比较字典中列表的值

Question

For background, I am trying to create a routine that stores my top 50 values in a dictionary and once at 50 values, start replacing lowest values with any better ones that I find.

I am trying to figure out how to compare values of a tuple, used inside of a dictionary.

The dictionary comprises of an int as the key, and then a tuple of x, y, z-coordinates and a power value associated to that position. so something like this:

int = 1              #all of these values will change with every iteration or new element
x_pos = 10.0
y_pos = 20.0
z_pos = 30.0
power = 9000.0

dictTop50[int] = (x_pos,y_pos,z_pos,power)

What I tried was to use a heap to get the lowest value out of my dictionary but I think I might be doing that wrong. Furthermore, I'm not completely sure how I will be able to compare the new values that I find to the ones already inside my dictionary, and then replace those. Here is my attempt at that:

if ( len(dictTop50) <= 50 ): 
    dictTop50[int] = (x_pos, y_pos, z_pos, power)
elif (new power value bigger than any of the 50 other power values): 
    del dictTop50[heapq.nsmallest(1, dictTop50(), key=itemgetter(1))] 
    dictTop50[int] = (x_pos, y_pos, z_pos, power)

in pseudo code:

if dictionary has 50 or less entries:
    just add new value to the dictionary
else if dictionary has more than 50 entries already and new value is greater than present value:
    delete the dictionary entry with smallest power value
    add new dictionary element with the highest power value

Sorry this was long but I've combed through so many other questions on stack overflow and couldn't find a solution yet. Any suggestions will be greatly appreciated!

Answer 1

If you are comparing the power values, then you could use a heap of tuples where the first value of each tuple is the power value. Something like

import heapq

def push_keep_fifty(heap, new):
    if len(heap) < 50:
        heapq.heappush(heap, new)
    elif new > heap[0]:
        heapq.heapreplace(heap, new)

i = 1 #don't use int as a variable name
x_pos = 10.0
y_pos = 20.0
z_pos = 30.0
power = 9000.0

heap = []
push_keep_fifty(heap, (power, i, x_pos, y_pos))

This works because Python compares tuples by comparing the first element, which is what we want compared. Keep in mind that It will then try and compare the second element if two of these tuples share a power value.

Answer 2

The code below uses a similar strategy to Patrick Haugh's answer, except that it also maintains your dictionary. As in Patrick's answer, we put the power value at the start of the tuple, so that it gets sorted correctly by the heap. We append the integer key to the end of the tuple that goes onto the heap because we need it to delete dict items that have been superseded.

Once the desired size limit has been reached we just need to check the current smallest item. If that smallest item is smaller than the new item, it gets replaced.

To test the update code I've created a generator, datagen , that makes fake data. To keep the output manageable, I've reduced the size to a Top 20 list rather than a Top 50.

from random import seed, randint
from heapq import heappush, heapreplace

seed(42)

# Make some fake data, in this form: (power, x_pos, y_pos, z_pos)
def datagen():
    m = (100., 1., 1., 1.)
    while True:
        yield tuple(randint(1, 100) * u for u in m)

top20 = {}
heap = []

# Update the heap & the dict
def update(k, tup):
    if len(top20) < 20:
        heappush(heap, tup + (k,))
        top20[k] = tup
    elif tup[0] > heap[0][0]:
        old = heapreplace(heap, tup + (k,))
        top20[k] = tup
        del top20[old[-1]]
        print('replaced', old[0])

# Test

for k, tup in zip(range(50), datagen()):
    print(k, tup)
    update(k, tup)

output

0 (8200.0, 15.0, 4.0, 95.0)
1 (3600.0, 32.0, 29.0, 18.0)
2 (9500.0, 14.0, 87.0, 95.0)
3 (7000.0, 12.0, 76.0, 55.0)
4 (500.0, 4.0, 12.0, 28.0)
5 (3000.0, 65.0, 78.0, 4.0)
6 (7200.0, 26.0, 92.0, 84.0)
7 (9000.0, 70.0, 54.0, 29.0)
8 (5800.0, 76.0, 36.0, 1.0)
9 (9800.0, 21.0, 90.0, 55.0)
10 (4400.0, 36.0, 20.0, 28.0)
11 (9800.0, 44.0, 14.0, 12.0)
12 (4900.0, 13.0, 46.0, 45.0)
13 (7800.0, 34.0, 6.0, 94.0)
14 (5900.0, 69.0, 16.0, 49.0)
15 (1100.0, 71.0, 38.0, 81.0)
16 (8000.0, 47.0, 74.0, 25.0)
17 (9100.0, 9.0, 6.0, 85.0)
18 (3000.0, 99.0, 38.0, 11.0)
19 (3000.0, 13.0, 49.0, 36.0)
20 (5900.0, 82.0, 47.0, 21.0)
replaced 500.0
21 (4800.0, 46.0, 27.0, 86.0)
replaced 1100.0
22 (3500.0, 90.0, 88.0, 83.0)
replaced 3000.0
23 (1000.0, 78.0, 82.0, 22.0)
24 (6900.0, 94.0, 32.0, 21.0)
replaced 3000.0
25 (6000.0, 49.0, 35.0, 82.0)
replaced 3000.0
26 (8900.0, 72.0, 29.0, 88.0)
replaced 3500.0
27 (4200.0, 99.0, 100.0, 8.0)
replaced 3600.0
28 (3000.0, 5.0, 41.0, 52.0)
29 (3500.0, 9.0, 28.0, 73.0)
30 (9200.0, 41.0, 28.0, 84.0)
replaced 4200.0
31 (6400.0, 51.0, 83.0, 59.0)
replaced 4400.0
32 (1900.0, 34.0, 18.0, 32.0)
33 (9600.0, 72.0, 69.0, 34.0)
replaced 4800.0
34 (9600.0, 75.0, 55.0, 75.0)
replaced 4900.0
35 (5200.0, 47.0, 29.0, 18.0)
36 (6600.0, 64.0, 12.0, 97.0)
replaced 5800.0
37 (700.0, 15.0, 20.0, 81.0)
38 (2100.0, 88.0, 55.0, 77.0)
39 (900.0, 50.0, 49.0, 77.0)
40 (6000.0, 68.0, 33.0, 71.0)
replaced 5900.0
41 (200.0, 88.0, 93.0, 15.0)
42 (8800.0, 69.0, 97.0, 35.0)
replaced 5900.0
43 (9900.0, 83.0, 44.0, 15.0)
replaced 6000.0
44 (3800.0, 56.0, 21.0, 59.0)
45 (100.0, 93.0, 93.0, 34.0)
46 (6500.0, 98.0, 23.0, 65.0)
replaced 6000.0
47 (1400.0, 81.0, 39.0, 82.0)
48 (6500.0, 78.0, 26.0, 20.0)
replaced 6400.0
49 (4800.0, 98.0, 21.0, 70.0)

The print('replaced', old[0]) call isn't necessary, but it's useful for testing and debugging.