Chichi - 1 year ago 66
Python Question

# How to define section number that 2d-array is divided to using Python?

I have this data structure:

It is 2d-array that is divided on 3 sections. For each letter in the array I need to define Section number. For example, letters

`a,b,c,d`
are in Section 1;
`e,f,g,h`
are in Section 2.

My code. Firstly, this 2d-array preparation:

``````from itertools import cycle
letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l']

#2d-array initialization
width, height = 3, 6
repRange = cycle(range(1, 3))
values = [0] * (width - 1)
array2d = [[next(repRange)] + values for y in range(height)]

#Filling array with letters:
m = 0
for i in range(height):
for j in range(1,width):
array2d[i][j] = letters[m]
m+=1

#Printing:
for row in array2d:
print(row)
``````

Output:

``````[1, 'a', 'b']
[2, 'c', 'd']
[1, 'e', 'f']
[2, 'g', 'h']
[1, 'i', 'j']
[2, 'k', 'l']
``````

Now I need to determine section number of each letter and save it along with the letter itself. I use
`defineSection`
function and save values in dictionary:

``````def defineSection(i, division, height):
if i <= division:
return 1
elif division*2 >= i > division :
return 2
elif division*3 >= i > division*2 :
return 3

dic = {}
for i in range(height):
for j in range(1,width):
section = defineSection(i+1, 2, height)
dic.update({array2d[i][j] : section})

for item in dic.items():
print(item)
``````

Output:

``````('f', 2)
('b', 1)
('c', 1)
('e', 2)
('k', 3)
('g', 2)
('d', 1)
('a', 1)
('l', 3)
('h', 2)
('i', 3)
('j', 3)
``````

It defined all section numbers for each letter correctly. But
`defineSection`
method is primitive and will not work if number of rows is bigger than 6.
I don't know how to implement
`defineSection`
method so that it defines Section number automatically taking into account only current Row number, division and number of rows in total.

Question: Is there some way I can simply determine section number without so many
`if-elif`
conditions and independently of total number of rows?

You can simplify your matrix creation code immensely. All you need is a `letters` iterator, which returns itself so you can iterate 2-letters at a time using zip.

``````In [3]: from itertools import cycle

In [4]: letters = "abcdefghijkl"

In [5]: ranges = cycle(range(1,3))

In [6]: iter_letters = iter(letters)

In [7]: matrix = [[i,a,b] for i,a,b in zip(ranges,iter_letters,iter_letters)]

In [8]: matrix
Out[8]:
[[1, 'a', 'b'],
[2, 'c', 'd'],
[1, 'e', 'f'],
[2, 'g', 'h'],
[1, 'i', 'j'],
[2, 'k', 'l']]
``````

As for assigning sections, note that a section is every two rows, which is four letters, so you can use simple floor division to "skip" counts.

``````In [9]: sections = {letter:(i//4 + 1) for i,letter in enumerate(letters)}

In [10]: sections
Out[10]:
{'a': 1,
'b': 1,
'c': 1,
'd': 1,
'e': 2,
'f': 2,
'g': 2,
'h': 2,
'i': 3,
'j': 3,
'k': 3,
'l': 3}
``````
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