Transform all floats ending in .0 in a dict to int

Question

How can I transform all float values in a dict ending in .0 to an int? Is there an elegant solution for this?

 {
   '4061_dummy': {
     'mae': 278.86288346930246,
     'best_par': {
       'metric': 'minkowski',
       'weights': 'distance',
       'n_neighbors': 6.0
     }
   },
   '4181_fourier': {
     'mae': 604.05,
     'best_par': {
       'metric': 'l1', 
       'weights': 'distance', 
       'n_neighbors': 1.0
     }
   },
   '4905_fourier': {
     'mae': 610.8769965277778,
     'best_par': {
       'metric': 'minkowski',
       'weights': 'uniform',
       'n_neighbors': 24.0
     }
   },
   'time': 1.47
 }

My solution for this problem you can see in the following.

for i in a:
    if isinstance(a[i], dict):
        for key in a[i]["best_par"]:
            if isinstance(a[i]["best_par"][key], float):
                if int(a[i]["best_par"][key]) == a[i]["best_par"][key] and isinstance(a[i]["best_par"][key], float):
                    a[i]["best_par"][key] = int(a[i]["best_par"][key])
                    print(a[i]["best_par"][key])

Iterate over your dictionary, check if `x == int(x)` and if so, replace `x` with `int(x)`? — Chris, Jan 26 '22 at 18:48
There seem to be three different steps needed: 1) walk the dict and find float values; 2) [check if the float is a whole number](/q/21583758/4518341); 3) convert it to int. Which part(s) do you need help with? Please [edit] to clarify. For more tips, see [ask]. — wjandrea, Jan 26 '22 at 18:52
@12944qwerty Brilliant solution! I'll transform the dict to str and then I can replace it like you said — Julio Reckin, Jan 26 '22 at 18:56
@Chris the comparison would not work in all cases. Perhaps, it's safer to change the data to string and check the pattern `[0-9]+.0`. — Amin, Jan 26 '22 at 19:18
Replacing all `".0"` with `""` would turn `604.05` into `6045`, would it not? — Chris, Jan 26 '22 at 19:19

score 3 · Accepted Answer · answered Jan 26 '22 at 19:05

3

Recursive solution:

def convert_integer_floats_to_ints(d):
    for key, val in d.items():
        if isinstance(val, float) and val == int(val):                
            d[key] = int(val)
        elif isinstance(val, dict):
            convert_integer_floats_to_ints(val)

answered Jan 26 '22 at 19:05

MikeM

13,156
2
34
47

Use `if isinstance(val, float) and val.is_integer():` – juanpa.arrivillaga Jan 26 '22 at 19:23
@juanpa.arrivillaga Yes, agreed, that would be better. – MikeM Jan 26 '22 at 19:31

score 1 · Answer 2 · answered Jan 26 '22 at 18:59

the only solution i can think of:

for i in data.keys():
    if i!='time': data[i]['best_par']['n_neighbors'] = int(data[i]['best_par']['n_neighbors'])

Output:

{'4061_dummy': {'best_par': {'metric': 'minkowski',
   'n_neighbors': 6,
   'weights': 'distance'},
  'mae': 278.86288346930246},
 '4144_dummy': {'best_par': {'metric': 'l1',
   'n_neighbors': 4,
   'weights': 'uniform'},
  'mae': 116.66041666666666},
 '4181_fourier': {'best_par': {'metric': 'l1',
   'n_neighbors': 1,
   'weights': 'distance'},
  'mae': 604.05},
 '4331_fourier': {'best_par': {'metric': 'minkowski',
   'n_neighbors': 1,
   'weights': 'distance'},
  'mae': 305.13958333333335},
 '4555_fourier': {'best_par': {'metric': 'minkowski',
   'n_neighbors': 1,
   'weights': 'distance'},
  'mae': 888.1520833333333},
 '4583_fourier': {'best_par': {'metric': 'l1',
   'n_neighbors': 1,
   'weights': 'distance'},
  'mae': 601.28125},
 '4853_dummy': {'best_par': {'metric': 'minkowski',
   'n_neighbors': 2,
   'weights': 'distance'},
  'mae': 77.25801122256406},
 '4905_fourier': {'best_par': {'metric': 'minkowski',
   'n_neighbors': 24,
   'weights': 'uniform'},
  'mae': 610.8769965277778},
 'time': 1.47}

Tim Roberts · Answer 3 · 2022-01-26T19:08:48.290

-2

It's quite straightforward.

for v in data.values():
    if isinstance(v,dict):
        v['best_par']['n_neighbors'] = int(v['best_par']['n_neighbors'])

This assumes you don't really need to troll through the deep structure LOOKING for such values. That's harder.

edited Jan 26 '22 at 19:08

answered Jan 26 '22 at 18:49

Tim Roberts

48,973
4
21
30

2

This converts all floats to integers. Not the ones that are ending in `.0` – 12944qwerty Jan 26 '22 at 18:50
1

Look at his data. The variable is called `n_neighbors`. By its name, you can tell that's always going to be an integral value. It's easy enough to detect, but in this particular case, unnecessary. The OP has described his problem incorrectly. – Tim Roberts Jan 26 '22 at 18:52
It'll throw an error when reaching the 'time' key. – Jorge L. Jan 26 '22 at 18:56
That's a valid concern. I'll change that. – Tim Roberts Jan 26 '22 at 19:08

Transform all floats ending in .0 in a dict to int

3 Answers3