I have a list:
somelist = [500, 600, 200, 1000]
I want to generate the rank order of that list:
rankorderofsomelist = [3, 2, 4, 1]
There are some complex solutions, but does anyone have any simple methods?
Since you've tagged this question scipy, you could use scipy.stats.rankdata:
>>> rankdata(somelist)
array([ 2., 3., 1., 4.])
>>> len(somelist) - rankdata(somelist)
array([ 2., 1., 3., 0.])
>>> len(somelist) - rankdata(somelist) + 1
array([ 3., 2., 4., 1.])
The real advantage is that you can specify how you want the corner cases to be treated:
>>> rankdata([0,1,1,2])
array([ 1. , 2.5, 2.5, 4. ])
>>> rankdata([0,1,1,2], method='min')
array([ 1, 2, 2, 4])
>>> rankdata([0,1,1,2], method='dense')
array([ 1, 2, 2, 3])
Simplest I can think of:
rankorder = sorted(range(len(thelist)), key=thelist.__getitem__)
This will, of course, produce [2, 1, 3, 0], because Python indexing is always zero-based -- if for some absolutely weird reason you need to add one to each index you can of course easily do so:
rankorder_weird = [1+x for x in rankorder]
Try this one-liner:
rankorderofsomelist = [sorted(somelist).index(x) for x in somelist]
Note that it'll behave as expected for a list with multiple entries of the same value (e.g. four instances of the same value, all of them the second-largest in the list, will all be ranked 2). Also note that Pythonic sorting is ascending (smallest to largest) and zero-based, so you may have to apply a final pass over the list to increment the ranks, reverse them, etc.
You can include that pass in the one-liner. To yield your desired result, just use:
rankorderofsomelist = [len(somelist)-(sorted(somelist).index(x)) for x in somelist]
Related
I have the following question. Is there somekind of method with numpy or scipy , which I can use to get an given unsorted array like this
a = np.array([0,0,1,1,4,4,4,4,5,1891,7]) #could be any number here
to something where the numbers are interpolated/mapped , there is no gap between the values and they are in the same order like before?:
[0,0,1,1,2,2,2,2,3,5,4]
EDIT
Is it furthermore possible to swap/shuffle the numbers after the mapping, so that
[0,0,1,1,2,2,2,2,3,5,4]
become something like:
[0,0,3,3,5,5,5,5,4,1,2]
Edit: I'm not sure what the etiquette is here (should this be a separate answer?), but this is actually directly obtainable from np.unique.
>>> u, indices = np.unique(a, return_inverse=True)
>>> indices
array([0, 0, 1, 1, 2, 2, 2, 2, 3, 5, 4])
Original answer: This isn't too hard to do in plain python by building a dictionary of what index each value of the array would map to:
x = np.sort(np.unique(a))
index_dict = {j: i for i, j in enumerate(x)}
[index_dict[i] for i in a]
Seems you need to rank (dense) your array, in which case use scipy.stats.rankdata:
from scipy.stats import rankdata
rankdata(a, 'dense')-1
# array([ 0., 0., 1., 1., 2., 2., 2., 2., 3., 5., 4.])
My situation: i have a pandas dataframe so that, for each row, I have to compute the following.
1) Get the first valute na excluded (df.apply(lambda x: x.dropna().iloc[0]))
2) Get the last valute na excluded (df.apply(lambda x: x.dropna().iloc[-1]))
3) Count the non na values (df.apply(lambda x: len(x.dropna()))
Sample case and expected output :
x = np.array([[1,2,np.nan], [4,5,6], [np.nan, 8,9]])
1) [1, 4, 8]
2) [2, 6, 9]
3) [2, 3, 2]
And i need to keep it optimized. So i turned to numpy and looked for a way to apply y = x[~numpy.isnan(x)] on a NxK array as a first step. Then,i would use what was shown here (Vectorized way of accessing row specific elements in a numpy array) for 1) and 2) but i am still empty handed for 3)
Here's one way -
In [756]: x
Out[756]:
array([[ 1., 2., nan],
[ 4., 5., 6.],
[ nan, 8., 9.]])
