将字典的字符串转换为int

Question

I have a dictionary loaded:

{'17.0x6.0x7.0 6.0': ['4.60x4.30x4.30 4.00, 4.60x4.30x4.30 1.00, 4.60x4.30x4.30 3.00'],
 '9.5x5.5x5.5 1.0': ['4.60x4.30x4.30 5.00'],
 '22.0x7.5x8.0 10.0': ['6.60x6.00x5.16 8.00, 9.00x6.00x5.60 6.00'],
 '17.0x6.0x7.0 6.0_1': ['8.75x6.60x5.60 7.00'],
 '9.5x5.5x5.5 2.0': ['4.60x4.30x4.30 2.00']}

I want to extract the float values after dimensions

Desired output:

{6:[4,1,3],
 1:[5],
 10:[8,6],
 6:[7]
 2:[2]}

I tried this to extract the dimensions

loaded_items = {int(float(k.split()[1])): [int(float(j.split()[1])) for i in v for j in i.split(",")] for k,v in loaded.items()}

but this gives me output as such:

{6: [7], 1: [5], 10: [8, 6], 2: [2]}

ie when it encounters the second similar object it replaces the value.

I tried one more method:

all_items_loaded = {}
for k,v in loaded.items():
    a = k.split()[1]
    print(a)
    for i in v:
        x = i.split()
        x = x[1::2]
        all_items_loaded[a] = x

and the output is:

{'6.0': ['4.00,', '1.00,', '3.00'],
 '1.0': ['5.00'],
 '10.0': ['8.00,', '6.00'],
 '6.0_1': ['7.00'],
 '2.0': ['2.00']}

Answer 1

There are problems with your data and the required output.

The input dict contains a key '17.0x6.0x7.0 6.0_1 '. While calling int(float()) on '6.0_1' works (it is cast to 6.01 and subsequently to 6) I would not recommend this and rather investigate the reason why this data is supplied in this way.

The desired output dict contains duplicate keys. This is not possible and causes the oerwriting you are describing. Change your output data structure from dict to eg list of dicts.

[{int(float(key.split()[1])): [int(float(val.split()[1])) for val in value[0].split(",")]} for (key,value) in a.items() ]
--> [{6: [4, 1, 3]}, {1: [5]}, {10: [8, 6]}, {6: [7]}, {2: [2]}]

Answer 2

You either need to decide on a different way of associating the data or choose a different data structure to use. One example, a defaultdict , is below. This is not the only solution and getting a better input would be the first priority.

from collections import defaultdict

d = defaultdict(list)
d['6.0'].append([7])
d['6.0'].append([4,1,3])

print(d['6.0'])

# result [[7], [4, 1, 3]]

Answer 3

The only suggestion I can make (aside from getting a better input) is to create tuples. Create a simple function to extract values from strings:

>>> def extract(v): return int(float(v.split()[1]))

And apply this function to dictionary key-values pairs and produce a list of pairs:

>>> loaded_items = {
...     '17.0x6.0x7.0 6.0': ['4.60x4.30x4.30 4.00, 4.60x4.30x4.30 1.00, 4.60x4.30x4.30 3.00'],
...     '9.5x5.5x5.5 1.0': ['4.60x4.30x4.30 5.00'],
...     '22.0x7.5x8.0 10.0': ['6.60x6.00x5.16 8.00, 9.00x6.00x5.60 6.00'],
...     '17.0x6.0x7.0 6.0_1': ['8.75x6.60x5.60 7.00'],
...     '9.5x5.5x5.5 2.0': ['4.60x4.30x4.30 2.00']
... }
>>> data = [(extract(k), [extract(v) for v in vs[0].split(",")])  for k, vs in loaded_items.items()]
>>> data
[(6, [4, 1, 3]), (1, [5]), (10, [8, 6]), (6, [7]), (2, [2])]

You can then group the values if needed:

>>> d1 = {}
>>> for k,v in data: d1.setdefault(k, []).extend(v)
>>> d1
{6: [4, 1, 3, 7], 1: [5], 10: [8, 6], 2: [2]}

Or:

>>> d2 = {}
>>> for k,v in data: d2.setdefault(k, []).append(v)
>>> d2
{6: [[4, 1, 3], [7]], 1: [[5]], 10: [[8, 6]], 2: [[2]]}