I am having two lists as follows:
list_1
['A-1','A-1','A-1','A-2','A-2','A-3']
list_2
['iPad','iPod','iPhone','Windows','X-box','Kindle']
I would like to split the list_2 based on the index values in list_1. For instance,
list_a1
['iPad','iPod','iPhone']
list_a2
['Windows','X-box']
list_a3
['Kindle']
I know index method, but it needs the value to be matched to be passed along with. In this case, I would like to dynamically find the indexes of the values in list_1 with the same value. Is this possible? Any tips/hints would be deeply appreciated.
Thanks.
There are a few ways to do this.
I'd do it by using zip and groupby.
First:
>>> list(zip(list_1, list_2))
[('A-1', 'iPad'),
('A-1', 'iPod'),
('A-1', 'iPhone'),
('A-2', 'Windows'),
('A-2', 'X-box'),
('A-3', 'Kindle')]
Now:
>>> import itertools, operator
>>> [(key, list(group)) for key, group in
... itertools.groupby(zip(list_1, list_2), operator.itemgetter(0))]
[('A-1', [('A-1', 'iPad'), ('A-1', 'iPod'), ('A-1', 'iPhone')]),
('A-2', [('A-2', 'Windows'), ('A-2', 'X-box')]),
('A-3', [('A-3', 'Kindle')])]
So, you just want each group, ignoring the key, and you only want the second element of each element in the group. You can get the second element of each group with another comprehension, or just by unzipping:
>>> [list(zip(*group))[1] for key, group in
... itertools.groupby(zip(list_1, list_2), operator.itemgetter(0))]
[('iPad', 'iPod', 'iPhone'), ('Windows', 'X-box'), ('Kindle',)]
I would personally find this more readable as a sequence of separate iterator transformations than as one long expression. Taken to the extreme:
>>> ziplists = zip(list_1, list_2)
>>> pairs = itertools.groupby(ziplists, operator.itemgetter(0))
>>> groups = (group for key, group in pairs)
>>> values = (zip(*group)[1] for group in groups)
>>> [list(value) for value in values]
… but a happy medium of maybe 2 or 3 lines is usually better than either extreme.
Usually I'm the one rushing to a groupby solution ;^) but here I'll go the other way and manually insert into an OrderedDict:
list_1 = ['A-1','A-1','A-1','A-2','A-2','A-3']
list_2 = ['iPad','iPod','iPhone','Windows','X-box','Kindle']
from collections import OrderedDict
d = OrderedDict()
for code, product in zip(list_1, list_2):
d.setdefault(code, []).append(product)
produces a d looking like
>>> d
OrderedDict([('A-1', ['iPad', 'iPod', 'iPhone']),
('A-2', ['Windows', 'X-box']), ('A-3', ['Kindle'])])
with easy access:
>>> d["A-2"]
['Windows', 'X-box']
and we can get the list-of-lists in list_1 order using .values():
>>> d.values()
[['iPad', 'iPod', 'iPhone'], ['Windows', 'X-box'], ['Kindle']]
If you've noticed that no one is telling you how to make a bunch of independent lists with names like list_a1 and so on-- that's because that's a bad idea. You want to keep the data together in something which you can (at a minimum) iterate over easily, and both dictionaries and list of lists qualify.
Maybe something like this?
#!/usr/local/cpython-3.3/bin/python
import pprint
import collections
def main():
list_1 = ['A-1','A-1','A-1','A-2','A-2','A-3']
list_2 = ['iPad','iPod','iPhone','Windows','X-box','Kindle']
result = collections.defaultdict(list)
for list_1_element, list_2_element in zip(list_1, list_2):
result[list_1_element].append(list_2_element)
pprint.pprint(result)
main()
Using itertools.izip_longest and itertools.groupby:
>>> from itertools import groupby, izip_longest
>>> inds = [next(g)[0] for k, g in groupby(enumerate(list_1), key=lambda x:x[1])]
First group items of list_1 and find the starting index of each group:
>>> inds
[0, 3, 5]
Now use slicing and izip_longest as we need pairs list_2[0:3], list_2[3:5], list_2[5:]:
>>> [list_2[x:y] for x, y in izip_longest(inds, inds[1:])]
[['iPad', 'iPod', 'iPhone'], ['Windows', 'X-box'], ['Kindle']]
To get a list of dicts you can something like:
>>> inds = [next(g) for k, g in groupby(enumerate(list_1), key=lambda x:x[1])]
>>> {k: list_2[ind1: ind2[0]] for (ind1, k), ind2 in
zip_longest(inds, inds[1:], fillvalue=[None])}
{'A-1': ['iPad', 'iPod', 'iPhone'], 'A-3': ['Kindle'], 'A-2': ['Windows', 'X-box']}
You could do this if you want simple code, it's not pretty, but gets the job done.
