检查列表值并将其分配给python中的字典键

Question

I have a list of words as below.

mylist = ['cat', 'yellow', 'car', 'red', 'green', 'jeep', 'rat','lorry']

I also have a list of lists for each essay in the dataset that contain values for the 'mylist' as given in the examples below (ie, if 'mylist' word appears in essay I make it 1, otherwise 0).

[[0,1,0,0,0,1,0,1], [1,0,0,0,0,1,0,0]]

In other words,

[0,1,0,0,0,1,0,1] says that this only has values 'yellow', 'jeep', 'lorry'

Now I have a dictionary of categories as below.

mydictionary = {'colour': ['red', 'yellow', 'green'], 'animal': ['rat','cat'], 
'vehicle': ['car', 'jeep']}

Now by using 'mydictionary' key values I want to transform the list of lists as follows (That is, if one or more values of the 'mylist' is 1, I mark the key as 1, else 0).

[[1,0,1], [0,1,0]]

In other words,

[1,0,1] says that;
1 - one or more '1's for elements in 'colours'
0 - no elements in 'animals'
0 - one or more '1's for elements in 'vehicles'

So my output should be a list of lists as mentioned above -> [[1,0,1], [0,1,0]]

I am new to pandas, Hence, I am interested in knowing if this is possible to do using pandas dataframes.

Answer 1

Setup

a = np.array(['cat', 'yellow', 'car', 'red', 'green', 'jeep', 'rat','lorry'])
b = np.array([[0,1,0,0,0,1,0,1], [1,0,0,0,0,1,0,0]], dtype=bool)

mydictionary = {
    'colour': ['red', 'yellow', 'green'],
    'animal': ['rat','cat'], 
    'vehicle': ['car', 'jeep']
}

---

Solution
Some minor additional setup
I just needed to get an array of sets in the correct order.

o = ['colour', 'animal', 'vehicle']
s = pd.Series(mydictionary).apply(set).loc[o]

s

colour     {green, red, yellow}
animal               {cat, rat}
vehicle             {jeep, car}
dtype: object

Use set intersection with numpy broadcasting

(s.values & [[set(a[l])] for l in b]).astype(bool).astype(int)

array([[1, 0, 1],
       [0, 1, 1]])

---

Additional Explanation

If I'm to use numpy broadcasting and I already have a series with values

s.values

[{'green', 'red', 'yellow'} {'cat', 'rat'} {'jeep', 'car'}]

Then I need a 2-D array with the other sets

[[set(a[l])] for l in b]

[[{'jeep', 'lorry', 'yellow'}], [{'cat', 'jeep'}]]

When I broadcast the & operation

s.values & [[set(a[l])] for l in b]

[[{'yellow'}  set()    {'jeep'}]
 [set()       {'cat'}  {'jeep'}]]

Conveniently, empty sets evaluate to False and non-empty sets to True in a bool context. Follow that with an int context and we have our solution.

(s.values & [[set(a[l])] for l in b]).astype(bool).astype(int)

array([[1, 0, 1],
       [0, 1, 1]])

Answer 2

I think you need:

mylist = ['cat', 'yellow', 'car', 'red', 'green', 'jeep', 'rat','lorry']
a = [[1,1,0,0,0,1,0,1], [1,0,0,0,0,1,0,0]]
mydictionary = {'colour': ['red', 'yellow', 'green'], 'animal': ['rat','cat', 'lorry'], 
'vehicle': ['car', 'jeep']}
#order of output categories
cols = ['colour','animal','vehicle']

df = pd.DataFrame(a, columns=mylist)
d = {k: oldk for oldk, oldv in mydictionary.items() for k in oldv}
df = df.rename(columns=d).groupby(axis=1, level=0).max().reindex(columns=cols)
print (df)
   colour  animal  vehicle
0       1       1        1
1       0       1        1

L = df.values.tolist()
print (L)
[[1, 1, 1], [0, 1, 1]]

Answer 3

Here is another approach without pandas:

list_of_list = <whatever you have>
for i, list in enumerate(list_of_list):
     #  temp_list will hold lists such [yellow, jeep, lorry]
     temp_list = [mylist[j] for j in range(len(list)) if list[j] == 1]

     for t, item in enumerate(temp_list):
           for k, key in enumerate(mydictionary.keys()):
               if item in mydictionary[key]:
                  temp_list[t] = k

     # now override the list of list
     list_of_list[i] = temp_list[i]

I didn't run the code. So, there might be some minor bugs. But, I am hoping you get the idea