I have a two numpy.arrays, I want to get following result efficiently
1.add the element's of b to a's sub-array
a=numpy.array([(1,2,3),(1,2,3)])
b=numpy.array([0,0])
->
c=[(0,1,2,3),(0,1,2,3)]
code in a loop
a=numpy.array([(1,2,3),(1,2,3)])
b=numpy.array([(0,0)])
c=numpy.zeros(2 , 4)
idx=0
for x in a:
c[idx]=(a[idx][0],a[idx][1],a[idx][2], b[idx])
idx = idx+1
and
2. Get an 2-D array with dimension(a.dim*b.dim, 2) from two 1-D arrays
a=numpy.array([(1,2)])
b=numpy.array([(3,4)])
->
c=[(1,3),(1,4),(2,3),(2,4)]
code in a loop
a=numpy.array([(1,2)])
b=numpy.array([(3,4)])
c=numpy.zeros(a.size*b.size , 2)
idx=0
for x in a:
for y in b:
c[idx]=(x,y)
idx = idx+1
For the first problem, you can define b differently and use numpy.hstack:
a = numpy.array([(1,2,3),(1,2,3)])
b = numpy.array([[0],[0]])
numpy.hstack((b,a))
Regarding the second problem, I would probably use sza's answer and create the numpy array from that result, if necessary. That technique was suggested in an old Stack Overflow question.
For the first one, you can do
>>> a=numpy.array([(1,2,3),(1,2,3)])
>>> b=numpy.array([0,0])
>>> [tuple(numpy.insert(x, 0, y)) for (x,y) in zip(a,b)]
[(0, 1, 2, 3), (0, 1, 2, 3)]
For the 2nd one, you can get the 2-D array like this
>>> a=numpy.array([(1,2)])
>>> b=numpy.array([(3,4)])
>>> import itertools
>>> c = list(itertools.product(a.tolist()[0], b.tolist()[0]))
[(1, 3), (1, 4), (2, 3), (2, 4)]
Related
With Python, I can assign multiple variables like so:
a, b = (1, 2)
print(a)
print(b)
# 1
# 2
Can I do something similar with the map function?
def myfunc(a):
return (a+1, a-1)
a_plus_one, a_minus_one = map(myfunc, (1, 2, 3))
# or
a_plus_one, a_minus_one = list(map(myfunc, (1,2,3)))
print(a_plus_one)
print(a_minus_one)
These attempts give me too many values to unpack error.
Edit:
Desired output is two new lists.
a_plus_one = (2, 3, 4)
a_minus_one = (0, 1, 2)
It looks like you're misunderstanding how map works. Look at list(map(myfunc, (1,2,3))):
[(2, 0), (3, 1), (4, 2)]
You want to transpose that using zip:
>>> a_plus_one, a_minus_one = zip(*map(myfunc, (1,2,3)))
>>> a_plus_one
(2, 3, 4)
>>> a_minus_one
(0, 1, 2)
For more info: Transpose list of lists
I have a "large" list of tuples:
thelist=[(1,2),(1,3),(2,3)]
I want to check whether any tuple in the list starts with a 1, and if it does, print "aaa":
for i in thelist:
templist.append((i[0],i))
for i in templist:
if i[0]==1:
print("aaa")
break
Which is rather ardurous as I have to create the templist. Is there any way I can do this:
if (1,_) in thelist:
print("aaa")
Where _ is the universal selector. Note that the list would be very large and thus it is very costly to implement another list.
There isn't, although you can just use any
any(i[0] == 1 for i in thelist) --> Returns true if the first element is 1
If you don’t actually need the actual tuple, like you do in your example, then you can actually use tuple unpacking for exactly that purpose:
>>> the_list = [(1, 2), (1, 3), (2, 3)]
>>> for x, y in the_list:
if x == 1:
print('aaa')
break
aaa
If you add a * in front of the y, you can also unpack tuples of different sizes, collecting the remainder of the tuple:
>>> other_list = [(1, 2, 3, 4, 5), (1, 3), (2, 3)]
>>> for x, *y in other_list:
if x == 1:
print(y)
break
[2, 3, 4, 5]
Otherwise, if you just want to filter your list based on some premise and then do something on those filtered items, you can use filter with a custom function:
>>> def startsWithOne(x):
return x[0] == 1
>>> thelist = [(1, 2), (1, 3), (2, 3)]
>>> for x in filter(starts_with_one, the_list):
print(x)
(1, 2)
(1, 3)
This is probably the most flexible way which also avoids creating a separate list in memory, as the elements are filtered lazily when you interate the list with your loop.
Finally, if you just want to figure out if any of your items starts with a 1, like you do in your example code, then you could just do it like this:
>>> if any(filter(starts_with_one, the_list)):
print('aaa')
aaa
But I assume that this was just an oversimplified example.
Mark Lutz in his book "Learning Python" gives an example:
>>> [(x,y) for x in range(5) if x%2==0 for y in range(5) if y%2==1]
[(0, 1), (0, 3), (2, 1), (2, 3), (4, 1), (4, 3)]
>>>
a bit later he remarks that 'a map and filter equivalent' of this is possible though complex and nested.
The closest one I ended up with is the following:
>>> list(map(lambda x:list(map(lambda y:(y,x),filter(lambda x:x%2==0,range(5)))), filter(lambda x:x%2==1,range(5))))
[[(0, 1), (2, 1), (4, 1)], [(0, 3), (2, 3), (4, 3)]]
>>>
The order of tuples is different and nested list had to be introduced. I'm curious what would be the equivalent.