list_1 = ['A-1','A-1','A-1','A-2','A-2','A-3']
list_2 = ['iPad','iPod','iPhone','Windows','X-box','Kindle']
list_1a = []
list_1b = []
list_1c = []
place = 0
for i in list_1[::1]:
if list_1[place] == 'A-1':
list_1a.append(list_2[place])
elif list_1[place] == 'A-2':
list_1b.append(list_2[place])
else:
list_1c.append(list_2[place])
place += 1
I haven't been able to find an understandable explanation of how to actually use Python's itertools.groupby() function. What I'm trying to do is this:
Take a list - in this case, the children of an objectified lxml element
Divide it into groups based on some criteria
Then later iterate over each of these groups separately.
I've reviewed the documentation, but I've had trouble trying to apply them beyond a simple list of numbers.
So, how do I use of itertools.groupby()? Is there another technique I should be using? Pointers to good "prerequisite" reading would also be appreciated.
IMPORTANT NOTE: You have to sort your data first.
The part I didn't get is that in the example construction
groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
groups.append(list(g)) # Store group iterator as a list
uniquekeys.append(k)
k is the current grouping key, and g is an iterator that you can use to iterate over the group defined by that grouping key. In other words, the groupby iterator itself returns iterators.
Here's an example of that, using clearer variable names:
from itertools import groupby
things = [("animal", "bear"), ("animal", "duck"), ("plant", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
for key, group in groupby(things, lambda x: x[0]):
for thing in group:
print("A %s is a %s." % (thing[1], key))
print("")
This will give you the output:
A bear is a animal.
A duck is a animal.
A cactus is a plant.
A speed boat is a vehicle.
A school bus is a vehicle.
In this example, things is a list of tuples where the first item in each tuple is the group the second item belongs to.
The groupby() function takes two arguments: (1) the data to group and (2) the function to group it with.
Here, lambda x: x[0] tells groupby() to use the first item in each tuple as the grouping key.
In the above for statement, groupby returns three (key, group iterator) pairs - once for each unique key. You can use the returned iterator to iterate over each individual item in that group.
Here's a slightly different example with the same data, using a list comprehension:
for key, group in groupby(things, lambda x: x[0]):
listOfThings = " and ".join([thing[1] for thing in group])
print(key + "s: " + listOfThings + ".")
This will give you the output:
animals: bear and duck.
plants: cactus.
vehicles: speed boat and school bus.
itertools.groupby is a tool for grouping items.
From the docs, we glean further what it might do:
# [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D
groupby objects yield key-group pairs where the group is a generator.
Features
A. Group consecutive items together
B. Group all occurrences of an item, given a sorted iterable
C. Specify how to group items with a key function *
Comparisons
# Define a printer for comparing outputs
>>> def print_groupby(iterable, keyfunc=None):
... for k, g in it.groupby(iterable, keyfunc):
... print("key: '{}'--> group: {}".format(k, list(g)))
# Feature A: group consecutive occurrences
>>> print_groupby("BCAACACAADBBB")
key: 'B'--> group: ['B']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'D'--> group: ['D']
key: 'B'--> group: ['B', 'B', 'B']
# Feature B: group all occurrences
>>> print_groupby(sorted("BCAACACAADBBB"))
key: 'A'--> group: ['A', 'A', 'A', 'A', 'A']
key: 'B'--> group: ['B', 'B', 'B', 'B']
key: 'C'--> group: ['C', 'C', 'C']
key: 'D'--> group: ['D']
# Feature C: group by a key function
>>> # islower = lambda s: s.islower() # equivalent
>>> def islower(s):
... """Return True if a string is lowercase, else False."""
... return s.islower()
>>> print_groupby(sorted("bCAaCacAADBbB"), keyfunc=islower)
key: 'False'--> group: ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'D']
key: 'True'--> group: ['a', 'a', 'b', 'b', 'c']
Uses
Anagrams (see notebook)
Binning
Group odd and even numbers
Group a list by values
Remove duplicate elements
Find indices of repeated elements in an array
Split an array into n-sized chunks
Find corresponding elements between two lists
Compression algorithm (see notebook)/Run Length Encoding
Grouping letters by length, key function (see notebook)
Consecutive values over a threshold (see notebook)
Find ranges of numbers in a list or continuous items (see docs)
Find all related longest sequences
Take consecutive sequences that meet a condition (see related post)
Note: Several of the latter examples derive from Víctor Terrón's PyCon (talk) (Spanish), "Kung Fu at Dawn with Itertools". See also the groupby source code written in C.
* A function where all items are passed through and compared, influencing the result. Other objects with key functions include sorted(), max() and min().
