Python: Assign Labels to values in an array

Question

I have an array which represents some time series data:

array([[[-0.59776013],
    [-0.59776013],
    [-0.59776013],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [ 0.31863936],
    [ 0.31863936],
    [ 0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [ 0.59776013],
    [ 0.59776013],
    [ 0.59776013],
    [ 0.93458929],
    [ 0.93458929],
    [ 0.93458929],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.06270678],
    [-0.06270678],
    [-0.06270678],
    [-0.06270678],
    [-0.06270678],
    [-0.06270678],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [ 0.75541503],
    [ 0.75541503],
    [ 0.75541503],
    [ 0.93458929],
    [ 0.93458929],
    [ 0.93458929],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [ 0.75541503],
    [ 0.75541503],
    [ 0.75541503],
    [-0.31863936],
    [-0.31863936],
    [-0.31863936],
    [ 0.31863936],
    [ 0.31863936],
    [ 0.31863936],
    [ 0.        ],
    [ 0.        ],
    [ 0.        ],
    [ 0.        ],
    [ 0.        ],
    [ 0.        ],
    [ 0.        ],
    [ 0.        ]]])

The unique values in this array are:

np.unique(sax_dataset_inv)
array([-0.59776013, -0.31863936, -0.06270678,  0.        ,  0.31863936,
    0.59776013,  0.75541503,  0.93458929])

My task

Assign either 'F' for fast, 'S' for slow or 'M' for medium to a given array value.

My attempt

I can do it for 2 assignments, 'F' or 'S':

sax_list = ['F' if element < 0 else 'S' for element in list(sax_dataset_inv.flatten())]

However I cannot understand how I can do the same expression above for 3 different labels.

Desired Output

Take an example array of [-3-2-1,0,1,2,3,4,5,6]

The values -3 to -1 inclusive should be assigned 'F'. Values 0 to 3 inclusive should be assigned 'M'. Values greater than 3 should be assigned 'S'.

Helpful: [ternary operator in Python](https://stackoverflow.com/questions/394809/does-python-have-a-ternary-conditional-operator) — blacksite, Feb 20 '19 at 13:19
also what to do with values greater than -1 and smaller than 0 ? — , Feb 20 '19 at 13:21
Have you considered initializing your resulting array with 'M' and then checking if the entry is bigger/smaller than your bounds? — DocDriven, Feb 20 '19 at 13:23
My attempt has been highlighted above. I edited the question to show that I can do the task for two labels. If element<0 assign 'F', else assign 'S'. Does If element<0 assign 'F', if element in range(0,0.4)assign 'M', else assign 'M' make sense? — Mazz, Feb 20 '19 at 14:07

Mohit Motwani · Answer 1 · 2019-02-20T13:39:05.727

You can do this:

arr = np.array([-3-2-1,0,1,2,3,4,5,6])
new_arr = np.zeros(shape = arr.shape, dtype=np.str)

new_arr[(arr>3)] = 'S'
new_arr[((arr>=-3) & (arr<=-1))] = 'F'
new_arr[((arr>=0)&(arr<=3))] = 'M'
new_arr
array(['', 'M', 'M', 'M', 'M', 'S', 'S', 'S'], dtype='<U1')

The values that don't match your condition will remain empty strings.

Also you can use numpy.empty to initialize an empty array:

new_arr = np.empty(shape = arr.shape, dtype=np.str)

jezrael · Answer 2 · 2019-02-20T14:41:26.953

Use numpy.select for vectorized solution:

new_arr = np.select([arr>3, (arr>=-3) & (arr<=-1), (arr>=0)&(arr<=3)],
                    ['S','F','M'], 
                    default='')
print (new_arr)

['F' 'F' 'F' 'M' 'M' 'M' 'M' 'S' 'S' 'S']

Performance:

arr = np.array([-3,-2,-1,0,1,2,3,4,5,6] * 1000)

my_list = [-3,-2,-1,0,1,2,3,4,5,6] * 1000

In [276]: %timeit my_list_mapping = ['F' if ((i >= -3) & (i <= -1)) else 'M' if ((i >= 0) & (i <= 3)) else 'S' for i in my_list]
1.14 ms ± 67.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [277]: %timeit np.select([arr>3, (arr>=-3) & (arr<=-1), (arr>=0)&(arr<=3)],['S','F','M'],  default='')
172 µs ± 7.35 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

cph_sto · Accepted Answer · 2019-02-20T13:32:19.510

1

Note: OP's intervals are [-3,-1], [0,3] & (3,..), so I am only assuming integral values. The conditions can be altered accordingly, but the design remains.

Using list comprehensions for if-elif-else:

my_list = [-3,-2,-1,0,1,2,3,4,5,6]

my_list_mapping = ['F' if ((i >= -3) & (i <= -1)) else 'M' if ((i >= 0) & (i <= 3)) else 'S' for i in my_list]
print(my_list_mapping)
    ['F', 'F', 'F', 'M', 'M', 'M', 'M', 'S', 'S', 'S']

edited Feb 20 '19 at 13:32

answered Feb 20 '19 at 13:26

cph_sto

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if value is `-0.5` the array element will be `'S'`, right ? – Feb 20 '19 at 13:28
Well, I assumed that OP is concerned with only integral values, because he specified the intervals as such. – cph_sto Feb 20 '19 at 13:28
no hard feelings, it's the OPs description of the problem that's flawed – Feb 20 '19 at 13:30
His intervals are `[-3,-1], [0,3], (3,..)` – cph_sto Feb 20 '19 at 13:30
This is what I was looking for. My if-elif-else condition was wrong. Thanks! – Mazz Feb 20 '19 at 14:16
1

@Murray Pleasure. You have redeemed me ;) – cph_sto Feb 20 '19 at 14:42

score 1 · Answer 4 · answered Feb 20 '19 at 13:38

You can neatly achieve your result by using the map function from Python core library Let's say your mapping function would look something like this:

    def mapping_function(value):
        if value >= -3 and value <= -1:
            return 'F'
        elif value >= 0 and value <= 3:
            return 'M'
        elif value > 3:
            return 'S'
        else:
            return 'U'  # eg. as undefined
    my_array = [-3, -2, -1, 0, 1, 2, 3, 4, 5, 6]
    mapped_vals = map(mapping_function, my_array)
    # if you want a list
    my_mapped_list = list(mapped_vals)

I think it is a readable and clear solution.

Since you have tagged it as a pandas question, take a look at pandas.Series.apply function from their docs. Instead of using it as map, you pass the mapping function as an argument to the apply function.

Python: Assign Labels to values in an array

4 Answers4