A note to append to #Kasramvd's explanation.
Readability is important in Python. It's one of the features of the language. Many will consider the list comprehension the only readable way.
Sometimes, however, especially when you are working with multiple iterations of conditions, it is clearer to separate your criteria from logic. In this case, using the functional method may be preferable.
from itertools import product
def even_and_odd(vals):
return (vals[0] % 2 == 0) and (vals[1] %2 == 1)
n = range(5)
res = list(filter(even_and_odd, product(n, n)))
One important point that you have to notice is that your nested list comprehension is of O(n2) order. Meaning that it's looping over a product of two ranges. If you want to use map and filter you have to create all the combinations. You can do that after or before filtering but what ever you do you can't have all those combinations with those two functions, unless you change the ranges and/or modify something else.
One completely functional approach is to use itertools.product() and filter as following:
In [16]: from itertools import product
In [17]: list(filter(lambda x: x[0]%2==0 and x[1]%2==1, product(range(5), range(5))))
Out[17]: [(0, 1), (0, 3), (2, 1), (2, 3), (4, 1), (4, 3)]
Also note that using a nested list comprehension with two iterations is basically more readable than multiple map/filter functions. And regarding the performance using built-in funcitons is faster than list comprehension when your function are merely built-in so that you can assure all of them are performing at C level. When you break teh chain with something like a lambda function which is Python/higher lever operation your code won't be faster than a list comprehension.
I think the only confusing part in the expression [(x, y) for x in range(5) if x % 2 == 0 for y in range(5) if y % 2 == 1] is that there an implicit flatten operation is hidden.
Let's consider the simplified version of the expression first:
def even(x):
return x % 2 == 0
def odd(x):
return not even(x)
c = map(lambda x: map(lambda y: [x, y],
filter(odd, range(5))),
filter(even, range(5)))
print(c)
# i.e. for each even X we have a list of odd Ys:
# [
# [[0, 1], [0, 3]],
# [[2, 1], [2, 3]],
# [[4, 1], [4, 3]]
# ]
However, we need pretty the same but flattened list [(0, 1), (0, 3), (2, 1), (2, 3), (4, 1), (4, 3)].
From the official python docs we can grab the example of flatten function:
from itertools import chain
flattened = list(chain.from_iterable(c)) # we need list() here to unroll an iterator
print(flattened)
Which is basically an equivalent for the following list comprehension expression:
flattened = [x for sublist in c for x in sublist]
print(flattened)
# ... which is basically an equivalent to:
# result = []
# for sublist in c:
# for x in sublist:
# result.append(x)
Range support step argument, so I come up with this solution using itertools.chain.from_iterable to flatten inner list:
from itertools import chain
list(chain.from_iterable(
map(
lambda x:
list(map(lambda y: (x, y), range(1, 5, 2))),
range(0, 5, 2)
)
))
Output:
Out[415]: [(0, 1), (0, 3), (2, 1), (2, 3), (4, 1), (4, 3)]
I'm trying to make a 3D histogram. Initially h = zeros((6,6,8)).
I'll explain my problem with an example. Suppose I have 3 lists of coordinates for h, each list for one dimension:
x = array([2,1,0,1,2,2])
y = array([1,3,0,3,2,1])
z = array([6,2,0,2,5,6]) (the coordinates (x[0],y[0],z[0]) and (x[6],y[6],z[6]) are duplicates, and (x[1],y[1],z[1]) and (x[3],y[3],z[3]) also are)
and also a list of corresponding quantities to accumulate into h:
q = array([1,2,5,9,8,7])
I tried and h[x,y,z] += q does not work because only q[5] = 7 is added to h[2,1,6] and q[0] = 1 is not.
How can I work around this? Thank you.
IIUC, you want np.add.at. To quote the docs: "For addition ufunc, this method is equivalent to a[indices] += b, except that results are accumulated for elements that are indexed more than once."
For example:
>>> np.add.at(h, [x,y,z], q)
>>> for i, val in np.ndenumerate(h):
... if val: print(i, val)
...
((0, 0, 0), 5.0)
((1, 3, 2), 11.0)
((2, 1, 6), 8.0)
((2, 2, 5), 8.0)
from itertools import combinations
a = [1,2,3]
combinations(a,2) #will give me ((1,2),(1,3),(2,3))
combinations(a,3) #will give me ((1,2,3),)
but what if I want results of different length which is in a array
e.g.
I want to find all combinations of given array[1,2,3] of length more than or equal to 2
so result should be ((1,2),(1,3),(2,3),(1,2,3))
something like c = combinations(a,>=2)
I tried to use lambda but its not working
c = combinations(a,lambda x: x for x in [2,3])
as well list comprehensive c = combinations(a,[x for x in [2,3]])
I know I can use a simple loop and then find out the combinations of diff length.
for l in [2,3]:
combinations(a,l)
But Is there any pythonic way to do this?
You could combine combinations and chain.from_iterable:
>>> from itertools import chain, combinations
>>> a = [1,2,3]
>>> n = 2
>>> cc = chain.from_iterable(combinations(a, i) for i in range(n, len(a)+1))
>>> list(cc)
[(1, 2), (1, 3), (2, 3), (1, 2, 3)]
chain.from_iterable here is flattening what the generator expression (combinations(a, i) for i in range(n, len(a)+1)) is producing. Otherwise you'd wind up with something like
>>> [list(combinations(a,i)) for i in range(n, len(a)+1)]
[[(1, 2), (1, 3), (2, 3)], [(1, 2, 3)]]
which is fine, but not quite in the format you were looking for.