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Need to refer to Value of a key of a dictionary nested within a dictionary - Printable Version

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Need to refer to Value of a key of a dictionary nested within a dictionary - xepicxmonkeyx - Oct-01-2016

You can find my code at the link below, and you can ignore any commented sections of code (Accept the one in make_grid() ):

link removed, code added below

from __future__ import print_function
from random import randint
list = {
    11: {'x': 1, 'y': 1, 'onScreen': 'X', 'occupied': False},
    12: {'x': 1, 'y': 2, 'onScreen': 'X', 'occupied': False},
    13: {'x': 1, 'y': 3, 'onScreen': 'O', 'occupied': True},
    14: {'x': 1, 'y': 4, 'onScreen': 'O', 'occupied': True},
    15: {'x': 1, 'y': 5, 'onScreen': 'O', 'occupied': True},
    16: {'x': 1, 'y': 6, 'onScreen': 'X', 'occupied': False},
    17: {'x': 1, 'y': 7, 'onScreen': 'X', 'occupied': False},
    21: {'x': 2, 'y': 1, 'onScreen': 'X', 'occupied': False},
    22: {'x': 2, 'y': 2, 'onScreen': 'X', 'occupied': False},
    23: {'x': 2, 'y': 3, 'onScreen': 'O', 'occupied': True},
    24: {'x': 2, 'y': 4, 'onScreen': 'O', 'occupied': True},
    25: {'x': 2, 'y': 5, 'onScreen': 'O', 'occupied': True},
    26: {'x': 2, 'y': 6, 'onScreen': 'X', 'occupied': False},
    27: {'x': 2, 'y': 7, 'onScreen': 'X', 'occupied': False},
    31: {'x': 3, 'y': 1, 'onScreen': 'X', 'occupied': True},
    32: {'x': 3, 'y': 2, 'onScreen': 'O', 'occupied': True},
    33: {'x': 3, 'y': 3, 'onScreen': 'O', 'occupied': True},
    34: {'x': 3, 'y': 4, 'onScreen': 'O', 'occupied': True},
    35: {'x': 3, 'y': 5, 'onScreen': 'O', 'occupied': True},
    36: {'x': 3, 'y': 6, 'onScreen': 'O', 'occupied': True},
    37: {'x': 3, 'y': 7, 'onScreen': 'O', 'occupied': True},
    41: {'x': 4, 'y': 1, 'onScreen': 'O', 'occupied': True},
    42: {'x': 4, 'y': 2, 'onScreen': 'O', 'occupied': True},
    43: {'x': 4, 'y': 3, 'onScreen': 'O', 'occupied': True},
    44: {'x': 4, 'y': 4, 'onScreen': 'O', 'occupied': True},
    45: {'x': 4, 'y': 5, 'onScreen': 'O', 'occupied': True},
    46: {'x': 4, 'y': 6, 'onScreen': 'O', 'occupied': True},
    47: {'x': 4, 'y': 7, 'onScreen': 'O', 'occupied': True},
    51: {'x': 5, 'y': 1, 'onScreen': 'O', 'occupied': True},
    52: {'x': 5, 'y': 2, 'onScreen': 'O', 'occupied': True},
    53: {'x': 5, 'y': 3, 'onScreen': 'O', 'occupied': True},
    54: {'x': 5, 'y': 4, 'onScreen': 'O', 'occupied': True},
    55: {'x': 5, 'y': 5, 'onScreen': 'O', 'occupied': True},
    56: {'x': 5, 'y': 6, 'onScreen': 'O', 'occupied': True},
    57: {'x': 5, 'y': 7, 'onScreen': 'O', 'occupied': True},
    61: {'x': 6, 'y': 1, 'onScreen': 'O', 'occupied': False},
    62: {'x': 6, 'y': 2, 'onScreen': 'X', 'occupied': False},
    63: {'x': 6, 'y': 3, 'onScreen': 'X', 'occupied': True},
    64: {'x': 6, 'y': 4, 'onScreen': 'O', 'occupied': True},
    65: {'x': 6, 'y': 5, 'onScreen': 'O', 'occupied': True},
    66: {'x': 6, 'y': 6, 'onScreen': 'O', 'occupied': False},
    67: {'x': 6, 'y': 7, 'onScreen': 'X', 'occupied': False},
    71: {'x': 7, 'y': 1, 'onScreen': 'X', 'occupied': False},
    72: {'x': 7, 'y': 2, 'onScreen': 'X', 'occupied': False},
    73: {'x': 7, 'y': 3, 'onScreen': 'X', 'occupied': True},
    74: {'x': 7, 'y': 4, 'onScreen': 'O', 'occupied': True},
    75: {'x': 7, 'y': 5, 'onScreen': 'O', 'occupied': True},
    76: {'x': 7, 'y': 6, 'onScreen': 'O', 'occupied': False},
    77: {'x': 7, 'y': 7, 'onScreen': 'O', 'occupied': False},
}


