For example:
alist=[['a','b'],[1,2]]
and the combination
('a',1)
Is there a way to get the index of this combination i.e. (0,0) because both are at the 0-th position in their respective list or a matrix like
[[1,1],[0,0]]
where the 1s indicate the position get selected to form the combination?
Well you can just create the indices as ruaridhw pointed out. You can do it like so:
from itertools import product
alist = [['a','b'],[1,2]]
print [a for a in product(*alist)]
print [list(a) for a in product(*[range(len(x)) for x in alist])]
Output:
[('a', 1), ('a', 2), ('b', 1), ('b', 2)]
[[0, 0], [0, 1], [1, 0], [1, 1]]
Since each element of given combination c is to be found in the corresponding row of alist, the following returns the tuple of their indices:
alist = [['a','b'],[1,2]]
c = ('a',1)
ix = tuple(row.index(elem) for row, elem in zip(alist, c))
Here ix becomes (0, 0)
I have a list of float points such as [x1,x2,x3,x4,....xn] that are plotted as a line graph. I would like to find the set of points where the slope is the steepest.
Right now, Im calculating the difference between a set of points in a loop and using the max() function to determine the maximum point.
Any other elegant way of doing this?
Assuming points is the list of your values, you can calculate the differences in a single line using:
max_slope = max([x - z for x, z in zip(points[:-1], points[1:])])
But what you gain in compactness, you probably lose in readability.
What happens in this list comprehension is the following:
Two lists are created based on the original one, namely points[:-1] & points[1:]. Points[:-1] starts from the beginning of the original list and goes to the second to last item (inclusive). Points[1:] starts from the second item and goes all the way to the last item (inclusive again.)
Example
example_list = [1, 2, 3, 4, 5]
ex_a = example_list[:-1] # [1, 2, 3, 4]
ex_b = example_list[1:] # [2, 3, 4, 5]
Then you zip the two lists creating an object from which you can draw x, z pairs to calculate your differences. Note that zip does not create a list in Python 3 so you need to pass it's return value to the list argument.
Like:
example_list = [1, 2, 3, 4, 5]
ex_a = example_list[:-1] # [1, 2, 3, 4]
ex_b = example_list[1:] # [2, 3, 4, 5]
print(list(zip(ex_a, ex_b))) # [(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
Finally, you calculate the differences using the created pairs, store the results in a list and get the maximum value.
If the location of the max slope is also interesting you can get the index from the created list by using the .index() method. In that case, though, it would probably be better to save the list created by the comprehension and not just use it.
Numpy has a number of tools for working with arrays. For example, you could:
import numpy as np
xx = np.array([x1, x2, x3, x4, ...]) # your list of values goes in there
print(np.argmax(xx[:-1] - xx[1:])) # for all python versions
So I have a tuple of tuples
a = ((1, 2), (7, 2), (5, 2), (3, 4), (8, 4))
I would like to remove all the tuples from 'a' which have the a common second element, except for one (any one of them).
For the example above I would like the new output a = ((1,2),(3,4))
In other words I would like to eliminate tuples which are considered duplicate elements in the second position of the tuple.
I would like to know the most efficient way to achieve this, and also like to know if I can do the same with lists instead of tuples?
You could create a dictionary from your elements, with whatever you wanted to be unique as the key, then extracting the values. This works for anything where the 'unique' sub-element is hashable. Integers are hashable:
def unique_by_key(elements, key=None):
if key is None:
# no key: the whole element must be unique
key = lambda e: e
return {key(el): el for el in elements}.values()
This function is pretty generic; it can be used to extract 'unique' elements by any trait, as long as whatever the key callable returns can be used as a key in a dictionary. Order will not be preserved, and currently the last element per key wins.
With the above function you can use a operator.itemgetter() object or a lambda to extract your second value from each element. This then works for both a sequence of tuples and a sequence of lists:
from operator import itemgetter
unique_by_second_element = unique_by_key(a, key=itemgetter(1))
Demo:
>>> from operator import itemgetter
>>> a = ((1, 2), (7, 2), (5, 2), (3, 4), (8, 4))
>>> unique_by_key(a, key=itemgetter(1))
[(5, 2), (8, 4)]
>>> b = [[1, 2], [7, 2], [5, 2], [3, 4], [8, 4]]
>>> unique_by_key(b, key=itemgetter(1))
[[5, 2], [8, 4]]
Note that the function always returns a list; you can always convert that back by calling tuple() on the result.
