Input:
dt = [6,7,8,9,10]
data = [1,2,3,4,5]
b = 8.0
b = np.require(b, dtype=np.float)
data += dt < b
data
Output:
array([2, 3, 3, 4, 5])
I tried to input different number but still couldn't figure out what's the "<" doing there....
Also, it seems to work only when b is np.float (hence the conversion).
The < with numpy arrays does an element-wise comparison. That means it returns an array where there is a True where the condition is true and False if not. The np.require line is necessary here so it actually uses NumPy arrays. You could drop the np.require if you converted your data and dt to np.arrays beforehand.
Then the result is added (element-wise) to the numeric array. In this context True is equal to 1 and False to zero.
>>> dt < b # which elements are smaller than b?
array([ True, True, False, False, False])
>>> 0 + (dt < b) # boolean arrays in arithmetic operations with numbers
array([1, 1, 0, 0, 0])
So it adds 1 to every element of data where the element in dt is smaller than 8.
dt is a list:
In [50]: dt = [6,7,8,9,10]
In [51]: dt < 8
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-51-3d06f93227f5> in <module>()
----> 1 dt < 8
TypeError: '<' not supported between instances of 'list' and 'int'
< (.__lt__) is not defined for lists.
But if one element of the comparison is an ndarray, then the numpy definition of __lt__ applies. dt is turned into an array, and it does an element by element comparison.
In [52]: dt < np.array(8)
Out[52]: array([ True, True, False, False, False])
In [53]: np.array(dt) < 8
Out[53]: array([ True, True, False, False, False])
numpy array operations also explain the data += part:
In [54]: data = [1,2,3,4,5] # a list
In [55]: data + (dt < np.array(8)) # list=>array, and boolean array to integer array
Out[55]: array([2, 3, 3, 4, 5])
In [56]: data
Out[56]: [1, 2, 3, 4, 5]
In [57]: data += (dt < np.array(8))
In [58]: data
Out[58]: array([2, 3, 3, 4, 5])
Actually I'm a bit surprised that with the += data has been changed from list to array. It means the data+=... has been implemented as an assignment:
data = data + (dt <np.array(8))
Normally + for a list is a concatenate:
In [61]: data += ['a','b','c']
In [62]: data
Out[62]: [1, 2, 3, 4, 5, 'a', 'b', 'c']
# equivalent of: data.extend(['a','b','c'])
You can often get away with using lists in array contexts, but it's better to make objects arrays, so you do get these implicit, and sometimes unexpected, conversions.
This is just an alias (or shortcut or convenience notation) to the equivalent function: numpy.less()
In [116]: arr1 = np.arange(8)
In [117]: scalar = 6.0
# comparison that generates a boolean mask
In [118]: arr1 < scalar
Out[118]: array([ True, True, True, True, True, True, False, False])
# same operation as above
In [119]: np.less(arr1, scalar)
Out[119]: array([ True, True, True, True, True, True, False, False])
Let's see how this boolean array can be added to a non-boolean array in this case. It is possible due to type coercion
# sample array
In [120]: some_arr = np.array([1, 1, 1, 1, 1, 1, 1, 1])
# addition after type coercion
In [122]: some_arr + (arr1 < scalar)
Out[122]: array([2, 2, 2, 2, 2, 2, 1, 1])
# same output achieved with `numpy.less()`
In [123]: some_arr + np.less(arr1, scalar)
Out[123]: array([2, 2, 2, 2, 2, 2, 1, 1])
So, type coercion happens on the boolean array and then addition is performed.
Related
Definition:Arranged array
an arranged array is an array of dim 2 , shape is square matrix (NXN) and for every cell in the matrix : A[I,J] > A[I,J+1] AND A[I,J] > A[I+1,J]
I have an assignment to write a func that:
gets a numpy array and returns
True if - the given array is an arranged array
False - otherwise
note: We CANNOT use loops, list comps OR recursion. the point of the task is to use numpy things.
assumptions: we can assume that the array isn't empty and has no NA's, also all of the cells are numerics
My code isn't very numpy oriented.. :
def is_square_ordered_matrix(A):
# Checking if the dimension is 2
if A.ndim != 2:
return False
# Checking if it is a squared matrix
if A.shape[0] != A.shape[1]:
return False
# Saving the original shape to reshape later
originalDim = A.shape
# Making it a dim of 1 to use it as a list
arrayAsList = list((A.reshape((1,originalDim[0]**2)))[0])
# Keeping original order before sorting
originalArray = arrayAsList[:]
# Using the values of the list as keys to see if there are doubles
valuesDictionary = dict.fromkeys(arrayAsList, 1)
# If len is different, means there are doubles and i should return False
if len(arrayAsList) != len(valuesDictionary):
return False
# If sorted list is equal to original list it means the original is already ordered and i should return True
arrayAsList.sort(reverse=True)
if originalArray == arrayAsList:
return True
else:
return False
True example:
is_square_ordered_matrix(np.arange(8,-1,-1).reshape((3,3)))
False example:
is_square_ordered_matrix(np.arange(9).reshape((3,3)))
is_square_ordered_matrix(np.arange(5,-1,-1).reshape((3,2)))
Simple comparison:
>>> def is_square_ordered_matrix(a):
... return a.shape[0] == a.shape[1] and np.all(a[:-1] > a[1:]) and np.all(a[:, :-1] > a[:, 1:])
...
