Appending a value to python dictionary value using the key if the value is a multiple array

Question

I declared a python dictionary with key and value as multiple array. Is it possible to append an array using the key and value index?

This is the way I initialized the python dictionary cvfoldacc

a = []
b = [] 
c = [] 
d = [] 
e = [] 
f = []
classifiers = [a,b,c,d,e,f]
cvfoldacc = dict.fromkeys(range(2,11), classifiers)

And the result of this initialization is as follows:

cvfoldacc>>

{2: [[], [], [], [], [], []],
 3: [[], [], [], [], [], []],
 4: [[], [], [], [], [], []],
 5: [[], [], [], [], [], []],
 6: [[], [], [], [], [], []],
 7: [[], [], [], [], [], []],
 8: [[], [], [], [], [], []],
 9: [[], [], [], [], [], []],
 10: [[], [], [], [], [], []]}

When I tried to append the first list of key(2) with the code cvfoldacc[2][0].append(8), I am getting the result as:

{2: [[8], [], [], [], [], []],
 3: [[8], [], [], [], [], []],
 4: [[8], [], [], [], [], []],
 5: [[8], [], [], [], [], []],
 6: [[8], [], [], [], [], []],
 7: [[8], [], [], [], [], []],
 8: [[8], [], [], [], [], []],
 9: [[8], [], [], [], [], []],
 10: [[8], [], [], [], [], []]}

But the expected answer should be this:

{2: [[8], [], [], [], [], []],
 3: [[], [], [], [], [], []],
 4: [[], [], [], [], [], []],
 5: [[], [], [], [], [], []],
 6: [[], [], [], [], [], []],
 7: [[], [], [], [], [], []],
 8: [[], [], [], [], [], []],
 9: [[], [], [], [], [], []],
 10: [[], [], [], [], [], []]}

Does this answer your question? ["Least Astonishment" and the Mutable Default Argument](https://stackoverflow.com/questions/1132941/least-astonishment-and-the-mutable-default-argument) — Green Cloak Guy, Nov 12 '19 at 06:08
The results are same even after creating separate list for every key. — user2159982, Nov 12 '19 at 06:22

Sayandip Dutta · Answer 1 · 2019-11-12T06:42:08.927

This is the same old deep copy vs shallow copy thing. Try this:

>>> a = []
>>> b = [] 
>>> c = [] 
>>> d = [] 
>>> e = [] 
>>> f = []
>>> classifiers = [a,b,c,d,e,f]
>>> import copy
>>> cvfoldacc = {k:copy.deepcopy(classifiers) for k in range(2,11)}
>>> cvfoldacc[2][0].append(8)
>>> cvfoldacc 
{2: [[8], [], [], [], [], []],
 3: [[], [], [], [], [], []],
 4: [[], [], [], [], [], []],
 5: [[], [], [], [], [], []],
 6: [[], [], [], [], [], []],
 7: [[], [], [], [], [], []],
 8: [[], [], [], [], [], []],
 9: [[], [], [], [], [], []],
 10: [[], [], [], [], [], []]}

You were doing:

a = []
b = [] 
c = [] 
d = [] 
e = [] 
f = []
classifiers = [a,b,c,d,e,f]
cvfoldacc = dict.fromkeys(range(2,11), classifiers)

Now, the dictionary that is getting created, has the same list in every key, not only they look same, those are the exact same objects, having same identity (memory location in CPython). Let's see:

>>> id(cvfoldacc[2])
171760008
>>> id(cvfoldacc[3])
171760008

And these will be same for other values as well. So dict.fromkeys() assigns the same value to all the keys in the dictionary, does not create a copy of it.

Now, another way, where you make a shallow copy, and this issue would be solved:

>>> cvfoldacc = {a:classifiers.copy() for a in range(2,11)}
>>> id(cvfoldacc[2])
171840616
>>> id(cvfoldacc[3])
171847688

Solved right?

>>> cvfoldacc[2][0].append(8)
{2: [[8], [], [], [], [], []],
 3: [[8], [], [], [], [], []],
 4: [[8], [], [], [], [], []],
 5: [[8], [], [], [], [], []],
 6: [[8], [], [], [], [], []],
 7: [[8], [], [], [], [], []],
 8: [[8], [], [], [], [], []],
 9: [[8], [], [], [], [], []],
 10: [[8], [], [], [], [], []]}

Apparently not!! Let's now look DEEPER. Let's look at the ids of the lists inside the lists:

>>> id(cvfoldacc[2][0])
171810120
>>> id(cvfoldacc[3][0])
171810120

So even though list.copy() created a copy of the outerlist, the inner lists are the same. So essentially you are appending in list: a which is present in all the keys so everything is getting modified.

Deepcopy copied all the objects recursively, avoiding the issue.

Could you please explain why the previous method did'nt worked.. — user2159982, Nov 12 '19 at 06:28
do ```[id(cvfoldacc[x]) for x in range(2,11)]``` for your original code. You'll see they are all the same ID - so you only made references to one list. Now do ```id(classifiers)```. Make a change like ```classifiers[0] = [1,2]```. See how it propagates through your dict? That means you made a dict whose values are all references to that original list. This is particularly an issue with any nested sequence. So you always need to make sure you're instantiating new objects in this situation. — neutrino_logic, Nov 12 '19 at 06:38
Glad to help. Also note, the other answer does the same thing, without importing `copy`. Make sure you understand the innerworkings. `{k:[x[:] for x in classifiers] for k in range(2,11)}` here `x[:]` serves as `deepcopy` as it creates a new copy of `a,b,c,d,e,f` in every iteration. — Sayandip Dutta, Nov 12 '19 at 06:45

han solo · Accepted Answer · 2019-11-13T06:58:48.487

You need to create new lists for values of each key like,

>>> a = []
>>> b = [] 
>>> c = [] 
>>> d = [] 
>>> e = [] 
>>> f = []
>>> classifiers = [a,b,c,d,e,f] 
>>> 
>>> d = {k:[x[:] for x in classifiers] for k in range(2,11)} # note i am creating a copy using the `list[:]` notation.
# either the comprehension or `k: copy.deepcopy(classifiers)` is fine
>>> d
{2: [[], [], [], [], [], []], 3: [[], [], [], [], [], []], 4: [[], [], [], [], [], []], 5: [[], [], [], [], [], []], 6: [[], [], [], [], [], []], 7: [[], [], [], [], [], []], 8: [[], [], [], [], [], []], 9: [[], [], [], [], [], []], 10: [[], [], [], [], [], []]}
>>> d[2][0].append(1)
>>> d
{2: [[1], [], [], [], [], []], 3: [[], [], [], [], [], []], 4: [[], [], [], [], [], []], 5: [[], [], [], [], [], []], 6: [[], [], [], [], [], []], 7: [[], [], [], [], [], []], 8: [[], [], [], [], [], []], 9: [[], [], [], [], [], []], 10: [[], [], [], [], [], []]}

Yes, you could do it either way you like, for your current structure — han solo, Nov 12 '19 at 06:30

Appending a value to python dictionary value using the key if the value is a multiple array

2 Answers2