What is the best way to categorize a list in python?
for example:
totalist is below
totalist[1] = ['A','B','C','D','E']
totalist[2] = ['A','B','X','Y','Z']
totalist[3] = ['A','F','T','U','V']
totalist[4] = ['A','F','M','N','O']
Say I want to get the list where the first two items are ['A','B'], basically list[1] and list[2]. Is there an easy way to get these without iterate one item at a time? Like something like this?
if ['A','B'] in totalist
I know that doesn't work.
You could check the first two elements of each list.
for totalist in all_lists:
if totalist[:2] == ['A', 'B']:
# Do something.
Note: The one-liner solutions suggested by Kasramvd are quite nice too. I found my solution more readable. Though I should say comprehensions are slightly faster than regular for loops. (Which I tested myself.)
Just for fun, itertools solution to push per-element work to the C layer:
from future_builtins import map # Py2 only; not needed on Py3
from itertools import compress
from operator import itemgetter
# Generator
prefixes = map(itemgetter(slice(2)), totalist)
selectors = map(['A','B'].__eq__, prefixes)
# If you need them one at a time, just skip list wrapping and iterate
# compress output directly
matches = list(compress(totalist, selectors))
This could all be one-lined to:
matches = list(compress(totalist, map(['A','B'].__eq__, map(itemgetter(slice(2)), totalist))))
but I wouldn't recommend it. Incidentally, if totalist might be a generator, not a re-iterable sequence, you'd want to use itertools.tee to double it, adding:
totalist, forselection = itertools.tee(totalist, 2)
and changing the definition of prefixes to map over forselection, not totalist; since compress iterates both iterators in parallel, tee won't have meaningful memory overhead.
Of course, as others have noted, even moving to C, this is a linear algorithm. Ideally, you'd use something like a collections.defaultdict(list) to map from two element prefixes of each list (converted to tuple to make them legal dict keys) to a list of all lists with that prefix. Then, instead of linear search over N lists to find those with matching prefixes, you just do totaldict['A', 'B'] and you get the results with O(1) lookup (and less fixed work too; no constant slicing).
Example precompute work:
from collections import defaultdict
totaldict = defaultdict(list)
for x in totalist:
totaldict[tuple(x[:2])].append(x)
# Optionally, to prevent autovivification later:
totaldict = dict(totaldict)
Then you can get matches effectively instantly for any two element prefix with just:
matches = totaldict['A', 'B']
You could do this.
>>> for i in totalist:
... if ['A','B']==i[:2]:
... print i
Basically you can't do this in python with a nested list. But if you are looking for an optimized approach here are some ways:
Use a simple list comprehension, by comparing the intended list with only first two items of sub lists:
>>> [sub for sub in totalist if sub[:2] == ['A', 'B']]
[['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'X', 'Y', 'Z']]
If you want the indices use enumerate:
>>> [ind for ind, sub in enumerate(totalist) if sub[:2] == ['A', 'B']]
[0, 1]
And here is a approach in Numpy which is pretty much optimized when you are dealing with large data sets:
>>> import numpy as np
>>>
>>> totalist = np.array([['A','B','C','D','E'],
... ['A','B','X','Y','Z'],
... ['A','F','T','U','V'],
... ['A','F','M','N','O']])
>>> totalist[(totalist[:,:2]==['A', 'B']).all(axis=1)]
array([['A', 'B', 'C', 'D', 'E'],
['A', 'B', 'X', 'Y', 'Z']],
dtype='|S1')
Also as an alternative to list comprehension in python if you don't want to use a loop and you are looking for a functional way, you can use filter function, which is not as optimized as a list comprehension:
>>> list(filter(lambda x: x[:2]==['A', 'B'], totalist))
[['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'X', 'Y', 'Z']]
You imply that you are concerned about performance (cost). If you need to do this, and if you are worried about performance, you need a different data-structure. This will add a little "cost" when you making the lists, but save you time when filtering them.
If the need to filter based on the first two elements is fixed (it doesn't generalise to the first n elements) then I would add the lists, as they are made, to a dict where the key is a tuple of the first two elements, and the item is a list of lists.
then you simply retrieve your list by doing a dict lookup. This is easy to do and will bring potentially large speed ups, at almost no cost in memory and time while making the lists.
Related
I want to get the subset of the list without repetition of any elements in python.
For example, Mylist=[['A','B'.'C','D','E']]
I want to get the output like [['A','B'.'C'],['D','E','A']]
I see that what you need is a list of permutations of your list.
But your list should not include nested lists, so change the way
you defined it to:
Mylist=['A','B','C','D','E']
(in single brackets).
To print a full list of permutations, you can use itertools.permutations,
(import itertools needed), e.g.
pLen = 3 # How many elements in each list
for prm in itertools.permutations(Mylist, pLen):
print(list(prm))
Note that I used casting to list type, as itertools.permutations
returns tuples.