>>> is_square_ordered_matrix(np.arange(8,-1,-1).reshape((3,3)))
True
>>> is_square_ordered_matrix(np.arange(9).reshape((3,3)))
False
>>> is_square_ordered_matrix(np.arange(5,-1,-1).reshape((3,2)))
False
First, compare a[:-1] with a[1:], which will compare the elements of each row with the elements of the next row, and then use np.all to judge:
>>> a = np.arange(8,-1,-1).reshape((3,3))
>>> a[:-1] # First and second lines
array([[8, 7, 6],
[5, 4, 3]])
>>> a[1:] # Second and third lines
array([[5, 4, 3],
[2, 1, 0]])
>>> a[:-1] > a[1:]
array([[ True, True, True],
[ True, True, True]])
>>> np.all(a[:-1] > a[1:])
True
Then compare a[:,:-1] with a[:, 1:], which will compare the columns:
>>> a[:, :-1] # First and second columns
array([[8, 7],
[5, 4],
[2, 1]])
>>> a[:, 1:] # Second and third columns
array([[7, 6],
[4, 3],
[1, 0]])
>>> a[:, :-1] > a[:, 1:]
array([[ True, True],
[ True, True],
[ True, True]])
The result of row comparison and column comparison is the result you want.
I am trying to use np.where() function with nested lists.
I would like to find an index with a given condition of the first layer of the nested list.
For example, if I have the following code
arr = [[1,1], [2,2],[3,3]]
a = np.where(arr == [2,2])
then ideally I would like code to return 'a' as 1.
Since [2,2] is in index 1 of the nested list.
However, I am just getting a empty array back as a result.
Of course, I can make it work easily by implementing external for loop such as
for n in range(len(arr)):
if arr[n] == [2,2]:
a = n
but I would like to implement this simply within the function np.where(write the entire code here).
Is there a way to do this?
Well you can write your own function to do so:
You'll need to
Find every line equal to what you looking for
Get indices of found rows (You can use where):
numpy compression
You can use compression operator to see if each line satisfies the condition. Such as:
np_arr = np.array(
[1, 2, 3, 4, 5]
)
print(np_arr < 3)
This will return a boolean where every element is True or False where the condition is satisfied:
[ True True False False False]
For a 2D array you'll get a 2D boolean array:
to_find = np.array([2, 2])
np_arr = np.array(
[
[1, 1],
[2, 2],
[3, 3],
[2, 2]
]
)
print(np_arr == to_find)
The result is:
[[False False]
[ True True]
[False False]
[ True True]]
Now we are looking for lines with all True values. So we can use all method of ndarray. And we will provide the axis we are looking to look to all. X, Y or Both. We want to look to x axis:
to_find = np.array([2, 2])
np_arr = np.array(
[
[1, 1],
[2, 2],
[3, 3],
[2, 2]
]
)
print((np_arr == to_find).all(axis=1))
The result is:
[False True False True]
Get indices of Trues
At the end you are looking for indices where the values are True:
np.where((np_arr == to_find).all(axis=1))
The result would be:
(array([1, 3]),)
The best solution is that mentioned by #Michael Szczesny, but using np.where you can do this too:
a = np.where(np.array(arr) == [2, 2])[0]
resulted_ind = np.where(np.bincount(a) == 2)[0] # --> [1]
numpy runs in Python, so you can use both the basic Python lists and numpy arrays (which are more like MATLAB matrices)
A list of lists:
In [43]: alist = [[1,1], [2,2],[3,3]]
A list has an index method, which tests against each element of the list (elements here are 2 element lists):
In [44]: alist.index([2,2])
Out[44]: 1
In [45]: alist.index([2,3])
Traceback (most recent call last):
Input In [45] in <cell line: 1>
alist.index([2,3])
ValueError: [2, 3] is not in list
alist==[2,2] returns False, because the list is not the same as the [2,2] list.
If we make an array from that list:
In [46]: arr = np.array(alist)
In [47]: arr
Out[47]:
array([[1, 1],
[2, 2],
[3, 3]])
we can do an == test - but it compares numeric elements.
In [48]: arr == np.array([2,2])
Out[48]:
array([[False, False],
[ True, True],
[False, False]])
Underlying this comparison is the concept of broadcasting, allow it to compare a (3,2) array with a (2,) (a 2d with a 1d). Here's its trivial, but it can be much more complicated.
