如果列表包含浮点数，如何在列表中选择相同的值

Question

In the following code I want to check how many unique values are in the list and this can be done in for loop. After knowing the number of unique values I want to see how many times a single unique values appear in a and then I want to count their number. Can someone please guide me how to do that. List contains floating points. What if I convert it in numpy array and then find same values.

`a= [1.0, 1.0, 1.0, 1.0, 1.5, 1.5, 1.5, 3.0, 3.0]
list = []
for i in a:
    if i not in list:
        list.append(i)

print(list)
for j in range(len(list))
    g= np.argwhere(a==list[j])
    print(g)`

Answer 1

You can use np.unique to get it done

np.unique(np.array(a),return_counts=True)

You can also do it using counters from collections

from collections import Counter
Var=dict(Counter(a))
print(Var)

The primitive way is to use loops

[[x,a.count(x)] for x in set(a)]

If you are not familiar with list comprehensions, this is its explaination

ls=[]
for x in set(a):
    ls.append([x,a.count(x)])
print(ls)

If you want it using if else,

counter = dict()
for k in a:
    if not k in counter:
        counter[k] = 1
    else:
        counter[k] += 1
print(counter)