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Trying to convert pandas df to np array, dtaidistance computes list instead

I am attempting to compute the distance matrix for an ndarray that I have converted from pandas. I tried to convert the pandas df currently in this format:
move_df =
movement
0 [4, 3, 6, 2]
1 [5, 2, 3, 6, 2]
2 [4, 7, 2, 3, 6, 1]
3 [4, 4, 4, 3]
... ...
33410 [2, 6, 3, 1, 8]
[33410 x 1 columns]
to a numpy ndarray by using the following:
1) m = move_df.to_numpy()
2) m = pd.DataFrame(move_df.tolist()).values
3) m = [move_df.tolist() for i in move_df.columns]
Each of these conversions resulted in a numpy array in this format:
[[list([4, 3, 6, 2])]
[list([5, 2, 3, 6, 2])]
[list([4, 7, 2, 3, 6, 1])]
[list([4, 4, 4, 3])]
...
[list([2, 6, 3, 1, 8])]]
So when I try to run dtaidistance matrix, I get the following error:
d_m = dtw.distance_matrix(m)
TypeError: unsupported operand type(s) for -: 'list' and 'list'
But when I create a list of lists by copying and pasting several of the numpy arrays created with any of the methods mentioned above, the code works. But this is not feasible in the long run since the arrays are over 30k rows. Is there something I am doing wrong in the conversion from pandas df to numpy array? I used
print(type(m))
and it outputs that it is a numpy array and I already know that I cannot subtract a list from a list, hence the error.
EDIT:
For move_df.head(10).to_dict()
{'movement': {0: [4, 3, 6, 2],
1: [5, 2, 3, 6, 2],
2: [4, 7, 2, 3, 6, 1],
3: [4, 4, 4, 3],
4: [3, 6, 2, 3, 3],
5: [6, 2, 1],
6: [1, 1, 1, 1],
7: [7, 2, 3, 1, 1],
8: [7, 2, 3, 2, 1],
9: [6, 2, 3, 1]}}

(one of the dtaidistance authors here)
The dtaidistance package expects one of three formats:
A 2D numpy array (where all sequences have the same length by definition)
A Python list of 1D numpy.array or array.array.
A Python list of Python lists
In your case you could do:
series = move_df['movement'].to_list()
dtw.distance_matrix(series)
which works then on a list of lists.
To use the fast C implementation an array is required (either Numpy or std lib array). If you want to keep different lengths you can do
series = move_df['movement'].apply(lambda a: np.array(a, dtype=np.double)).to_list()
dtw.distance_matrix_fast(series)
Note that it might make sense to do the apply operation inplace on your move_df datastructure such that you only have to do it once and not keep track of two nearly identical datastructures. After you do this, the to_list call is sufficient. Thus:
move_df['movement'] = move_df['movement'].apply(lambda a: np.array(a, dtype=np.double))
series = move_df['movement'].to_list()
dtw.distance_matrix_fast(series)
If you want to use a 2D numpy matrix, you would need to truncate or pad all series to be the same length as is explained in other answers (for dtw padding is more common to not lose information).
ps. This assumes you want to do univariate DTW, the ndim subpackage for multivariate time series expects a different datastructure.

Assuming you want to form an array with the lists of length 4:
m = df['movement'].str.len().eq(4)
a = np.array(df.loc[m, 'movement'].to_list())
output:
array([[4, 3, 6, 2],
[4, 4, 4, 3],
[1, 1, 1, 1],
[6, 2, 3, 1]])
used input:
df = pd.DataFrame({'movement': [[4, 3, 6, 2],
[5, 2, 3, 6, 2],
[4, 7, 2, 3, 6, 1],
[4, 4, 4, 3],
[3, 6, 2, 3, 3],
[6, 2, 1],
[1, 1, 1, 1],
[7, 2, 3, 1, 1],
[7, 2, 3, 2, 1],
[6, 2, 3, 1]]})