To find rows where all values are True, use:
In [50]: (arr == np.array([2,2])).all(axis=1)
Out[50]: array([False, True, False])
and where finds the True in that array (the result is a tuple with 1 array):
In [51]: np.where(_)
Out[51]: (array([1]),)
In Octave the equivalent is:
>> arr = [[1,1];[2,2];[3,3]]
arr =
1 1
2 2
3 3
>> all(arr == [2,2],2)
ans =
0
1
0
>> find(all(arr == [2,2],2))
ans = 2
Let's say I have an array a where I would like to calculate the mean across multiple slices defined in idx:
a = np.arange(10)
idx = np.random.choice([0,1], a.size).astype(bool)
a, idx
Out[1]: (array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
array([False, False, True, True, False, False, False, True, True,
True]))
With the desired output:
array([2.5, 8. ])
Of course I could write a simple for loop, but I would prefer a fully vectorized approach given that the arrays sizes can become quite big and 2D.
It's possible to do it completely vectorized:
edges = np.diff(idx.astype(np.int8), prepend=0, append=0)
rising = np.where(edges == 1)[0]
falling = np.where(edges == -1)[0]
cum = np.insert(np.cumsum(a), 0, 0)
means = (cum[falling] - cum[rising]) / (falling - rising)
This takes about 0.2 seconds on my machine with a = np.arange(10**7).
My goal is to take a list of np.arrays and create an associated list or array that classifies each as having a duplicate or not. Here's what I thought would work:
www = [np.array([1, 1, 1]), np.array([1, 1, 1]), np.array([2, 1, 1])]
uniques, counts = np.unique(www, axis = 0, return_counts = True)
counts = [1 if x > 1 else 0 for x in counts]
count_dict = dict(zip(uniques, counts))
[count_dict[i] for i in www]
The desired output for this case would be :
[1, 1, 0]
because the first and second element have another copy within the original list. It seems that the problem is that I cannot use a np.array as a key for a dictionary.
Suggestions?
First convert www to a 2D Numpy array then do the following:
In [18]: (counts[np.where((www[:,None] == uniques).all(2))[1]] > 1).astype(int)
Out[18]: array([1, 1, 0])
here we use broadcasting for check the equality of all www rows with uniques array and then using all() on last axis to find out which of its rows are completely equal to uniques rows.
Here's the elaborated results:
In [20]: (www[:,None] == uniques).all(2)
Out[20]:
array([[ True, False],
[ True, False],
[False, True]])
# Respective indices in `counts` array
In [21]: np.where((www[:,None] == uniques).all(2))[1]
Out[21]: array([0, 0, 1])
In [22]: counts[np.where((www[:,None] == uniques).all(2))[1]] > 1
Out[22]: array([ True, True, False])
In [23]: (counts[np.where((www[:,None] == uniques).all(2))[1]] > 1).astype(int)
Out[23]: array([1, 1, 0])
In Python, lists (and numpy arrays) cannot be hashed, so they can't be used as dictionary keys. But tuples can! So one option would be to convert your original list to a tuple, and to convert uniques to a tuple. The following works for me:
www = [np.array([1, 1, 1]), np.array([1, 1, 1]), np.array([2, 1, 1])]
www_tuples = [tuple(l) for l in www] # list of tuples
uniques, counts = np.unique(www, axis = 0, return_counts = True)
counts = [1 if x > 1 else 0 for x in counts]
# convert uniques to tuples
uniques_tuples = [tuple(l) for l in uniques]
count_dict = dict(zip(uniques_tuples, counts))
[count_dict[i] for i in www_tuples]
Just a heads-up: this will double your memory consumption, so it may not be the best solution if www is large.
You can mitigate the extra memory consumption by ingesting your data as tuples instead of numpy arrays if possible.
Suppose there are two arrays, vals contains values, and masks contains booleans indicating whether to use the values in vals, or nans. The goal is to build an array ret of the same length as masks, containing the values from vals in corresponding locations to True entries in masks.
For example, suppose
vals = np.array([1, 2])
masks = [True, False, False, True]
Then ret, the return value, should be
array([1, None, None, 2], dtype=object)
This is obviously very easy to do using loops:
import numpy as np
def unsparse(vals, masks):
vals_i = 0
ret = []
for m in masks:
if m:
ret.append(vals[vals_i])
vals_i += 1
else:
ret.append(None)
return np.array(ret)
>> unsparse(np.array([1, 2]), [True, False, False, True])
array([1, None, None, 2], dtype=object)
Is there a way to do so without loops and more concisely?
You could do something like this -
out = np.empty(masks.shape,dtype=object)
out[masks] = vals[:masks.sum()]
Please note that :masks.sum() selects first N elements from vals, where N is the number of TRUE elements in masks.
If it's guaranteed that the number of TRUE elements is same as number of elements in vals, then you could just simply do -
out[masks] = vals
Sample run -
In [34]: vals = np.array([1, 2, 6, 8, 9])
...: masks = np.array([True, False, False, True, False, True])
...:
In [35]: out = np.empty(masks.shape,dtype=object)
...: out[masks] = vals[:masks.sum()]
...:
In [36]: out
Out[36]: array([1, None, None, 2, None, 6], dtype=object)