In [768]: m = ~np.isnan(x)
In [769]: first_idx = m.argmax(1)
In [770]: last_idx = m.shape[1] - m[:,::-1].argmax(1) - 1
In [771]: x[np.arange(len(first_idx)), first_idx]
Out[771]: array([ 1., 4., 8.])
In [772]: x[np.arange(len(last_idx)), last_idx]
Out[772]: array([ 2., 6., 9.])
In [773]: m.sum(1)
Out[773]: array([2, 3, 2])
Alternatively, we could make use of cumulative-summation to get those indices, like so -
In [787]: c = m.cumsum(1)
In [788]: first_idx = (c==1).argmax(1)
In [789]: last_idx = c.argmax(1)
In python's numpy, why does ogrid always produce int64 results?
For my application, I don't want to use int64 because of memory limitations (which come into play when the output components are broadcast together later). Is there any better alternative than recasting post-hoc:
y, x = np.ogrid[:9000,:9000]
y = y.astype(np.int16)
x = x.astype(np.int16)
For most other numpy calls a cleaner solution would be to use a dtype=... optional argument, but ogrid isn't invoked as a function. Instead it seems comparable to operators like a+b, except that those usually have alternatives like np.add(a,b,dtype=np.int8).
You can generate the same shapes with ix_, and have full control of dtype:
In [476]: np.ix_(np.arange(5,dtype=float),np.arange(5,dtype=np.int16))
Out[476]:
(array([[ 0.],
[ 1.],
[ 2.],
[ 3.],
[ 4.]]), array([[0, 1, 2, 3, 4]], dtype=int16))
In [477]: np.ogrid[:5,:5]
Out[477]:
[array([[0],
[1],
[2],
[3],
[4]]), array([[0, 1, 2, 3, 4]])]
meshgrid as well:
In [488]: np.meshgrid(np.arange(5, dtype=float), np.arange(5, dtype=np.int16), sparse=True, indexing='ij')
Out[488]:
[array([[ 0.],
[ 1.],
[ 2.],
[ 3.],
[ 4.]]), array([[0, 1, 2, 3, 4]], dtype=int16)]
Another alternative is to use np.newaxis more directly:
y = np.arange(9000, dtype=np.int16)[:,None]
x = np.arange(9000, dtype=np.int16)[None,:]
For example, given:
import numpy as np
data = np.array(
[[0, 0, 0],
[0, 1, 1],
[1, 0, 1],
[1, 0, 1],
[0, 1, 1],
[0, 0, 0]])
I want to get a 3-dimensional array, looking like:
result = array([[[ 2., 0.],
[ 0., 2.]],
[[ 0., 2.],
[ 0., 0.]]])
One way is:
for row in data
newArray[ row[0] ][ row[1] ][ row[2] ] += 1
What I'm trying to do is the following:
for i in dimension1
for j in dimension2
for k in dimension3
result[i,j,k] = (data[data[data[:,0]==i, 1]==j, 2]==k).sum()
This doesn't seem to work and I would like to achieve the desired result by sticking to my implementation rather than the one mentioned in the beginning (or using any extra imports, eg counter).
Thanks.
You can also use numpy.histogramdd for this:
>>> np.histogramdd(data, bins=(2, 2, 2))[0]
array([[[ 2., 0.],
[ 0., 2.]],
[[ 0., 2.],
[ 0., 0.]]])
The problem is that data[data[data[:,0]==i, 1]==j, 2]==k is not what you expect it to be.
Let's take this apart for the case (i,j,k) == (0,0,0)
data[:,0]==0 is [True, True, False, False, True, True], and data[data[:,0]==0] correctly gives us the lines where the first number is 0.
Now from those lines we get the lines where the second number is 0: data[data[:,0]==0, 1]==0, which gives us [True, False, False, True]. And this is the problem. Because if we take those indices from data, i.e., data[data[data[:,0]==0, 1]==0] we do not get the rows where the first and second number are 0, but the 0th and 3rd row instead:
In [51]: data[data[data[:,0]==0, 1]==0]
Out[51]: array([[0, 0, 0],
[1, 0, 1]])
And if we now filter for the rows where the third number is 0, we get the wrong result w.r.t. the orignal data.
