Custom Dictionary | 易学教程

问题

I have a problem when adding value inside the nested dictionary using the same keys and the value is always shown the same value, The fact is, i want update the value event the keys is same. This algorithm is the basic of Artificial Fish Swarm Algorithm

# example >> fish_template = {0:{'weight':3.1,'visual':2,'step':1},1:'weight':3,'visual':4,'step':2}}

fish = {}
fish_value = {}
weight = [3.1, 3, 4.1, 10]
visual = [2, 4, 10, 3]
step = [1, 2, 5, 1.5]

len_fish = 4

for i in range(0,len_fish):
  for w, v, s in zip(weight, visual, step):
     fish_value["weight"] = w
     fish_value["visual"] = v
     fish_value["step"] = s
     fish[i] = fish_value

  print("show fish",fish)

I expect the result to be like fish_template, but it isn't. The values for the keys 'weight', 'visual', 'step' are always the same with values of 0, 1, 2, and 3. Any solution?

回答1:

The issue is with fish[i], you simply created a dict with the same element: fish_value. Python does not generate a new memory for the same variable name, so all your dict keys point to the same value=fish_value, which gets overwritten and all your dict values take the last state of fish_value. To overcome this, you can do the following:

fish   = {}
weight = [3.1, 3, 4.1, 10]
visual = [2, 4, 10, 3]
step   = [1, 2, 5, 1.5]

len_fish = 4

for i in range(0, len_fish):
     fish[i]= {"weight": weight[i], "visual": visual[i], "step": step[i]}

print("show fish", fish)

As @Error mentioned, the for loop can be replaced by this one-liner:

fish = dict((i, {"weight": weight[i], "visual": visual[i], "step": step[i]}) for i in range(len_fish))

回答2:

Not sure I fully understand what you're trying to do here, but the problem is the last line of your inner for loop. You're looping over i in the main loop, then then inner loop is setting fish[i] multiple times. As a result all your fish_value will look identical.

回答3:

Because of aliasing; the line fish[i] = fish_value is bad practice, fish_value gets overwritten each time you loop; then fish[i] = fish_value just assigns a shallow copy into fish[i], which is not what you want. But really you can avoid the loop with a dict comprehension.

Anyway, better coding practice is to declare your own Fish class with members weight, visual, step, as below. Note how:

we use zip() function to combine the separate w,v,s lists into a tuple-of-list.
Then the syntax *wvs unpacks each tuple into three separate values ('weight', 'visual', 'step'). This is called tuple unpacking, it saves you needing another loop, or indexing.
a custom __repr__() method (with optional ASCII art) makes each object user-legible. (Strictly we should be overriding __str__ rather than __repr__, but this works)

Code:

class Fish():
    def __init__(self, weight=None, visual=None, step=None):
        self.weight = weight
        self.visual = visual
        self.step = step
    def __repr__(self):
        """Custom fishy __repr__ method, with ASCII picture"""
        return f'<º)))< 🐟 [ Weight: {self.weight}, visual: {self.visual}, step: {self.step} ]'
    # define whatever other methods you need on 'Fish' object...

# Now create several Fish'es...
swarm = [ Fish(*wvs) for wvs in zip([3.1, 3, 4.1, 10], [2, 4, 10, 3], [1, 2, 5, 1.5]) ]
# zip() function combines the lists into a tuple-of-list. `*wvs` unpacks each tuple into three separate values ('weight', 'visual', 'step')

# See what we created...
>>> swarm
[<º)))< 🐟 [ Weight: 3.1, visual: 2, step: 1 ], <º)))< 🐟 [ Weight: 3, visual: 4, step: 2 ], <º)))< 🐟 [ Weight: 4.1, visual: 10, step: 5 ], <º)))< 🐟 [ Weight: 10, visual: 3, step: 1.5 ]]

# ... or for prettier output...
>>> for f in swarm: print(f)

<º)))< 🐟 [ Weight: 3.1, visual: 2, step: 1 ]
<º)))< 🐟 [ Weight: 3, visual: 4, step: 2 ]
<º)))< 🐟 [ Weight: 4.1, visual: 10, step: 5 ]
<º)))< 🐟 [ Weight: 10, visual: 3, step: 1.5 ]

来源：https://stackoverflow.com/questions/55885842/custom-dictionary

标签

python-3.x

list

dictionary

optimization

swarm