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How do I replace elements in two-dimensional array in python?

My task is to replace all the elements whose both indexes are odd with 1, and all the elements whose both indexes are even with -1.

You can replace elements by using a double index like: array[y][x] if array is list of lists.
This is a example:
array = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
length = len(array)
for y in range(length):
for x in range(length):
if y % 2 and x % 2:
array[y][x] = 1
elif y % 2 == 0 and x % 2 == 0:
array[y][x] = 0
print(array)
This will output: [[0, 2, 0], [4, 1, 6], [0, 8, 0]]

Answer:
for i in range(row):
for j in range(col):
if (i%2==0 and j%2==0):
array[i][j] = -1
elif (i%2 and j%2):
array[I][j] = 1
row => length of nested array
col => length of the array

test = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
def replace(arr):
for i in range(len(arr)):
for j in range(len(arr)):
if i % 2 and j % 2: # checks if both indexes are odd
test[i][j] = 1
elif not i % 2 and not j % 2: # checks if both indexes are even
test[i][j] = -1
return arr
print(replace(test)) # [[-1, 2, -1], [4, 1, 6], [-1, 8, -1]]

Move Zeroes to the end of the string

I was just doing some leetcode
but this blew my mind.
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
nums = [0, 1, 0, 3 ,12]
k = 0
i=0
while i < len(nums):
print(i)
if nums[i] == 0:
print('inside if=',i)
print('before',nums)
k += 1
print('removed',nums[i])
nums.remove(i)
print('after',nums)
else:
i += 1
print('no change',nums)
print('final',nums)
for j in range (k):
nums.append(0)
print(nums)
and the output is:
0
inside if= 0
before [0, 1, 0, 3, 12]
element to be removed 0
after [1, 0, 3, 12]
0
no change [1, 0, 3, 12]
->1
->inside if= 1
->before [1, 0, 3, 12]
->element to be removed 0
->after [0, 3, 12]
1
no change [0, 3, 12]
2
no change [0, 3, 12]
final [0, 3, 12]
[0, 3, 12, 0, 0]
How can it remove element 1 if it can read that it has to remove 0?

So I got to know my dumb mistake.
Actually remove() is used to remove the element that is mentioned.
Like if I type remove(6)from a list a = [1,2,3,4,5,6]. it removes element 6, not the element that is at 6th position.
so use pop() to do so.

Python create multi dimensional array of of normal array

I want to create a multi dimensional Array out of normal array. When there is a 0 in my start array i want to change the row number by 1.
My start array looks like this:
arr = [1, 2, 3, 0, 4, 5, 6, 0, 7, 8, 9]
My futre multi array should look like this:
multi_arr = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
My code looks like this so far:
multi_arr = []
end = len(arr)
i = 0 #row
j = 0 #column
for h in range(end):
if arr[h] != 0:
j += 1
multi_arr[i][j]= arr[h]
elif arr[i] != 0:
i += 1
I always get a list index error with this code.

If you really want to work with lists, you dont need a column indexer. You can simply append the value to the correct list.
multi_arr = [[]] # start with a nested list, since your vector doesnt start with a 0
endi = len(arr)
i = 0 #row
for h in range(endi):
if arr[h] != 0:
multi_arr[i].append(arr[h])
else:
i += 1
multi_arr.insert(i,[]) # open new list
output:
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

This should do the job in a simpler way:
arr = [1, 2, 3, 0, 4, 5, 6, 0, 7, 8, 9]
multi_arr = []
temp_arr = []
for h in arr:
if h != 0:
temp_arr.append(h)
else:
multi_arr.append(temp_arr)
temp_arr = []
if h != 0: # Insert the last array it the last number wasn't 0
multi_arr.append(temp_arr)
print(multi_arr)