Response
# OP: Yes, you can use `groupby`, e.g.
[do_something(list(g)) for _, g in groupby(lxml_elements, criteria_func)]
The example on the Python docs is quite straightforward:
groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
groups.append(list(g)) # Store group iterator as a list
uniquekeys.append(k)
So in your case, data is a list of nodes, keyfunc is where the logic of your criteria function goes and then groupby() groups the data.
You must be careful to sort the data by the criteria before you call groupby or it won't work. groupby method actually just iterates through a list and whenever the key changes it creates a new group.
A neato trick with groupby is to run length encoding in one line:
[(c,len(list(cgen))) for c,cgen in groupby(some_string)]
will give you a list of 2-tuples where the first element is the char and the 2nd is the number of repetitions.
Edit: Note that this is what separates itertools.groupby from the SQL GROUP BY semantics: itertools doesn't (and in general can't) sort the iterator in advance, so groups with the same "key" aren't merged.
Another example:
for key, igroup in itertools.groupby(xrange(12), lambda x: x // 5):
print key, list(igroup)
results in
0 [0, 1, 2, 3, 4]
1 [5, 6, 7, 8, 9]
2 [10, 11]
Note that igroup is an iterator (a sub-iterator as the documentation calls it).
This is useful for chunking a generator:
def chunker(items, chunk_size):
'''Group items in chunks of chunk_size'''
for _key, group in itertools.groupby(enumerate(items), lambda x: x[0] // chunk_size):
yield (g[1] for g in group)
with open('file.txt') as fobj:
for chunk in chunker(fobj):
process(chunk)
Another example of groupby - when the keys are not sorted. In the following example, items in xx are grouped by values in yy. In this case, one set of zeros is output first, followed by a set of ones, followed again by a set of zeros.
xx = range(10)
yy = [0, 0, 0, 1, 1, 1, 0, 0, 0, 0]
for group in itertools.groupby(iter(xx), lambda x: yy[x]):
print group[0], list(group[1])
Produces:
0 [0, 1, 2]
1 [3, 4, 5]
0 [6, 7, 8, 9]
WARNING:
The syntax list(groupby(...)) won't work the way that you intend. It seems to destroy the internal iterator objects, so using
for x in list(groupby(range(10))):
print(list(x[1]))
will produce:
[]
[]
[]
[]
[]
[]
[]
[]
[]
[9]
Instead, of list(groupby(...)), try [(k, list(g)) for k,g in groupby(...)], or if you use that syntax often,
def groupbylist(*args, **kwargs):
return [(k, list(g)) for k, g in groupby(*args, **kwargs)]
and get access to the groupby functionality while avoiding those pesky (for small data) iterators all together.
I would like to give another example where groupby without sort is not working. Adapted from example by James Sulak
from itertools import groupby
things = [("vehicle", "bear"), ("animal", "duck"), ("animal", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
for key, group in groupby(things, lambda x: x[0]):
for thing in group:
print "A %s is a %s." % (thing[1], key)
print " "
output is
A bear is a vehicle.
A duck is a animal.
A cactus is a animal.
A speed boat is a vehicle.
A school bus is a vehicle.
there are two groups with vehicule, whereas one could expect only one group
#CaptSolo, I tried your example, but it didn't work.
from itertools import groupby
[(c,len(list(cs))) for c,cs in groupby('Pedro Manoel')]
Output:
[('P', 1), ('e', 1), ('d', 1), ('r', 1), ('o', 1), (' ', 1), ('M', 1), ('a', 1), ('n', 1), ('o', 1), ('e', 1), ('l', 1)]
As you can see, there are two o's and two e's, but they got into separate groups. That's when I realized you need to sort the list passed to the groupby function. So, the correct usage would be:
name = list('Pedro Manoel')
name.sort()
[(c,len(list(cs))) for c,cs in groupby(name)]
Output:
[(' ', 1), ('M', 1), ('P', 1), ('a', 1), ('d', 1), ('e', 2), ('l', 1), ('n', 1), ('o', 2), ('r', 1)]
Just remembering, if the list is not sorted, the groupby function will not work!
Sorting and groupby
from itertools import groupby
val = [{'name': 'satyajit', 'address': 'btm', 'pin': 560076},
{'name': 'Mukul', 'address': 'Silk board', 'pin': 560078},
{'name': 'Preetam', 'address': 'btm', 'pin': 560076}]
for pin, list_data in groupby(sorted(val, key=lambda k: k['pin']),lambda x: x['pin']):
... print pin
... for rec in list_data:
... print rec
...
o/p:
560076
{'name': 'satyajit', 'pin': 560076, 'address': 'btm'}
{'name': 'Preetam', 'pin': 560076, 'address': 'btm'}
560078
{'name': 'Mukul', 'pin': 560078, 'address': 'Silk board'}
Sadly I don’t think it’s advisable to use itertools.groupby(). It’s just too hard to use safely, and it’s only a handful of lines to write something that works as expected.
def my_group_by(iterable, keyfunc):
"""Because itertools.groupby is tricky to use
The stdlib method requires sorting in advance, and returns iterators not
lists, and those iterators get consumed as you try to use them, throwing
everything off if you try to look at something more than once.
