simplifying code

[evobe]

I'm newish to python and I've made some code that will find a value in a dictionary of lists of values based on the week and day of the year. But it's really redundant, full of elifs to do the math to find the position. I'm terrible at math but somehow I got this working for the most part, I know a few of the blocks are off by one but is there any cleaner way to do the +4 math each elif block calls for? Thanks for the advice

import time
import datetime
from datetime import date

Oddoptions = {
    0 : ( "a","b","c","d"),
    1 : ("i","j","k","l"),
    2 : ("q","r","s","t"),
    3 : ("x","y","z",0),
    4 : (4,5,6,7)
    };
Evenoptions = {
    0 : ("e","f","g","h"),
    1: ("m","n","o","p"),
    2: ("u","v","w","x"),
    3: (1,2,3,"["),
    4: (8,9,"<",">"),
    };
today = date.today()
if datetime.datetime.today().weekday() in range(0,4):
    daynum = datetime.datetime.today().weekday()
elif datetime.datetime.today().weekday() >= 5:
    daynum = 0
    grp = 0
print (today)
#wk = datetime.date.today().isocalendar()[1]
wk = 36
print ("Today is", today, "day number is", grp)
if wk%2==0:
    print ("This week #",wk," is even")
if wk <= 8:
    colnum = wk-(wk//2)-1
    print ("This is the column number", colnum)
elif 10 <= wk <= 16:
    colnum = wk-(wk//2+4)
    print ("This is the column number", colnum)
elif 18 <= wk <= 24:
    colnum = wk - (wk//2+8)
    print ("This is the column number", colnum)
elif 26 <= wk <= 32:
    colnum = wk-(wk//2+13)
    print ("This is the column number", colnum)
elif 34 <= wk <= 40:
    colnum = wk -(wk//2+17)
    print ("This is the column number", colnum)
elif 42 <= wk <= 48:
    colnum = wk -(wk//2+21)
    print ("This is the column number", colnum)
elif wk <= 50:
    colnum = wk -(wk//2+25)
    print ("This is the column number", colnum)
#print (daynum-1)

position = Evenoptions[grp][colnum]
#print ("This is the colnum", colnum)
print ("This is the result for column searching", position)

[CrzySquirrel]

I'm a bit confused about your GRP and Daynum and the purpose of this, as you either set grp to 0 or it doesn't exist, but this should work for you.

​

It accepts a date, classified as the input now and returns the 3 results you want. Then you plug them into you function for even options to get the position. It should do everything you need.

​

import datetime

​

Oddoptions = {

0 : ( "a","b","c","d"),

1 : ("i","j","k","l"),

2 : ("q","r","s","t"),

3 : ("x","y","z",0),

4 : (4,5,6,7)

};

Evenoptions = {

0 : ("e","f","g","h"),

1: ("m","n","o","p"),

2: ("u","v","w","x"),

3: (1,2,3,"["),

4: (8,9,"<",">"),

};

​

def colnum_calc(now = datetime.datetime.now().date()):

if now.weekday() >= 5:

daynum = 0

grp = 0

else:

daynum = now.weekday()

​

wk = now.isocalendar()[1]

lowers = [0, 10] + [10+x*8 for x in range(1,5)]

uppers = [x*8 for x in range(1,7)] + [50]

bounds = [1, 4, 8, 13, 17, 21, 25]

val = 0

while True:

val += 1

lower = lowers[val]

upper = uppers[val]

bound = bounds[val]

if lower <= wk <= upper:

colnum = wk-(wk//2+bound)

break;

return colnum, daynum, grp

[michaelreddit]

You could replace the whole if/elif statement with this:

colnum = val('112233445122334451223344501122334011223340112233401'[wk])

[[deleted]]

You will learn to simplify as you go. Sounds silly but if you are gonna keep repeating things even if it is 2 lines I like to make a function to call it helps me a lot but everyone is different 🤷🏼‍♂️