A dataframe created with:
In [112]: df = pd.DataFrame({'movement': {0: [4, 3, 6, 2],
...: 1: [5, 2, 3, 6, 2],
...: 2: [4, 7, 2, 3, 6, 1],
...: 3: [4, 4, 4, 3],
...: 4: [3, 6, 2, 3, 3],
...: 5: [6, 2, 1],
...: 6: [1, 1, 1, 1],
...: 7: [7, 2, 3, 1, 1],
...: 8: [7, 2, 3, 2, 1],
...: 9: [6, 2, 3, 1]}})
has an object dtype column that contains lists. The array derived from that column is object dtype:
In [121]: arr = df['movement'].to_numpy()
In [122]: arr
Out[122]:
array([list([4, 3, 6, 2]), list([5, 2, 3, 6, 2]),
list([4, 7, 2, 3, 6, 1]), list([4, 4, 4, 3]),
list([3, 6, 2, 3, 3]), list([6, 2, 1]), list([1, 1, 1, 1]),
list([7, 2, 3, 1, 1]), list([7, 2, 3, 2, 1]), list([6, 2, 3, 1])],
dtype=object)
By selecting the column I get a 1d array, not the 2d you get. Otherwise it's the same
This cannot be converted into a 2d numeric dtype array. For most purposes we can think of this as a list of lists.
In [123]: arr.tolist()
Out[123]:
[[4, 3, 6, 2],
[5, 2, 3, 6, 2],
[4, 7, 2, 3, 6, 1],
[4, 4, 4, 3],
[3, 6, 2, 3, 3],
[6, 2, 1],
[1, 1, 1, 1],
[7, 2, 3, 1, 1],
[7, 2, 3, 2, 1],
[6, 2, 3, 1]]
If the lists were all the same length, or if we pick a subset, it is possible to construct a 2d array:
In [125]: arr[[0,3,6,9]]
Out[125]:
array([list([4, 3, 6, 2]), list([4, 4, 4, 3]), list([1, 1, 1, 1]),
list([6, 2, 3, 1])], dtype=object)
In [126]:
In [126]: np.stack(arr[[0,3,6,9]])
Out[126]:
array([[4, 3, 6, 2],
[4, 4, 4, 3],
[1, 1, 1, 1],
[6, 2, 3, 1]])
Padding and slicing could also be used to force the lists to matching lengths - but that could mean losing information.
But without knowing what dtw.distance_matrix expects (looks like it wants a 2d numeric array), or what these lists represent, I can't go further.
The fundamental point is that your dataframe contains lists that vary in length.

numpy.concatenate float64(101,1) and float64(101,)

I'm a MatLab user who recently converted to python. I am running a for loop that cuts a longer signal into individual trials, normalizes them to 100% trial and then would like to have the trials listed horizontally in a single variable. My code is
RHipFE=np.empty([101, 1])
newlength = 101
for i in range(0,len(R0X)-1,2):
iHipFE=redataf.RHipFE[R0X[i]:R0X[i+1]]
x=np.arange(0,len(iHipFE),1)
new_x = np.linspace(x.min(), x.max(), newlength)
iHipFEn = interpolate.interp1d(x, iHipFE)(new_x)
RHipFE=np.concatenate((RHipFE,iHipFEn),axis=1)
When I run this, I get the error "ValueError: all the input arrays must have same number of dimensions". Which I assume is because RHipFE is (101,1) while iHipFEn is (101,). Is the best solution to make iHipFEn (101,1)? If so, how does one do this in the above for loop?

Generally it's faster to collect arrays in a list, and use some form of concatenate once. List append is faster than concatenate:
In [51]: alist = []
In [52]: for i in range(3):
...: alist.append(np.arange(i,i+5))
...:
In [53]: alist
Out[53]: [array([0, 1, 2, 3, 4]), array([1, 2, 3, 4, 5]), array([2, 3, 4, 5, 6])]
Various ways of joining
In [54]: np.vstack(alist)
Out[54]:
array([[0, 1, 2, 3, 4],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6]])
In [55]: np.column_stack(alist)
Out[55]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])
In [56]: np.stack(alist, axis=1)
Out[56]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])
In [57]: np.array(alist)
Out[57]:
array([[0, 1, 2, 3, 4],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6]])
Internally, vstack, column_stack, stack expand the dimension of the components, and concatenate on the appropriate axis:
In [58]: np.concatenate([l[:,None] for l in alist],axis=1)
Out[58]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])

Combinatorics: list with alternative elements

I have a python list with 1D numpy arrays as elements, which have one or more elements. Consider each of the array elements as alternatives for the respective list element.
An example:
[array([1]),array([2]),array([2,3]),array([3]),array([4]),array([3,4,5])]
I want a two things:
1) All combinations regarding the alternatives:
array([[1,2,2,3,4,3],
[1,2,3,3,4,3],
[1,2,2,3,4,4],
[1,2,3,3,4,4],
[1,2,2,3,4,5],
[1,2,3,3,4,5]])
2. The combination that has the least amount repetitions:
array([1,2,2,3,4,5])
or
array([1,2,3,3,4,5]).
The second should not be so hard to get, once one has the first thing, but I am not sure.
I would also like to use my more complex setups like
datasets_complete = [("iris1", iris1), ("iris2", iris2)]
percentages = [0.05, 0.1, 0.2, 0.5]
imputers = [SimpleFill(), KNN(k=3), SoftImpute(), MICE()]
gridWidths = [0.1, 0.2]
seq = [datasets_complete, percentages, imputers, gridWidths]
testgrid = all_combinations(seq)
where iris1 and iris2 are pandas DataFrames.

