorting algorithm. It is one of the simplest. There exist many other sorting algorithms.
This seems trivial (again) but has me stumped.
I need to find the indexes of multiple values in a numpy array. I can do this with where and isin but the resulting answer always has a length of 1 regardless of how many indexes are found. Example
import numpy as np
a = [1,3,5,7,9,11,13,15]
b = [1,7,13]
x = np.where(np.isin(a,b))
print(x)
print(len(x))
this returns
(array([0, 3, 6]),)
1
I think its because the array is a single item inside a tuple. How do I return just the array?
Just use
x = np.where(np.isin(a,b))[0]
to get what you expect.
As hpaulj points out in the comments where returns a tuple with one array for each input array dimension, in this case there is only one, which is why x is a tuple of length one.
How to find the minimum number of ways in which elements taken from a list can sum towards a given number(N)
For example if list = [1,3,7,4] and N=14 function should return 2 as 7+7=14
Again if N= 11, function should return 2 as 7+4 =11. I think I have figured out the algorithm but unable to implement it in code.
Pls use Python, as that is the only language I understand(at present)
Sorry!!!
Since you mention dynamic programming in your question, and you say that you have figured out the algorithm, i will just include an implementation of the basic tabular method written in Python without too much theory.
The idea is to have a tabular structure we will use to compute all possible values we need without having to doing the same computations many times.
The basic formula will try to sum values in the list till we reach the target value, for every target value.
It should work, but you can of course make some optimization like trying to order the list and/or find dividends in order to construct a smaller table and have faster termination.
Here is the code:
import sys
# num_list : list of numbers
# value: value for which we want to get the minimum number of addends
def min_sum(num_list, value):
list_len = len(num_list)
# We will use the tipycal dynamic programming table construct
# the key of the list will be the sum value we want,
# and the value will be the
# minimum number of items to sum
# Base case value = 0, first element of the list is zero
value_table = [0]
# Initialize all table values to MAX
# for range i use value+1 because python range doesn't include the end
# number
for i in range(1, value+1):
value_table.append(sys.maxsize);
# try every combination that is smaller than <value>
for i in range(1, value+1):
for j in range(0, list_len):
if (num_list[j] <= i):
tmp = value_table[i-num_list[j]]
if ((tmp != sys.maxsize) and (tmp + 1 < value_table[i])):
value_table[i] = tmp + 1
return value_table[value]
## TEST ##
num_list = [1,3,16,5,3]
value = 22
print("Min Sum: ",min_sum(num_list,value)) # Outputs 3
it would be helpful if you include your Algorithm in Pseudocode - it will very much look like Python :-)
Another aspect: your first operation is a multiplication with one item from the list (7) and one outside of the list (2), whereas for the second opration it is 7+4 - both values in the list.
Is there a limitation for which operation or which items to use (from within or without the list)?
I have a list of values... some are valid, some not, the invalid contain None. I test each element for validity. When I find an invalid element, I need to find indexes of the closest previous and the closest next valid value. I need to treat the list as a cycle, so if there are no valid elements before/after the initial element, it should search from the end/start of the list... This way, if there is at least one valid value in the list, both the previous and the next valid value will be found (it can be the same index, if there is only one valid value in the list).
Ok, it is complicated to explain and it gets very complicated when I try to code it. But I am sure there is some simple and idiomatical way to do it. Can you help me?
Neither I am a programmer, but I use something like this for analysing data around the 24-hour clock (that means, for example 23:00 is before 01:00 etc.). Then I compute the unknown values as a weighted averages of the previous and next known values. The parameters are the position of the unknown element in the list and the list itself:
def get_prev_known(i, some_list):
for j, item in enumerate(some_list[i::-1]+some_list[:i:-1]):
if j > 0 and item is not None:
return (i-j) % len(some_list)
def get_next_known(i, some_list):
for j, item in enumerate(some_list[i:]+some_list[:i]):
if j > 0 and item is not None:
return (i+j) % len(some_list)
it should solve your problem.
I am currently trying to implement Conway's Game of Life in a Code, and therefore built a function which generates the coordinates depending of the size of the window.
def coords_maker(num_x, num_y):
num_x += 1
num_y += 1
coords = []
for i in range (0,num_y, 1):
for n in range (0,num_x,1):
coords.append ('x'+str(n)+'y'+str(i))
return coords
Yet, I would like to randomly assign values to the resulting strings, to mark them either as alive (1) or dead (0). However they only way to convert a string to a variable name known to me is via a dict and var(), but however, it is essential for the further code that the coordinates stay sorted, as I want to be able to iterate over the ordered items and put the cursor accordingly to the coordinates name. Something like:
print ('\033['+X_COORD+';'+Y_COORD+'f'+ x1y5)
if e.g. x1y5 is the corresponding value (0 or 1) of the variable
Is there a convenient method how to either do this via a dict or how to convert the name of the strings to variable names?
Or probably. If I keep one dict and one list and store the coordinate names in the list and the values in the dict?
Thank you in advance!
kyril
You use a dictionary:
def coords_maker(num_x, num_y):
num_x += 1
num_y += 1
coords = {}
for i in range (0,num_y, 1):
for n in range (0,num_x,1):
coords['x'+str(n)+'y'+str(i)] = 0
return coords
You then access the value with
coords[x][y]
And change it like so:
coords[x][y] = 1
Now, of course this converting of coordinates to strings is completely pointless. Simply use a list of lists:
def coords_maker(num_x, num_y):
num_x += 1
num_y += 1
coords = [[0]*num_x for x in range(num_y)]
return coords
And I don't know why you add 1 to the coordinates either:
def coords_maker(num_x, num_y):
return [[0]*num_x for x in range(num_y)]