And that's why your approach does not work. For better methods, see the other answers.
You can do something like the following
#Get output dimension and construct output array.
>>> dshape = tuple(data.max(axis=0)+1)
>>> dshape
(2, 2, 2)
>>> out = np.zeros(shape)
If you have numpy 1.8+:
out.flat[np.ravel_multi_index(data.T, dshape)]+=1
Else:
#Get indices and unique the resulting array
>>> inds = np.ravel_multi_index(data.T, dshape)
>>> inds, inverse = np.unique(inds, return_inverse=True)
>>> values = np.bincount(inverse)
>>> values
array([2, 2, 2])
>>> out.flat[inds] = values
>>> out
array([[[ 2., 0.],
[ 0., 2.]],
[[ 0., 2.],
[ 0., 0.]]])
Numpy versions before numpy 1.7 do not have a add.at attribute and the top code will not work without it. As ravel_multi_index may not be the fastest algorithm ever you can look into taking the unique rows of a numpy array. In effect these two operations should be equivalent.
Don't fear the imports. They're what make Python awesome.
If question assumes that you already have the result matrix.
import numpy as np
data = np.array(
[[0, 0, 0],
[0, 1, 1],
[1, 0, 1],
[1, 0, 1],
[0, 1, 1],
[0, 0, 0]]
)
result = np.zeros((2,2,2))
# range of each dim, aka allowable values for each dim
dim_ranges = zip(np.zeros(result.ndim), np.array(result.shape)-1)
dim_ranges
# Out[]:
# [(0.0, 2), (0.0, 2), (0.0, 2)]
# Multidimentional histogram will effectively "count" along each dim
sums,_ = np.histogramdd(data,bins=result.shape,range=dim_ranges)
result += sums
result
# Out[]:
# array([[[ 2., 0.],
# [ 0., 2.]],
#
# [[ 0., 2.],
# [ 0., 0.]]])
This solution solves for any "result" ndarray, no matter what the shape. Additionally, it works fine even if your "data" ndarray has indices which are out-of-bounds for your result matrix.
I am trying to fill an empty(not np.empty!) array with values using append but I am gettin error:
My code is as follows:
import numpy as np
result=np.asarray([np.asarray([]),np.asarray([])])
result[0]=np.append([result[0]],[1,2])
And I am getting:
ValueError: could not broadcast input array from shape (2) into shape (0)
I might understand the question incorrectly, but if you want to declare an array of a certain shape but with nothing inside, the following might be helpful:
Initialise empty array:
>>> a = np.zeros((0,3)) #or np.empty((0,3)) or np.array([]).reshape(0,3)
>>> a
array([], shape=(0, 3), dtype=float64)
Now you can use this array to append rows of similar shape to it. Remember that a numpy array is immutable, so a new array is created for each iteration:
>>> for i in range(3):
... a = np.vstack([a, [i,i,i]])
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
np.vstack and np.hstack is the most common method for combining numpy arrays, but coming from Matlab I prefer np.r_ and np.c_:
Concatenate 1d:
>>> a = np.zeros(0)
>>> for i in range(3):
... a = np.r_[a, [i, i, i]]
...
>>> a
array([ 0., 0., 0., 1., 1., 1., 2., 2., 2.])
Concatenate rows:
>>> a = np.zeros((0,3))
>>> for i in range(3):
... a = np.r_[a, [[i,i,i]]]
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
Concatenate columns:
>>> a = np.zeros((3,0))
>>> for i in range(3):
... a = np.c_[a, [[i],[i],[i]]]
...