Here's a NumPy based approach -
def all_combs(a): # Parte-1
num_combs = np.prod(list(map(len,a)))
return np.array(np.meshgrid(*a)).reshape(-1,num_combs).T
def get_minrep_combs(a): # Parte-2
out = all_combs(a)
counts = (np.diff(np.sort(out,axis=1),axis=1)==0).sum(1)
return out[counts == counts.min()]
Sample run -
In [161]: a = [np.array([1]),np.array([2]),np.array([2,3]),np.array([3]),\
...: np.array([4]),np.array([3,4,5])]
In [162]: all_combs(a) # Part-1 results
Out[162]:
array([[1, 2, 2, 3, 4, 3],
[1, 2, 2, 3, 4, 4],
[1, 2, 2, 3, 4, 5],
[1, 2, 3, 3, 4, 3],
[1, 2, 3, 3, 4, 4],
[1, 2, 3, 3, 4, 5]])
In [163]: get_minrep_combs(a) # Part-2 results
Out[163]:
array([[1, 2, 2, 3, 4, 5],
[1, 2, 3, 3, 4, 5]])
Just to give you guys a sense of all_combs, here's a bit more "normal" sample case runs -
In [166]: a = [np.array([2,3]), np.array([5,6,7])]
In [167]: all_combs(a)
Out[167]:
array([[2, 5],
[3, 5],
[2, 6],
[3, 6],
[2, 7],
[3, 7]])
In [164]: a = [np.array([2,3,4]), np.array([5,6,7,9])]
In [165]: all_combs(a)
Out[165]:
array([[2, 5],
[3, 5],
[4, 5],
[2, 6],
[3, 6],
[4, 6],
[2, 7],
[3, 7],
[4, 7],
[2, 9],
[3, 9],
[4, 9]])
For performance
For performance, we can avoid the transpose in part-1 and perform the operations in part-2 along the columns (axis=0) and also use slicing to avoid np.diff and thus have one optimized version, like so -
def get_minrep_combs_optimized(a): # Parte-1,2
num_combs = np.prod(list(map(len,a)))
out = np.array(np.meshgrid(*a)).reshape(-1,num_combs)
sorted_out = np.sort(out,axis=0)
counts = (sorted_out[1:] == sorted_out[:-1]).sum(0)
return out[:,counts == counts.min()].T
Sample run -
In [188]: a = [np.array([1]),np.array([2]),np.array([2,3]),np.array([3]),\
...: np.array([4]),np.array([3,4,5])]
In [189]: get_minrep_combs_optimized(a)
Out[189]:
array([[1, 2, 2, 3, 4, 5],
[1, 2, 3, 3, 4, 5]])
Runtime test
Here's one way to create a sample input data, which has upto 3 elems and each sub-list has some matches across elements in other sub-lists -
In [42]: lens = np.random.randint(1,4,(20))
In [43]: a = [np.random.randint(1,10,L) for L in lens]
In [44]: lens
Out[44]: array([1, 1, 2, 2, 2, 2, 1, 2, 3, 2, 1, 2, 2, 3, 1, 1, 3, 2, 2, 3])
In [45]: a
Out[45]:
[array([8]),
array([8]),
array([7, 9]),
array([5, 5]),
array([6, 4]),
array([3, 1]),
array([8]),
array([1, 9]),
array([9, 5, 7]),
array([1, 1]),
array([3]),
array([1, 5]),
array([5, 5]),
array([7, 9, 2]),
array([5]),
array([1]),
array([3, 2, 9]),
array([3, 7]),
array([5, 3]),
array([2, 7, 3])]
Timings -
In [46]: %timeit leastReps(combinations(a)) ##Daniel Forsman's soln
1 loops, best of 3: 330 ms per loop
In [47]: %timeit get_minrep_combs_optimized(a)
10 loops, best of 3: 28.7 ms per loop
Let's have more matches -
In [50]: a = [np.random.randint(1,4,L) for L in lens]
In [51]: %timeit leastReps(combinations(a)) ##Daniel Forsman's soln
1 loops, best of 3: 328 ms per loop
In [52]: %timeit get_minrep_combs_optimized(a)
10 loops, best of 3: 29.5 ms per loop
Doesn't change much for the performance difference.