# Sends data
class readGrid(object):
    def __init__(self, x, y, boolean, occu):
        self.x = x
        self.y = y
        self.boolean = boolean


# produces the Peg Grid, and outputs it
grid_list = []

list
def make_grid():
    for key, item in list[11]:
        print(item)

        #my_var = readGrid(sub_list[0], sub_list[1], sub_list[2], sub_list[3])
        #grid_list.append(my_var.boolean)
    #print(*grid_list, sep=', ')


make_grid()



#for item in list:
#    if item[11]['onScreen']:
#
#        twoOverKey = str(item['x'] + 2) + str(item['y']);
#        twoOver = list[twoOverKey];
#        if twoOver['occupied'] == False:
#            print("h")
            # do something
What I'm trying to do is print out the Character held in 'onScreen' by for-looping through the dictionary in the make_grid() function. Of course this would be easy if it was a list, but... it's not!


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - metulburr - Oct-02-2016

Quote:What I'm trying to do is print out the Character held in 'onScreen' by for-looping through the dictionary in the make_grid() function. Of course this would be easy if it was a list, but... it's not!
You can loop through a dictionary too

for k,v in list.items():
	print(v['onScreen'])
You also do not need to import future print on 2.7.10

and you should not overwrite "list" built-in

https://repl.it/X3G/4836


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - xepicxmonkeyx - Oct-02-2016

(Oct-02-2016, 01:00 AM)metulburr Wrote:
Quote:What I'm trying to do is print out the Character held in 'onScreen' by for-looping through the dictionary in the make_grid() function. Of course this would be easy if it was a list, but... it's not!
You can loop through a dictionary too

for k,v in list.items():
	print(v['onScreen'])
You also do not need to import future print on 2.7.10

and you should not overwrite "list" built-in

https://repl.it/X3G/4836

This worked like a charm, had it printing out randomly at the start (don't know why), but got it working in the end!


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - Skaperen - Oct-02-2016

(Oct-02-2016, 01:00 AM)metulburr Wrote: You also do not need to import future print on 2.7.10

but if you are giving away (or selling) code for others to just run, it might still be needed for a system that has 2.6 :(


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - metulburr - Oct-02-2016

(Oct-02-2016, 05:40 AM)Skaperen Wrote:
(Oct-02-2016, 01:00 AM)metulburr Wrote: You also do not need to import future print on 2.7.10

but if you are giving away (or selling) code for others to just run, it might still be needed for a system that has 2.6 :(

Thats why i specifically said 2.7.10. 
2.6 is quite old. 

And if you are giving away or selling code for others to run you should build an executable with it so that they dont even need python. Then you can attach whichever python interpreter your program needs with it.


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - xepicxmonkeyx - Oct-02-2016

(Oct-02-2016, 02:26 AM)xepicxmonkeyx Wrote:
(Oct-02-2016, 01:00 AM)metulburr Wrote:
Quote:What I'm trying to do is print out the Character held in 'onScreen' by for-looping through the dictionary in the make_grid() function. Of course this would be easy if it was a list, but... it's not!
You can loop through a dictionary too

for k,v in list.items():
	print(v['onScreen'])
You also do not need to import future print on 2.7.10

and you should not overwrite "list" built-in

https://repl.it/X3G/4836

This worked like a charm, had it printing out randomly at the start (don't know why), but got it working in the end!

I mentioned it printed out the dict randomly. For the printing of the grid that wasn't something I needed. But how do I make it do so when I do?


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - Yoriz - Oct-02-2016

dict have no order(unless using ordered dict), so will be printed randomly, to print the values in an order the key will have to be used. A list will always print in the same order.


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - metulburr - Oct-02-2016

You can use collections.OrderdDict to maintain the order
https://pymotw.com/2/collections/ordereddict.html


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - wavic - Oct-02-2016

In version 3.6 dicts will be ordered


RE: Need to refer to Value of a key of a dictionary nested within a dictionary - j.crater - Oct-02-2016

(Oct-02-2016, 03:42 PM)wavic Wrote: In version 3.6 dicts will be ordered

Oh my, that's some news!
Sorry for being ignorant and not up to date with news... do you have a source for this info?