This is happening because at the point where you are trying set multi_arr[i][j], it is still a 1-d array. You can verify this with a print statement or a debugger - like so:
for h in range(end):
if arr[h] != 0:
print(multi_arr)
j += 1
multi_arr[i][j]= arr[h]
elif arr[i] != 0:
i += 1
I would definitely try something more pythonic for what you are trying to accomplish, reading the elements into a temp array and appending that temp array to multi array.
arr = [1, 2, 3, 0, 4, 5, 6, 0, 7, 8, 9]
multi_arr = []
temp_array = []
for item in arr:
if item == 0:
multi_arr.append(temp_array)
temp_array = []
else:
temp_array.append(item)
multi_arr.append(temp_array)
print(multi_arr)

You are declaring a 1D array and appending like 2D array
One solution two your Problem is :
arr=[1,2,3,4,5,6,7,8,9]
multi_arr=[]
end = len(arr)
a = []
counter =0
for i in arr :
if counter < 3 :
a.append(i)
counter += 1
else:
multi_arr.append(a)
a=[]
counter = 0
a.append(i)
counter += 1
multi_arr.append(a)
print(multi_arr)

Here is a yield approach:
def f(arr):
res = []
for x in arr:
if x == 0:
yield res
res = []
else:
res.append(x)
yield res
list(f(arr))
#> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

A different approach for you: super simple and no loop.
import numpy as np
arr = [1, 2, 3, 0, 4, 5, 6, 0, 7, 8, 9]
a = [i for i in arr if i]
x = arr[arr!=0] + 1
np.array(a).reshape([-1, x])
Output:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
And if you need lists, just use np.array(a).reshape([x, y]).tolist() instead.
One-liner:
After converting your list to a numpy array a = np.array(arr):
a[a!=0].reshape([-1, len(a) - np.count_nonzero(a)+1])

Finding the number of substrings which sum is equal to m

I'm trying some python and I got this:
I have a string S='3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2' and m=9.
I want to know how many substrings with with sum equals m there are.
So with S and m above i whould get 7 as a result as:
'3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2'
_'0,4,0,3,1,0,1,0'_____________
_'0,4,0,3,1,0,1'_______________
___'4,0,3,1,0,1,0'_____________
____'4,0,3,1,0,1'_______________
____________________'0,0,5,0,4'_
______________________'0,5,0,4'_
_______________________'5,0,4'_
Now, the code i came up with does something like that
def es1(S,m):
c = 0
M = 0
ls = StringToInt(S)
for x in ls:
i= ls.index(x)
for y in ls[i+1:]:
M = x + y
if M == m:
c += 1
M = 0
break
if M > m:
M = 0
break
else:
continue
return c
def StringToInt(ls):
s = [int(x) for x in ls.split(',')]
return s
Where StringToInt obv gives me a list of int to work with.
The thing I don't get is where my concept is wrong since es1 returns 3