You could use the itertools.product function like this:
import itertools
from numpy import array
test = [array([1]),array([2]),array([2,3]),array([3]),array([4]),array([3,4,5])]
combinations = [list(tup) for tup in itertools.product(*test)]
print(combinations)
This returns:
[[1, 2, 2, 3, 4, 3],
[1, 2, 2, 3, 4, 4],
[1, 2, 2, 3, 4, 5],
[1, 2, 3, 3, 4, 3],
[1, 2, 3, 3, 4, 4],
[1, 2, 3, 3, 4, 5]]
Part number 2 is not solvable, as there can be non-unique solutions...

def combinations(arrList):
mesh=np.meshgrid(*arrList)
mesh=[arr.ravel() for arr in mesh]
return np.array(mesh).T
def leastReps(combs):
uniques=np.array([np.unique(arr).size for arr in list(combs)])
mostUni = (uniques == np.max(uniques))
return combs[mostUni]
Only difference between mine and Divakar's is that I don't need to calculate the number of products in advance, and I use most uniques and he uses least repetitions, which should be equivalent.

Is there a way to loop through the return value of np.where?

Is there a way to loop-through this tuple(?) where the left array are positions in an array and the right array is the value I would like to insert into the given positions:
(array([ 0, 4, 6, ..., 9992, 9996, 9997]), array([3, 3, 3, ..., 3, 3, 3]))
The output above is generated from the following piece of code:
np.where(h2 == h2[i,:].max())[1]
I would like the result to be like this:
array[0] = 3
array[4] = 3
...
array[9997] = 3

Just use a simple indexing:
indices, values = my_tuple
array[indices] = values
If you don't have the final array yet you can create it using a desire function like np.zeros, np.ones, etc. with a size as the size of maximum index.

I think you want the transpose of the where tuple:
In [204]: x=np.arange(1,13).reshape(3,4)
In [205]: x
Out[205]:
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
In [206]: idx=np.where(x)
In [207]: idx
Out[207]:
(array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2], dtype=int32),
array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3], dtype=int32))
In [208]: ij=np.transpose(idx)
In [209]: ij
Out[209]:
array([[0, 0],
[0, 1],
[0, 2],
[0, 3],
[1, 0],
[1, 1],
[1, 2],
[1, 3],
[2, 0],
[2, 1],
[2, 2],
[2, 3]], dtype=int32)
In fact there's a function that does just that:
np.argwhere(x)
Iterating on ij, I can print:
In [213]: for i,j in ij:
...: print('array[{}]={}'.format(i,j))
...:
array[0]=0
array[0]=1
array[0]=2
zip(*) is a list version of transpose:
for i,j in zip(*idx):
print(i,j)

Make a list of ranges in numpy

I want to make a list of integer sequences with random start points. The way I would do this in pure python is
x = np.zeros(1000, 10) # 1000 sequences of 10 elements each
starts = np.random.randint(1, 1000, 1000)
for i in range(len(x)):
x[i] = np.arange(starts[i], starts[i] + 10)
I wonder if there is a more elegant way of doing this using Numpy functionality.

You can use broadcasting after extending starts to a 2D version and adding in the 1D range array, like so -
x = starts[:,None] + np.arange(10)
Explanation
Let's take a small example for starts to see what that broadcasting does in this case.
In [382]: starts
Out[382]: array([3, 1, 3, 2])
In [383]: starts.shape
Out[383]: (4,)
In [384]: starts[:,None]
Out[384]:
array([[3],
[1],
[3],
[2]])
In [385]: starts[:,None].shape
Out[385]: (4, 1)
In [386]: np.arange(10).shape
Out[386]: (10,)
Thus, looking at the shapes and putting those together, a schematic diagram of the same would look something like this -
starts : 4
np.arange(10) : 10
After extending starts :
starts[:,None] : 4 x 1
np.arange(10) : 10
Thus, when we add starts[:,None] with np.arange(10), the elems of starts[:,None] would be broadcasted along its second axis 10 times corresponding to the length of the other array along that axis. For np.arange(10), it would be converted to 2D with its first dim being a singleton dim and its elems being broadcasted along it 4 times correspoinding to the length of 4 for the other array starts[:,None] along that axis. Please note that there aren't explicit replications, as under the hood the elems are broadcasted and added on the fly.
Thus, functionally we would have the replications, like so -
In [391]: np.repeat(starts[:,None],10,axis=1)
Out[391]:
array([[3, 3, 3, 3, 3, 3, 3, 3, 3, 3],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[3, 3, 3, 3, 3, 3, 3, 3, 3, 3],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2]])
In [392]: np.repeat(np.arange(10)[None],4,axis=0)
Out[392]:
array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]])
These broadcasted elems are then added to give us the desired output x.