Sorry if this has already been asked, I couldn't find it anywhere. Basically how do I get 2 separate ranges within a list in Python.
If I want the 1st, 2nd, 5th and 6th elements of a list I know I can do this,
l = range(0,15)
l[1:3]+l[5:7]
but this assumes that l is easy to write. However I am scrapping something from a webpage using BeautifulSoup4, so I'm using soup.find_all (which gives me a list), so I can't simply write out 2 lists, l and concatenate them.
I want an answer that is something like
l = range(0,15)
l[1:3,5:7]
(but of course without the error) :)
This might be what you want. itemgetter creates a function that retrieves the listed indices:
>>> import operator
>>> snip = operator.itemgetter(1,2,5,6)
>>> snip(range(15))
(1, 2, 5, 6)
>>> snip('abcdefg')
('b', 'c', 'f', 'g')
>>> snip([1,2,3,4,5,6,7,8])
(2, 3, 6, 7)
I would do this with a function:
def multi_range(l, *args):
output = []
for indices in args:
output += l[indices[0]:indices[1]]
return output
So the first argument would be the list, and the rest of the parameters are tuples with the indices you're looking to pull. It would work fine with a long list name:
long_list_name = range(0, 15)
print multi_range(long_list_name, (1, 3), (5, 7))
>>> [1, 2, 5, 6]
l = range(0, 15)
print([l[i] for i in [1,2, 5,6]])
Not sure why you think l[1:3]+l[5:7] is hard, find_all returns a normal python list like any other.
Or using map:
l = range(0, 15)
print(list(map(l.__getitem__,(1,2,5,6))))
Is this OK?
indices = [1, 2, 5, 6]
selected = [l[i] for i in indices]
The Python 2 documentation says:
Built-in Functions: map(function, iterable, ...)
Apply function to every item of iterable and return a list of the
results. If additional iterable arguments are passed, function must
take that many arguments and is applied to the items from all
iterables in parallel.
If one iterable is shorter than another it is assumed to be extended
with None items.
If function is None, the identity function is assumed; if there are
multiple arguments, map() returns a list consisting of tuples
containing the corresponding items from all iterables (a kind of
transpose operation).
The iterable arguments may be a sequence or any iterable object; the
result is always a list.
What role does this play in making a Cartesian product?
content = map(tuple, array)
What effect does putting a tuple anywhere in there have? I also noticed that without the map function the output is abc and with it, it's a, b, c.
I want to fully understand this function. The reference definitions is also hard to understand. Too much fancy fluff.
map isn't particularly pythonic. I would recommend using list comprehensions instead:
map(f, iterable)
is basically equivalent to:
[f(x) for x in iterable]
map on its own can't do a Cartesian product, because the length of its output list is always the same as its input list. You can trivially do a Cartesian product with a list comprehension though:
[(a, b) for a in iterable_a for b in iterable_b]
The syntax is a little confusing -- that's basically equivalent to:
result = []
for a in iterable_a:
for b in iterable_b:
result.append((a, b))
map doesn't relate to a Cartesian product at all, although I imagine someone well versed in functional programming could come up with some impossible to understand way of generating a one using map.
map in Python 3 is equivalent to this:
def map(func, iterable):
for i in iterable:
yield func(i)
and the only difference in Python 2 is that it will build up a full list of results to return all at once instead of yielding.
Although Python convention usually prefers list comprehensions (or generator expressions) to achieve the same result as a call to map, particularly if you're using a lambda expression as the first argument:
[func(i) for i in iterable]
As an example of what you asked for in the comments on the question - "turn a string into an array", by 'array' you probably want either a tuple or a list (both of them behave a little like arrays from other languages) -
>>> a = "hello, world"
>>> list(a)
['h', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd']
>>> tuple(a)
('h', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd')
A use of map here would be if you start with a list of strings instead of a single string - map can listify all of them individually:
>>> a = ["foo", "bar", "baz"]
>>> list(map(list, a))
[['f', 'o', 'o'], ['b', 'a', 'r'], ['b', 'a', 'z']]
Note that map(list, a) is equivalent in Python 2, but in Python 3 you need the list call if you want to do anything other than feed it into a for loop (or a processing function such as sum that only needs an iterable, and not a sequence). But also note again that a list comprehension is usually preferred:
>>> [list(b) for b in a]
[['f', 'o', 'o'], ['b', 'a', 'r'], ['b', 'a', 'z']]
map creates a new list by applying a function to every element of the source:
xs = [1, 2, 3]
# all of those are equivalent — the output is [2, 4, 6]
# 1. map
ys = map(lambda x: x * 2, xs)
# 2. list comprehension
ys = [x * 2 for x in xs]
# 3. explicit loop
ys = []
for x in xs:
ys.append(x * 2)
n-ary map is equivalent to zipping input iterables together and then applying the transformation function on every element of that intermediate zipped list. It's not a Cartesian product:
xs = [1, 2, 3]
ys = [2, 4, 6]
def f(x, y):
return (x * 2, y // 2)
# output: [(2, 1), (4, 2), (6, 3)]
# 1. map
zs = map(f, xs, ys)
# 2. list comp
zs = [f(x, y) for x, y in zip(xs, ys)]
# 3. explicit loop
zs = []
for x, y in zip(xs, ys):
zs.append(f(x, y))
I've used zip here, but map behaviour actually differs slightly when iterables aren't the same size — as noted in its documentation, it extends iterables to contain None.