But if you want only a limited number of such permutations,
you can achieve it using another itertools function, namely islice:
pLen = 3 # How many elements in each list
listLen = 4 # How many lists
result = [ list(prm) for prm in (itertools.islice(
itertools.permutations(Mylist, pLen), listLen)) ]
The result is:
[['A', 'B', 'C'], ['A', 'B', 'D'], ['A', 'B', 'E'], ['A', 'C', 'B']]
Other approach, without itertools:
import random
pLen = 3 # How many elements in each list
listLen = 4 # How many lists
result = []
wrk = Mylist.copy()
for i in range(listLen):
random.shuffle(wrk)
result.append(wrk[:pLen])
I am attempting to deduplicate some pandas DataFrames, and I have a function that does this pair-wise (i.e. two dfs at a time). I want to write another function that takes a list of DataFrames of arbitrary length and combines the first two elements in the list, then combines the result with the third element in the list until we reach the end of the list.
For simplicity, I'll assume my deduplication function is simply string concatenation.
I tried some recursive functions, but it's not quite correct.
def dedupe_recursive(input_list):
if input_list == []:
return
else:
for i in range(0, len(input_list)-1):
new_list = input_list[i+1:]
deduped = dedupe(new_list[i], new_list[i+1])
print(deduped, new_list)
return dedupe_recursive(new_list)
Input (list): ['a', 'b', 'c', 'd']
Output (list of lists): [['ab'], ['ab', 'c'], ['abc', 'd']]
There's a function for exactly this kind of thing, it's called reduce. You would use it like this:
from functools import reduce
final_df = reduce(dedupe, list_of_dataframes)
I want to get an intersection of lists where duplication is not eliminated.
And I hope that the method is a fast way not to use loops.
Below was my attempt, but this method failed because duplicates were removed.
a = ['a','b','c','f']
b = ['a','b','b','o','k']
tmp = list(set(a) & set(b))
>>>tmp
>>>['b','a']
I want the result to be ['a', 'b', 'b'].
In this method, 'a' is a fixed value and 'b' is a variable value.
And the concept of extracting 'a' value from 'b'.
Is there a way to extract a list of cross-values that do not remove duplicate values?
A solution could be
good = set(a)
result = [x for x in b if x in good]
there are two loops here; one is the set-building loop of set (that is implemented in C, a hundred of times faster than whatever you can do in Python) the other is the comprehension and runs in the interpreter.
The first loop is done to avoid a linear search in a for each element of b (if a becomes big this can be a serious problem).
Note that using filter instead is probably not going to gain much (if anything) because despite the filter loop being in C, for each element it will have to get back to the interpreter to call the filtering function.
Note that if you care about speed then probably Python is not a good choice... for example may be PyPy would be better here and in this case just writing an optimal algorithm explicitly should be ok (avoiding re-searching a for duplicates when they are consecutive in b like happens in your example)
good = set(a)
res = []
i = 0
while i < len(b):
x = b[i]
if x in good:
while i < len(b) and b[i] == x: # is?
res.append(x)
i += 1
else:
i += 1
Of course in performance optimization the only real way is try and measure with real data on the real system... guessing works less and less as technology advances and becomes more complicated.
If you insist on not using for explicitly then this will work:
>>> list(filter(a.__contains__, b))
['a', 'b', 'b']
But directly calling magic methods like __contains__ is not a recommended practice to the best of my knowledge, so consider this instead:
>>> list(filter(lambda x: x in a, b))
['a', 'b', 'b']
And if you want to improve the lookup in a from O(n) to O(1) then create a set of it first:
>>> a_set = set(a)
>>> list(filter(lambda x: x in a_set, b))
['a', 'b', 'b']
>>a = ['a','b','c','f']
>>b = ['a','b','b','o','k']
>>items = set(a)
>>found = [i for i in b if i in items]
>>items
{'f', 'a', 'c', 'b'}
>>found
['a', 'b', 'b']
This should do your work.
I guess it's not faster than a loop and finally you probably still need a loop to extract the result. Anyway...
from collections import Counter
a = ['a','a','b','c','f']
b = ['a','b','b','o','k']
count_b = Counter(b)
count_ab = Counter(set(b)-set(a))
count_b - count_ab
#=> Counter({'a': 1, 'b': 2})
I mean, if res holds the result, you need to:
[ val for sublist in [ [s] * n for s, n in res.items() ] for val in sublist ]
#=> ['a', 'b', 'b']
It isn't clear how duplicates are handled when performing an intersection of lists which contain duplicate elements, as you have given only one test case and its expected result, and you did not explain duplicate handling.
According to how keeping duplicates work currently, the common elements are 'a' and 'b', and the intersection list lists 'a' with multiplicity 1 and 'b' with multiplicity 2. Note 'a' occurs once on both lists a and b, but 'b' occurs twice on b. The intersection list lists the common element with multiplicity equal to the list having that element at the maximum multiplicity.
The answer is yes. However, a loop may implicitly be called - though you want your code to not explicitly use any loop statements. This algorithm, however, will always be iterative.
Step 1: Create the intersection set, Intersect that does not contain duplicates (You already done that). Convert to list to keep indexing.