You could use zip to progressively add numbers to a list of sums and count how many 9s you have at each pass:
S = '3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2'
m = 9
numbers = list(map(int,S.split(",")))
result = 0
sums = numbers
for i in range(len(numbers)):
result += sums.count(m)
sums = [a+b for a,b in zip(sums,numbers[i+1:]) ]
print(result)
For a more "functional programming" approach, you can use accumulate from itertools:
from itertools import accumulate
numbers = list(map(int,S.split(",")))
ranges = (numbers[i:] for i in range(len(numbers)))
sums = (accumulate(r) for r in ranges)
result = sum( list(s).count(m) for s in sums )
print(result)
To explain how this works, let's first look at the content of ranges, which are substrings from each position up to the end of the list:
[3, 0, 4, 0, 3, 1, 0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[0, 4, 0, 3, 1, 0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[4, 0, 3, 1, 0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[0, 3, 1, 0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[3, 1, 0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[1, 0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[0, 1, 0, 1, 0, 0, 5, 0, 4, 2]
[1, 0, 1, 0, 0, 5, 0, 4, 2]
[0, 1, 0, 0, 5, 0, 4, 2]
[1, 0, 0, 5, 0, 4, 2]
[0, 0, 5, 0, 4, 2]
[0, 5, 0, 4, 2]
[5, 0, 4, 2]
[0, 4, 2]
[4, 2]
[2]
When we make a cumulative sum of the rows (sums), we obtain the total of values starting at the position defined by the row number and for a length defined by the column number. e.g. line 5, column 3 represents the sum of 3 values starting at the fifth position:
[3, 3, 7, 7, 10, 11, 11, 12, 12, 13, 13, 13, 18, 18, 22, 24]
[0, 4, 4, 7, 8, 8, 9, 9, 10, 10, 10, 15, 15, 19, 21]
[4, 4, 7, 8, 8, 9, 9, 10, 10, 10, 15, 15, 19, 21]
[0, 3, 4, 4, 5, 5, 6, 6, 6, 11, 11, 15, 17]
[3, 4, 4, 5, 5, 6, 6, 6, 11, 11, 15, 17]
[1, 1, 2, 2, 3, 3, 3, 8, 8, 12, 14]
[0, 1, 1, 2, 2, 2, 7, 7, 11, 13]
[1, 1, 2, 2, 2, 7, 7, 11, 13]
[0, 1, 1, 1, 6, 6, 10, 12]
[1, 1, 1, 6, 6, 10, 12]
[0, 0, 5, 5, 9, 11]
[0, 5, 5, 9, 11]
[5, 5, 9, 11]
[0, 4, 6]
[4, 6]
[2]
In this triangular matrix each position corresponds to the sum of one of the possible substrings. We simply need to count the number of 9s in there to get the result.
The above solutions will perform in O(N^2) time but, if you are concerned with performance, there is a way to obtain the result in O(N) time using a dictionary. Rather than build the whole sub arrays in the above logic, you could simply count the number of positions that add up to each sum. Then, for the sum at each position, go directly to a previous sum total that is exactly m less to get the number of substrings for that position.
from itertools import accumulate
from collections import Counter
numbers = map(int,S.split(","))
result = 0
sums = Counter([0])
for s in accumulate(numbers):
result += sums[s-m]
sums[s] += 1
print(result)
Note that all these solutions support negative numbers in the list as well as a negative or zero target.

As mentioned by others, your code only looks at sums of pairs of elements from the list. You need to look at sublists.
Here is a O(n) complexity solution (i.e. it's efficient since it only scans though the list once):
def es2(s, m):
s = string_to_int(s)
c = 0
# index of left of sub-list
left = 0
# index of right of sub-list
right = 0
# running total of sublist sum
current_sum = 0
while True:
# if the sub-list has the correct sum
if current_sum == m:
# add as many zeros on the end as works
temp_current_sum = current_sum
for temp_right in range(right, len(s) + 1):
if temp_current_sum == m:
c += 1
if temp_right<len(s):
temp_current_sum += s[temp_right]
else:
break
if current_sum >= m:
# move the left end along and update running total
current_sum -= s[left]
left += 1
else:
# move the right end along and update running total
if right == len(s):
# if the end of the list is reached, exit
return c
current_sum += s[right]
right += 1
def string_to_int(ls):
s = [int(x) for x in ls.split(',')]
return s
if __name__ == '__main__':
print(es2('3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2', 9))

This is the code you are looking for man. i felt looking by position was better for this problem so I did it and it worked.
def es1(S,m):
c = 0
M = 0
ls = StringToInt(S)
for i in range(0, len(ls)):
M = 0
for x in range(i, len(ls)):
M += ls[x]
if M == m:
c += 1
elif M >= m:
break
return c
def StringToInt(ls):
s = [int(x) for x in ls.split(',')]
return s
print(es1("3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2", 9))
OUTPUT:
7