Simplifying a bit, you can imagine map() doing something like this:
def mymap(func, lst):
result = []
for e in lst:
result.append(func(e))
return result
As you can see, it takes a function and a list, and returns a new list with the result of applying the function to each of the elements in the input list. I said "simplifying a bit" because in reality map() can process more than one iterable:
If additional iterable arguments are passed, function must take that many arguments and is applied to the items from all iterables in parallel. If one iterable is shorter than another it is assumed to be extended with None items.
For the second part in the question: What role does this play in making a Cartesian product? well, map() could be used for generating the cartesian product of a list like this:
lst = [1, 2, 3, 4, 5]
from operator import add
reduce(add, map(lambda i: map(lambda j: (i, j), lst), lst))
... But to tell the truth, using product() is a much simpler and natural way to solve the problem:
from itertools import product
list(product(lst, lst))
Either way, the result is the cartesian product of lst as defined above:
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5)]
The map() function is there to apply the same procedure to every item in an iterable data structure, like lists, generators, strings, and other stuff.
Let's look at an example:
map() can iterate over every item in a list and apply a function to each item, than it will return (give you back) the new list.
Imagine you have a function that takes a number, adds 1 to that number and returns it:
def add_one(num):
new_num = num + 1
return new_num
You also have a list of numbers:
my_list = [1, 3, 6, 7, 8, 10]
if you want to increment every number in the list, you can do the following:
>>> map(add_one, my_list)
[2, 4, 7, 8, 9, 11]
Note: At minimum map() needs two arguments. First a function name and second something like a list.
Let's see some other cool things map() can do.
map() can take multiple iterables (lists, strings, etc.) and pass an element from each iterable to a function as an argument.
We have three lists:
list_one = [1, 2, 3, 4, 5]
list_two = [11, 12, 13, 14, 15]
list_three = [21, 22, 23, 24, 25]
map() can make you a new list that holds the addition of elements at a specific index.
Now remember map(), needs a function. This time we'll use the builtin sum() function. Running map() gives the following result:
>>> map(sum, list_one, list_two, list_three)
[33, 36, 39, 42, 45]
REMEMBER:
In Python 2 map(), will iterate (go through the elements of the lists) according to the longest list, and pass None to the function for the shorter lists, so your function should look for None and handle them, otherwise you will get errors. In Python 3 map() will stop after finishing with the shortest list. Also, in Python 3, map() returns an iterator, not a list.
Python3 - map(func, iterable)
One thing that wasn't mentioned completely (although #BlooB kinda mentioned it) is that map returns a map object NOT a list. This is a big difference when it comes to time performance on initialization and iteration. Consider these two tests.
import time
def test1(iterable):
a = time.clock()
map(str, iterable)
a = time.clock() - a
b = time.clock()
[ str(x) for x in iterable ]
b = time.clock() - b
print(a,b)
def test2(iterable):
a = time.clock()
[ x for x in map(str, iterable)]
a = time.clock() - a
b = time.clock()
[ str(x) for x in iterable ]
b = time.clock() - b
print(a,b)
test1(range(2000000)) # Prints ~1.7e-5s ~8s
test2(range(2000000)) # Prints ~9s ~8s
As you can see initializing the map function takes almost no time at all. However iterating through the map object takes longer than simply iterating through the iterable. This means that the function passed to map() is not applied to each element until the element is reached in the iteration. If you want a list use list comprehension. If you plan to iterate through in a for loop and will break at some point, then use map.