Step 2: Create a second array, IntersectD. Create a new variable Freq which counts the maximum number of occurrences for that common element, using count. Use Intersect and Freq to append the element Intersect[k] a number of times depending on its corresponding Freq[k].
An example code with 3 lists would be
a = ['a','b','c','1','1','1','1','2','3','o']
b = ['a','b','b','o','1','o','1']
c = ['a','a','a','b','1','2']
intersect = list(set(a) & set(b) & set(c)) # 3-set case
intersectD = []
for k in range(len(intersect)):
cmn = intersect[k]
freq = max(a.count(cmn), b.count(cmn), c.count(cmn)) # 3-set case
for i in range(freq): # Can be done with itertools
intersectD.append(cmn)
>>> intersectD
>>> ['b', 'b', 'a', 'a', 'a', '1', '1', '1', '1']
For cases involving more than two lists, freq for this common element can be computed using a more complex set intersection and max expression. If using a list of lists, freq can be computed using an inner loop. You can also replace the inner i-loop with an itertools expression from How can I count the occurrences of a list item?.
I was wondering what would be an efficient an elegant way of slicing a python list based on the index. In order to provide a minimal example:
temp = ['a','b','c','d']
index_needed=[0,2]
How can I slice the list without the loop?
expected output
output_list =['a','c']
I have a sense that there would be a way but haven't figured out any. Any suggestions?
First, note that indexing in Python begins at 0. So the indices you need will be [0, 2].
You can then use a list comprehension:
temp = ['a', 'b', 'c', 'd']
idx = [0, 2]
res = [temp[i] for i in idx] # ['a', 'c']
With built-ins, you may find map performs better:
res = map(temp.__getitem__, idx) # ['a', 'c']
Since you are using Python 2.7, this returns a list. For Python 3.x, you would need to pass the map object to list.
If you are looking to avoid a Python-level loop altogether, you may wish to use a 3rd party library such as NumPy:
import numpy as np
temp = np.array(['a', 'b', 'c', 'd'])
res = temp[idx]
# array(['a', 'c'],
# dtype='<U1')
res2 = np.delete(temp, idx)
# array(['b', 'd'],
# dtype='<U1')
This returns a NumPy array, which you can then be converted to a list via res.tolist().
Use this :
temp = ['a','b','c','d']
temp[0:4:2]
#Output
['a', 'c']
Here first value is starting index number which is (Included) second value is ending index number which is (Excluded) and third value is (steps) to be taken.
Happy Learning...:)
An alternative that pushes the work to the C layer on CPython (the reference interpreter):
from operator import itemgetter
temp = ['a','b','c','d']
index_needed=[0,2]
output_list = itemgetter(*index_needed)(temp)
That returns tuple of the values; if list is necessary, just wrap in the list constructor:
output_list = list(itemgetter(*index_needed)(temp))
Note that this only works properly if you need at least two indices; itemgetter is variable return type based on how it's initialized, returning the value directly when it's passed a single key to pull, and a tuple of values when passed more than one key.
It's also not particularly efficient for one-off uses. A more common use case would be if you had an iterable of sequences (typically tuples, but any sequence works), and don't care about them. For example, with an input list of:
allvalues = [(1, 2, 3, 4),
(5, 6, 7, 8)]
if you only wanted the values from index 1 and 3, you could write a loop like:
for _, x, _, y in allvalues:
where you unpack all the values but send the ones you don't care about to _ to indicate the lack of interest, or you can use itemgetter and map to strip them down to what you care about before the unpack:
from future_builtins import map # Because Py2's map is terrible; not needed on Py3
for x, y in map(itemgetter(1, 3), allvalues):
The itemgetter based approach doesn't care if you have more than four items in a given element of allvalues, while manual unpacking would always require exactly four; which is better is largely based on your use case.
If I have sublist A: ['E','C', 'W'], what is the most pythonic way to order the sublist according to the order of master list M: ['C','B','W','E','K']
My solution is seems rather rudimentary. I am curious if there is a more 'pythonic' way to get the same result.
ORDER = ['C','B','W','E','K']
possibilities = ['E','C', 'W']
possibilities_in_order = []
for x in ORDER:
if x in possibilities: possibilities_in_order.append(x)
>>> order = ['C','B','W','E','K']
>>> possibilities = ['E','C','W']
>>> possibilities_in_order = sorted(possibilities, key=order.index)
>>> possibilities_in_order
['C', 'W', 'E']
How this works: for each element in possibilities, order.index(element) is called, and the list is simply sorted by those respective positions.
More details: Built-in Functions → sorted.
possibilities.sort(key=lambda x : ORDER.index(x))
Here's a linear-time solution:
posset = set(possibilities)
[letter for letter in order if letter in posset]
This filters the master list for only the members of the sublist. It's O(n) because it only traverses the master list once, and will perform well if the sublist is close in size to the master list.
This also assumes that possibilities has no duplicates. You can handle that if necessary, however, although it will make the code more complex.