Your code counts how many pairs of numbers there are in the String S which together give m while you actually want to test all possible substrings.
You could do something like:
numbers = [3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2]
m = 9
c = 0
for i in range(len(numbers)):
for j in range(len(numbers)-i):
sum = 0
for k in numbers[i:i+j]:
sum += k
if sum == m:
c += 1
print(c)
Output:
7

EDIT! ->This Code is actually all possible subsets, not sublists. I am going to leave this here though in case this solution is helpful to anyone who visits this question.
This code gets every solution. If you take a look in the function es1() the result variable is huge list of arrays with all the possible solutions.
import itertools
def es1(S,m):
result = [seq for i in range(len(StringToInt(S)), 0, -1) for seq in itertools.combinations(StringToInt(S), i) if sum(seq) == m]
return len(result)
def StringToInt(ls):
s = [int(x) for x in ls.split(',')]
return s
print(es1("3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2", 9))
OUTPUT:
4608
There are 4608 possible sets that add to the value 9.

s = "3,0,4,0,3,1,0,1,0,1,0,0,5,0,4,2"
m = 9
sn = s.replace(",","+") # replace "," with "+"
d = {} # create a dictionary
# create list of strings:
# s2 = ["3+0","0+4","4+0".............]
# s3 = ["3+0+4","0+4+0","4+0+3".............]
# .
# .
# .
for n in range(2,len(s.split(","))):
d["s%s"%n] = [sn[i:i+(2*n-1)] for i in range(0,len(sn),2)][:-n+1]
# evaluate whether sum of individual lists equal to m or not, then find out how many such lists are there
l = sum([eval(x)==m for y in d.values() for x in y] )
# remember we didnot add s1, i,e check whether individual string == m or not
l = l+sum([x==m for x in s.split(",")])
print(l)
7

How to replace K-values of list by skipping 9 from startend

I have a number lets make it a list s = [1,6,9,2,3,2,7,3,6,8,4,1,9,0,0,3,6,8]
and I have k = 4
What I want is to replace 4-values of list s with number 9 which dont have number 9.
means at position 2 ,we have 9,so skip that and replace next one.
Output should be like: [9,9,9,9,9,2,7,3,6,8,4,1,9,0,0,3,6,8]
With this code I am unable to skip 9's in it:
x= [1,6,9,2,3,2,7,3,6,8,4,1,9,0,0,3,6,8]
k = 4
def elements_replaced(lst, new_element, indices):
return [new_element if i in indices else e for i, e in enumerate(lst)]
output = elements_replaced(x,9,range(k))
print output

you can try:
>>> s = [1,6,9,2,3,2,7,3,6,8,4,1,9,0,0,3,6,8]
>>> k = 4
>>> for index,number in enumerate(s):
if k > 0:
if number != 9:
s[index] = 9
k = k-1
else :
break
>>> s
[9, 9, 9, 9, 9, 2, 7, 3, 6, 8, 4, 1, 9, 0, 0, 3, 6, 8]

You can also use list comprehensions. This will get inefficient if your input list is large relative to the number of nines.
from itertools import chain, repeat
s = [1,6,9,2,3,2,7,3,6,8,4,1,9,0,0,3,6,8]
nines = chain([9] * 4, repeat(None))
result = [x if x == 9 else next(nines) or x for x in s]
print(result)
# [9, 9, 9, 9, 9, 2, 7, 3, 6, 8, 4, 1, 9, 0, 0, 3, 6, 8]

x = [1,6,9,2,3,2,7,3,6,8,4,1,9,0,0,3,6,8]
k = 4
a = 0
while k and a < len(x):
if x[a] != 9:
x[a] = 9
k -= 1
a += 1

x= [1,6,9,2,3,2,7,3,6,8,4,1,9,0,0,3,6,8]
k = 4
def elements_replaced(lst, new_element, indices):
for index, value in enumerate(lst):
if index in indices and value != new_element:
lst[index] = new_element
return lst
output = elements_replaced(x,9,range(k+1))
print (output)