What's the best way to find diagonals of a grid?-CodePudding

So I'm trying to find diagonals on a grid. I've tried looking at some stack overflow pages that seemed to provide a clue but most of them includes itertools or numpy and a lot of built in functions.

Note: U may have seen this question somewhere else. This is basically @Bruffff's second account

So I'm just asking on behalf of all the beginners here, what's the most basic way to find all the diagonals in a given grid.

Here is what I have so far:

def columns(test):
    ans = []
    for x, lst in enumerate(test):
        if x < len(test)-1:
           ans.append(lst[x 1])
    return ans
print(columns(test))

and given this grid,

test = [["r","a","w","b","i","t"],
        ["x","a","y","z","c","h"],
        ["p","q","b","e","i","e"],
        ["t","r","s","b","o","g"],
        ["u","w","x","v","i","t"]
        ["u","w","x","v","i","t"],

the output returns

['a', 'y', 'e', 'o', 't']

but my expected output is

[(u), (uw) (twx), (prxv), (xqsvi), (rabbit), (ayeot), (wzig), (bce),(ih), (t)]

How can I do this without using the complex built in functions and import, numpy or itertools??

I know they may be easier, but I want to learn things using the basics first, so yea, please use the most basic of approaches to do this. Thank you:)

CodePudding user response：

def Diagonals(test):
    l = len(test)
    temp = [] # Add whitespaces to end
    result = []
    for row in test:
        temp.append([" "]*(l-1) row [" "]*(l-1))
    test = temp
    for i in range(l):
        current = ""
        for j in range(2*l-1):
            current  = test[i j][j]
        result.append(current.strip())
    return result

What this code is doing is adding whitespaces onto the end of the list, so that it can just get all (6) length diagonals, then trims off the whitespaces at the end.

i.e. for 3

*.bob..
.*obo..
..*ob..

.*bob..
..*bo..
..b*b..

..*ob..
..o*o..
..bo*..

..b*b..
..ob*..
..bob*.

..bo*..
..obo*.
..bob.*

CodePudding user response：

a very long and not so efficient implementation

# your code goes here
def get_row_from_grid(l):
    """ take 2D matrix and give, diagonal elements"""

    return "".join([v2 for i, v in enumerate(l) for j, v2 in enumerate(v) if i == j])




l = [      ["r", "a", "w", "b", "i", "t"],
            ["x", "a", "y", "z", "c", "h"],
            ["p", "q", "b", "e", "i", "e"],
            ["t", "r", "s", "b", "o", "g"],
            ["u", "w", "x", "v", "i", "t"],
            ["u", "w", "x", "v", "i", "t"]]

val1, val2 = len(l)-1, 0
result = []

# traversing lower triangular matrix
for i in range(len(l)-1, -1, -1):
        tmp = get_row_from_grid([j[:len(l)-i] for j in l[i:]])
        result.append(tmp)

#  traversing upper triangluar matrix
for i in range(0, len(l)):
    tmp = get_row_from_grid( [ j[i 1:] for j in  l[:][i:]])
    if tmp:
        result.append(tmp)
print(result)

output

['u', 'uw', 'twx', 'prxv', 'xqsvi', 'rabbit', 'ayeot', 'yeot', 'eot', 'ot', 't']

CodePudding user response：

Here you go, but it isn't ordered though!

   def columns(test):
      results = set()
      for i in range(len(test)):
        tmp1=""
        tmp2=""
        for j in range(0,len(test)-i):
            tmp1 =test[i j][j]
            tmp2 =test[j][i j]
        results.add(tmp1)
        results.add(tmp2)
      return list(results) # if you need to return it like a list

Returned value:

['ih', 'prxv', 'u', 't', 'bce', 'twx', 'rabbit', 'ayeot', 'wzig', 'uw', 'xqsvi']

Edit:

for "other diagonal side":

def columns(test):
    results = set()
    n=len(test)
    for i in range(n):
        tmp1=""
        tmp2=""
        for j in range(i 1):
            tmp1 =test[i-j][j]
            tmp2 =test[n-1-j][n-1-i j]
        results.add(tmp1)
        results.add(tmp2)
    return list(results)