>>> a
array([[ 0., 1., 2.],
[ 0., 1., 2.],
[ 0., 1., 2.]])
numpy.append is pretty different from list.append in python. I know that's thrown off a few programers new to numpy. numpy.append is more like concatenate, it makes a new array and fills it with the values from the old array and the new value(s) to be appended. For example:
import numpy
old = numpy.array([1, 2, 3, 4])
new = numpy.append(old, 5)
print old
# [1, 2, 3, 4]
print new
# [1, 2, 3, 4, 5]
new = numpy.append(new, [6, 7])
print new
# [1, 2, 3, 4, 5, 6, 7]
I think you might be able to achieve your goal by doing something like:
result = numpy.zeros((10,))
result[0:2] = [1, 2]
# Or
result = numpy.zeros((10, 2))
result[0, :] = [1, 2]
Update:
If you need to create a numpy array using loop, and you don't know ahead of time what the final size of the array will be, you can do something like:
import numpy as np
a = np.array([0., 1.])
b = np.array([2., 3.])
temp = []
while True:
rnd = random.randint(0, 100)
if rnd > 50:
temp.append(a)
else:
temp.append(b)
if rnd == 0:
break
result = np.array(temp)
In my example result will be an (N, 2) array, where N is the number of times the loop ran, but obviously you can adjust it to your needs.
new update
The error you're seeing has nothing to do with types, it has to do with the shape of the numpy arrays you're trying to concatenate. If you do np.append(a, b) the shapes of a and b need to match. If you append an (2, n) and (n,) you'll get a (3, n) array. Your code is trying to append a (1, 0) to a (2,). Those shapes don't match so you get an error.
This error arise from the fact that you are trying to define an object of shape (0,) as an object of shape (2,). If you append what you want without forcing it to be equal to result[0] there is no any issue:
b = np.append([result[0]], [1,2])
But when you define result[0] = b you are equating objects of different shapes, and you can not do this. What are you trying to do?
Here's the result of running your code in Ipython. Note that result is a (2,0) array, 2 rows, 0 columns, 0 elements. The append produces a (2,) array. result[0] is (0,) array. Your error message has to do with trying to assign that 2 item array into a size 0 slot. Since result is dtype=float64, only scalars can be assigned to its elements.
In [65]: result=np.asarray([np.asarray([]),np.asarray([])])
In [66]: result
Out[66]: array([], shape=(2, 0), dtype=float64)
In [67]: result[0]
Out[67]: array([], dtype=float64)
In [68]: np.append(result[0],[1,2])
Out[68]: array([ 1., 2.])
np.array is not a Python list. All elements of an array are the same type (as specified by the dtype). Notice also that result is not an array of arrays.
Result could also have been built as
ll = [[],[]]
result = np.array(ll)
while
ll[0] = [1,2]
# ll = [[1,2],[]]
the same is not true for result.
np.zeros((2,0)) also produces your result.
Actually there's another quirk to result.
result[0] = 1
does not change the values of result. It accepts the assignment, but since it has 0 columns, there is no place to put the 1. This assignment would work in result was created as np.zeros((2,1)). But that still can't accept a list.
But if result has 2 columns, then you can assign a 2 element list to one of its rows.
result = np.zeros((2,2))
result[0] # == [0,0]
result[0] = [1,2]
What exactly do you want result to look like after the append operation?
numpy.append always copies the array before appending the new values. Your code is equivalent to the following:
import numpy as np
result = np.zeros((2,0))
new_result = np.append([result[0]],[1,2])
result[0] = new_result # ERROR: has shape (2,0), new_result has shape (2,)
Perhaps you mean to do this?
import numpy as np
result = np.zeros((2,0))
result = np.append([result[0]],[1,2])
SO thread 'Multiply two arrays element wise, where one of the arrays has arrays as elements' has an example of constructing an array from arrays. If the subarrays are the same size, numpy makes a 2d array. But if they differ in length, it makes an array with dtype=object, and the subarrays retain their identity.
Following that, you could do something like this:
In [5]: result=np.array([np.zeros((1)),np.zeros((2))])
In [6]: result
Out[6]: array([array([ 0.]), array([ 0., 0.])], dtype=object)
In [7]: np.append([result[0]],[1,2])
Out[7]: array([ 0., 1., 2.])
In [8]: result[0]
Out[8]: array([ 0.])
In [9]: result[0]=np.append([result[0]],[1,2])
In [10]: result
Out[10]: array([array([ 0., 1., 2.]), array([ 0., 0.])], dtype=object)
However, I don't offhand see what advantages this has over a pure Python list or lists. It does not work like a 2d array. For example I have to use result[0][1], not result[0,1]. If the subarrays are all the same length, I have to use np.array(result.tolist()) to produce a 2d array.