通过使用键作为列表索引将python字典转换/映射为列表

Question

I have a python dictionary as follows:

dict = {4:0.65,8:1.23,3:0.43}

I would like to convert this to a python list by using the key as the index to the list. The desired converted result would be:

listLength = 10
plist = [0,0,0,0.43,0.65,0,0,0,1.23,0]

I know how to do the above using a loop but that is not pythonic and it is not fast. What is the most pythonic way to do the above without using a loop.

I specially need to do the above with the best performance.

Answer 1

Since you tag pandas , solution from reindex

pd.Series(d).reindex(range(10),fill_value=0).tolist()
Out[369]: [0.0, 0.0, 0.0, 0.43, 0.65, 0.0, 0.0, 0.0, 1.23, 0.0]

Answer 2

Using numpy and numpy indexing is going to be the most performant solution:

out = np.zeros(10)
out[list(d.keys())] = list(d.values())

array([0.  , 0.  , 0.  , 0.43, 0.65, 0.  , 0.  , 0.  , 1.23, 0.  ])

Performance since you asked:

k = np.random.randint(1, 100000, 10000)
v = np.random.rand(10000)
d = dict(zip(k, v))

In [119]: %%timeit
     ...: out = np.zeros(100000)
     ...: out[list(d.keys())] = list(d.values())
     ...:
     ...:
1.86 ms ± 13.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [120]: %timeit [d.get(i, 0) for i in range(100000)]
17.4 ms ± 231 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [121]: %timeit pd.Series(d).reindex(range(100000),fill_value=0).tolist()
9.77 ms ± 148 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 3

For larger data sets you can gain some speed using np.fromiter directly on the key and value iterators instead of creating lists first.

Create test case

>>> d = dict(zip(np.random.randint(1, 10, 1_000_000).cumsum(), np.arange(1_000_000.)))
>>> out = np.zeros(10_000_000)

Define fromiter method

>>> def use_iter():
...     k, v = (np.fromiter(w, dtype=t, count=len(d)) for w, t in [(d.keys(), int), (d.values(), float)])
...     out[k] = v
...     return out

and list method for reference

>>> def use_list():
...     out[list(d.keys())] = list(d.values())
...     return out

and time them

>>> timeit(use_iter, number=100)
4.2583943260106025
>>> timeit(use_list, number=100)
17.10310926999955

Also, check correctness

>>> np.all(use_list() == use_iter())
True

Answer 4

You could to something like this:

list_length = 10
d = {4: 0.65, 8: 1.23, 3: 0.43}
plist = [d.get(i, 0) for i in range(list_length)]

print(plist)

Output

[0, 0, 0, 0.43, 0.65, 0, 0, 0, 1.23, 0]

Note: Don't use the name dict for your own variables, you will shadow the built-in name dict .

Answer 5

Avoid shadowing the built-in dict . Use some other name instead.

dict_ = {4:0.65,8:1.23,3:0.43}
length = max(dict_) + 1  # Get number of entries needed
list_ = [0] * length  # Initialize a list of zeroes
for i in dict_:
    list_[i] = dict_[i]

Answer 6

Using list comprehension

lst = [d[i] if i in d else 0 for i in range(10)]
print(lst)
# [0, 0, 0, 0.43, 0.65, 0, 0, 0, 1.23, 0]

Expanded:

lst = []
for i in range(10):
    if i in d:
        lst.append(d[i])
    else:
        lst.append(0)

Answer 7

You can just iterate over the dictionary and place them into a list. I am doing error checking to make sure that the key is within the specified list length.

  list = [0] * length
    for key, val in d.items():
        if key < length:
            list[key] = val

If you want the list to be as big as the max key, follow this bellow

maxKey = max(d, key=int)
list = [0] * maxKey
for key, val in d.items():
    list[key] = val