"""
ret = defaultdict(list)
for k in iterable:
ret[keyfunc(k)].append(k)
return dict(ret)
Use it like this:
def first_letter(x):
return x[0]
my_group_by('four score and seven years ago'.split(), first_letter)
to get
{'f': ['four'], 's': ['score', 'seven'], 'a': ['and', 'ago'], 'y': ['years']}
How do I use Python's itertools.groupby()?
You can use groupby to group things to iterate over. You give groupby an iterable, and a optional key function/callable by which to check the items as they come out of the iterable, and it returns an iterator that gives a two-tuple of the result of the key callable and the actual items in another iterable. From the help:
groupby(iterable[, keyfunc]) -> create an iterator which returns
(key, sub-iterator) grouped by each value of key(value).
Here's an example of groupby using a coroutine to group by a count, it uses a key callable (in this case, coroutine.send) to just spit out the count for however many iterations and a grouped sub-iterator of elements:
import itertools
def grouper(iterable, n):
def coroutine(n):
yield # queue up coroutine
for i in itertools.count():
for j in range(n):
yield i
groups = coroutine(n)
next(groups) # queue up coroutine
for c, objs in itertools.groupby(iterable, groups.send):
yield c, list(objs)
# or instead of materializing a list of objs, just:
# return itertools.groupby(iterable, groups.send)
list(grouper(range(10), 3))
prints
[(0, [0, 1, 2]), (1, [3, 4, 5]), (2, [6, 7, 8]), (3, [9])]
This basic implementation helped me understand this function. Hope it helps others as well:
arr = [(1, "A"), (1, "B"), (1, "C"), (2, "D"), (2, "E"), (3, "F")]
for k,g in groupby(arr, lambda x: x[0]):
print("--", k, "--")
for tup in g:
print(tup[1]) # tup[0] == k
-- 1 --
A
B
C
-- 2 --
D
E
-- 3 --
F
One useful example that I came across may be helpful:
from itertools import groupby
#user input
myinput = input()
#creating empty list to store output
myoutput = []
for k,g in groupby(myinput):
myoutput.append((len(list(g)),int(k)))
print(*myoutput)
Sample input: 14445221
Sample output: (1,1) (3,4) (1,5) (2,2) (1,1)
from random import randint
from itertools import groupby
l = [randint(1, 3) for _ in range(20)]
d = {}
for k, g in groupby(l, lambda x: x):
if not d.get(k, None):
d[k] = list(g)
else:
d[k] = d[k] + list(g)
the code above shows how groupby can be used to group a list based on the lambda function/key supplied. The only problem is that the output is not merged, this can be easily resolved using a dictionary.
Example:
l = [2, 1, 2, 3, 1, 3, 2, 1, 3, 3, 1, 3, 2, 3, 1, 2, 1, 3, 2, 3]
after applying groupby the result will be:
for k, g in groupby(l, lambda x:x):
print(k, list(g))
2 [2]
1 [1]
2 [2]
3 [3]
1 [1]
3 [3]
2 [2]
1 [1]
3 [3, 3]
1 [1]
3 [3]
2 [2]
3 [3]
1 [1]
2 [2]
1 [1]
3 [3]
2 [2]
3 [3]
Once a dictionary is used as shown above following result is derived which can be easily iterated over:
{2: [2, 2, 2, 2, 2, 2], 1: [1, 1, 1, 1, 1, 1], 3: [3, 3, 3, 3, 3, 3, 3, 3]}
The key thing to recognize with itertools.groupby is that items are only grouped together as long as they're sequential in the iterable. This is why sorting works, because basically you're rearranging the collection so that all of the items which satisfy callback(item) now appear in the sorted collection sequentially.
That being said, you don't need to sort the list, you just need a collection of key-value pairs, where the value can grow in accordance to each group iterable yielded by groupby. i.e. a dict of lists.
>>> things = [("vehicle", "bear"), ("animal", "duck"), ("animal", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
>>> coll = {}
>>> for k, g in itertools.groupby(things, lambda x: x[0]):
... coll.setdefault(k, []).extend(i for _, i in g)
...
{'vehicle': ['bear', 'speed boat', 'school bus'], 'animal': ['duck', 'cactus']}