Efficient method to calculate the rank vector of a list in Python

This doesn’t give the exact result you specify, but perhaps it would be useful anyways. The following snippet gives the first index for each element, yielding a final rank vector of [0, 1, 2, 2, 2, 5, 6]

def rank_index(vector):
    return [vector.index(x) for x in sorted(range(n), key=vector.__getitem__)]

Your own testing would have to prove the efficiency of this.

There is a really nice module called Ranking http://pythonhosted.org/ranking/ with an easy to follow instruction page. To download, simply use easy_install ranking

Here is a small variation of unutbu’s code, including an optional ‘method’ argument for the type of value of tied ranks.

def rank_simple(vector):
    return sorted(range(len(vector)), key=vector.__getitem__)

def rankdata(a, method='average'):
    n = len(a)
    ivec=rank_simple(a)
    svec=[a[rank] for rank in ivec]
    sumranks = 0
    dupcount = 0
    newarray = [0]*n
    for i in xrange(n):
        sumranks += i
        dupcount += 1
        if i==n-1 or svec[i] != svec[i+1]:
            for j in xrange(i-dupcount+1,i+1):
                if method=='average':
                    averank = sumranks / float(dupcount) + 1
                    newarray[ivec[j]] = averank
                elif method=='max':
                    newarray[ivec[j]] = i+1
                elif method=='min':
                    newarray[ivec[j]] = i+1 -dupcount+1
                else:
                    raise NameError('Unsupported method')

            sumranks = 0
            dupcount = 0


    return newarray

These codes give me a lot of inspiration, especially unutbu’s code.
However my needs are simpler, so I changed the code a little.

Hoping to help the guys with the same needs.

Here is the class to record the players’ scores and ranks.

class Player():
    def __init__(self, s, r):
        self.score = s
        self.rank = r

Some data.

l = [Player(90,0),Player(95,0),Player(85,0), Player(90,0),Player(95,0)]

Here is the code for calculation:

l.sort(key=lambda x:x.score, reverse=True)    
l[0].rank = 1
dupcount = 0
prev = l[0]
for e in l[1:]:
    if e.score == prev.score:
        e.rank = prev.rank
        dupcount += 1
    else:
        e.rank = prev.rank + dupcount + 1
        dupcount = 0
        prev = e

import numpy as np

def rankVec(arg):
    p = np.unique(arg) #take unique value
    k = (-p).argsort().argsort() #sort based on arguments in ascending order
    dd = defaultdict(int)
    for i in xrange(np.shape(p)[0]):
        dd[p[i]] = k[i]
    return np.array([dd[x] for x in arg])

timecomplexity is 46.2us

This is one of the functions that I wrote to calculate rank.

def calculate_rank(vector):
  a={}
  rank=1
  for num in sorted(vector):
    if num not in a:
      a[num]=rank
      rank=rank+1
  return[a[i] for i in vector]

input:

calculate_rank([1,3,4,8,7,5,4,6])

output:

[1, 2, 3, 7, 6, 4, 3, 5]

[sorted(l).index(x) for x in l]

sorted(l) will give the sorted version
index(x) will give the index in the sorted array

for example :

l = [-1, 3, 2, 0,0]
>>> [sorted(l).index(x) for x in l]
[0, 4, 3, 1, 1]

So.. this is 2019, and I have no idea why nobody suggested the following:

# Python-only
def rank_list( x, break_ties=False ):
    n = len(x)
    t = list(range(n))
    s = sorted( t, key=x.__getitem__ )

    if not break_ties:
        for k in range(n-1):
            t[k+1] = t[k] + (x[s[k+1]] != x[s[k]])

    r = s.copy()
    for i,k in enumerate(s):
        r[k] = t[i]

    return r

# Using Numpy, see also: np.argsort
def rank_vec( x, break_ties=False ):
    n = len(x)
    t = np.arange(n)
    s = sorted( t, key=x.__getitem__ )

    if not break_ties:
        t[1:] = np.cumsum(x[s[1:]] != x[s[:-1]])

    r = t.copy()
    np.put( r, s, t )
    return r

This approach has linear runtime complexity after the initial sort, it only stores 2 arrays of indices, and does not require values to be hashable (only pairwise comparison needed).

AFAICT, this is better than other approaches suggested so far:

• @unutbu’s approach is essentially similar, but (I would argue) too complicated for what the OP asked;
• All suggestions using .index() are terrible, with a runtime complexity of N^2;
• @Yuvraj Singh improves slightly upon the .index() search using a dictionary, however with search and insert operations at each iteration, this is still highly inefficient both in time (NlogN) and space, and it also requires the values to be hashable.

I really don’t get why all the existing solutions are so complex. This can be done just like this:

[index for element, index in sorted(zip(sequence, range(len(sequence))))]

You build tuples which contain the elements and a running index. Then you sort the whole thing, and tuples sort by their first element and during ties by their second element. This way one has a sorted list of these tuples and just need to pick out the indices from that afterwards. Also this removes the need to look up elements in the sequence afterwards, which likely makes it a O(N²) operation whereas this is O(N log(N)).

This works for the spearman correlation coefficient .

def get_rank(X, n):
    x_rank = dict((x, i+1) for i, x in enumerate(sorted(set(X))))
    return [x_rank[x] for x in X]

The rank function could be implemented in O(n log n) time and O(n) additional space using the following approach.

import bisect

def rank_list(lst: list[int]) -> list[int]:
    sorted_vals = sorted(set(lst))
    return [bisect.bisect_left(sorted_vals, val) for val in lst]

I use here bisect library, but for the pure independent code it is enough to implement binary search procedure on the sorted array with unique values for a query on existing (in this array) value.

The most pythonic style to find rank of an array:

a = [10.0, 9.8, 8.0, 7.8, 7.7, 7.0, 6.0, 5.0, 4.0, 2.0]
rank = lambda arr: list(map(lambda i: sorted(arr).index(i)+1, arr))
rank(a)