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我如何使用Python

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, and the examples, 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.

2016年12月03日43分35秒

As Sebastjan said, you first have to sort your data. This is important.

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.

2016年12月03日43分35秒

Can you show us your code?

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.

2016年12月03日43分35秒

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.

2016年12月03日43分35秒

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]

2016年12月03日43分35秒

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.

2016年12月03日43分35秒

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

2016年12月03日43分35秒

@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!

2016年12月03日43分35秒

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])]

2016年12月03日43分35秒

Thanks for lots of nice anwser, this is a hackerrank challenge about groupby, this is link.

args = "1234567890"

for key, group in groupby(args):
    # print group # you can look what group is.
    print((len(list(group)), int(key)), end=" ")

the output:

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) (1, 9) (1, 0)

just count the num occurrence.

2016年12月03日43分35秒