I have a numpy array with integer values, let's call this array x.
I want to create some sort of list where for each value, I have the indices of x that hold this value.
For example, for:
x = [1,2,2,4,7,1,1,7,16]
I want to get:
{1: [0,5,6], 2:[1,2], 4:[3], 7:[4,7], 16:[8]}
The parenthesis I used are arbitrary, I don't care which data structure I use as long as I can output my result to a file as quickly as possible. At the end I want a .txt file that reads:
0,5,6
1,2
3
4,7
8
Since you mentioned you're not picky about the data structure of your values,tTo get something like the dictionary you posted in your question, you could do a dictionary comprehension over the unique values in x with np.where for the values:
>>> {i:np.where(x == i)[0] for i in set(x)}
{1: array([0, 5, 6]),
2: array([1, 2]),
4: array([3]),
7: array([4, 7]),
16: array([8])}
Comparing this to a more vanilla loop through a list, this will be significantly faster for larger arrays:
def list_method(x):
res = {i:[] for i in set(x)}
for i, value in enumerate(x):
res[value].append(i)
return res
def np_method(x):
return {i:np.where(x == i)[0] for i in set(x)}
x = np.random.randint(1, 50, 1000000)
In [5]: %timeit list_method(x)
259 ms ± 4.03 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [6]: %timeit np_method(x)
120 ms ± 4.15 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Pure python will be like this:
result = {}
for idx,val in enumerate(x):
arr = result.get(val,[])
arr.append(idx)
result[val] = arr
Fastest method is to use defaultdict (Pure Python with O(n))
from collections import defaultdict
def list_to_unique_dict(data):
unique_dict = defaultdict(list)
for index, value in enumerate(data):
unique_dict[value].append(index)
return dict(unique_dict)
x = [1,2,2,4,7,1,1,7,16]
list_to_unique_dict(x)
Out[62]: {1: [0, 5, 6], 2: [1, 2], 4: [3], 7: [4, 7], 16: [8]}
x = np.random.randint(1, 50, 1000000)
%timeit list_to_unique_dict(x)
232 ms ± 13.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
x = [1,2,2,4,7,1,1,7,16]
numlist = []
numdict = {}
c = 0
for n in x:
if n not in numlist:
numlist.append(n)
numdict[n] = [c]
else:
numdict[n].append(c)
c += 1
print(numlist, numdict)
Output is:
[1, 2, 4, 7, 16] {1: [0, 5, 6], 2: [1, 2], 4: [3], 7: [4, 7], 16: [8]}
To write to file use:
with open('file.txt', 'w') as f:
f.write(str(numdict))
