Performance enhancement of ranking function by replacement of lambda x with vectorization

Question

I have a ranking function that I apply to a large number of columns of several million rows which takes minutes to run. By removing all of the logic preparing the data for application of the .rank( method, i.e., by doing this:

ranked = df[['period_id', 'sector_name'] + to_rank].groupby(['period_id', 'sector_name']).transform(lambda x: (x.rank(ascending = True) - 1)*100/len(x))        

I managed to get this down to seconds. However, I need to retain my logic, and am struggling to restructure my code: ultimately, the largest bottleneck is my double use of lambda x:, but clearly other aspects are slowing things down (see below). I have provided a sample data frame, together with my ranking functions below, i.e. an MCVE. Broadly, I think that my questions boil down to:

(i) How can one replace the .apply(lambda x usage in the code with a fast, vectorized equivalent? (ii) How can one loop over multi-indexed, grouped, data frames and apply a function? in my case, to each unique combination of the date_id and category columns.
(iii) What else can I do to speed up my ranking logic? the main overhead seems to be in .value_counts(). This overlaps with (i) above; perhaps one can do most of this logic on df, perhaps via construction of temporary columns, before sending for ranking. Similarly, can one rank the sub-dataframe in one call?
(iv) Why use pd.qcut() rather than df.rank()? the latter is cythonized and seems to have more flexible handling of ties, but I cannot see a comparison between the two, and pd.qcut() seems most widely used.

Sample input data is as follows:

import pandas as pd
import numpy as np
import random

to_rank = ['var_1', 'var_2', 'var_3']
df = pd.DataFrame({'var_1' : np.random.randn(1000), 'var_2' : np.random.randn(1000), 'var_3' : np.random.randn(1000)})
df['date_id'] = np.random.choice(range(2001, 2012), df.shape[0])
df['category'] = ','.join(chr(random.randrange(97, 97 + 4 + 1)).upper() for x in range(1,df.shape[0]+1)).split(',')

The two ranking functions are:

def rank_fun(df, to_rank): # calls ranking function f(x) to rank each category at each date
    #extra data tidying logic here beyond scope of question - can remove
    ranked = df[to_rank].apply(lambda x: f(x))
    return ranked


def f(x):
    nans = x[np.isnan(x)] # Remove nans as these will be ranked with 50
    sub_df = x.dropna() # 
    nans_ranked = nans.replace(np.nan, 50) # give nans rank of 50

    if len(sub_df.index) == 0: #check not all nan.  If no non-nan data, then return with rank 50
        return nans_ranked

    if len(sub_df.unique()) == 1: # if all data has same value, return rank 50
        sub_df[:] = 50
        return sub_df

    #Check that we don't have too many clustered values, such that we can't bin due to overlap of ties, and reduce bin size provided we can at least quintile rank.
    max_cluster = sub_df.value_counts().iloc[0] #value_counts sorts by counts, so first element will contain the max
    max_bins = len(sub_df) / max_cluster 

    if max_bins > 100: #if largest cluster <1% of available data, then we can percentile_rank
        max_bins = 100

    if max_bins < 5: #if we don't have the resolution to quintile rank then assume no data.
        sub_df[:] = 50
        return sub_df

    bins = int(max_bins) # bin using highest resolution that the data supports, subject to constraints above (max 100 bins, min 5 bins)

    sub_df_ranked = pd.qcut(sub_df, bins, labels=False) #currently using pd.qcut.  pd.rank( seems to have extra functionality, but overheads similar in practice
    sub_df_ranked *= (100 / bins) #Since we bin using the resolution specified in bins, to convert back to decile rank, we have to multiply by 100/bins.  E.g. with quintiles, we'll have scores 1 - 5, so have to multiply by 100 / 5 = 20 to convert to percentile ranking
    ranked_df = pd.concat([sub_df_ranked, nans_ranked])
    return ranked_df

And the code to call my ranking function and recombine with df is:

# ensure don't get duplicate columns if ranking already executed
ranked_cols = [col + '_ranked' for col in to_rank]

ranked = df[['date_id', 'category'] + to_rank].groupby(['date_id', 'category'], as_index = False).apply(lambda x: rank_fun(x, to_rank)) 
ranked.columns = ranked_cols        
ranked.reset_index(inplace = True)
ranked.set_index('level_1', inplace = True)    
df = df.join(ranked[ranked_cols])

I am trying to get this ranking logic as fast as I can, by removing both lambda x calls; I can remove the logic in rank_fun so that only f(x)'s logic is applicable, but I also don't know how to process multi-index dataframes in a vectorized fashion. An additional question would be on differences between pd.qcut( and df.rank(: it seems that both have different ways of dealing with ties, but the overheads seem similar, despite the fact that .rank( is cythonized; perhaps this is misleading, given the main overheads are due to my usage of lambda x.

I ran %lprun on f(x) which gave me the following results, although the main overhead is the use of .apply(lambda x rather than a vectorized approach:

Line # Hits Time Per Hit % Time Line Contents

 2                                           def tst_fun(df, field):
 3         1          685    685.0      0.2      x = df[field]
 4         1        20726  20726.0      5.8      nans = x[np.isnan(x)]
 5         1        28448  28448.0      8.0      sub_df = x.dropna()
 6         1          387    387.0      0.1      nans_ranked = nans.replace(np.nan, 50)
 7         1            5      5.0      0.0      if len(sub_df.index) == 0: 
 8                                                   pass #check not empty.  May be empty due to nans for first 5 years e.g. no revenue/operating margin data pre 1990
 9                                                   return nans_ranked
10                                           
11         1        65559  65559.0     18.4      if len(sub_df.unique()) == 1: 
12                                                   sub_df[:] = 50 #e.g. for subranks where all factors had nan so ranked as 50 e.g. in 1990
13                                                   return sub_df
14                                           
15                                               #Finally, check that we don't have too many clustered values, such that we can't bin, and reduce bin size provided we can at least quintile rank.
16         1        74610  74610.0     20.9      max_cluster = sub_df.value_counts().iloc[0] #value_counts sorts by counts, so first element will contain the max
17                                               # print(counts)
18         1            9      9.0      0.0      max_bins = len(sub_df) / max_cluster #
19                                           
20         1            3      3.0      0.0      if max_bins > 100: 
21         1            0      0.0      0.0          max_bins = 100 #if largest cluster <1% of available data, then we can percentile_rank
22                                           
23                                           
24         1            0      0.0      0.0      if max_bins < 5: 
25                                                   sub_df[:] = 50 #if we don't have the resolution to quintile rank then assume no data.
26                                           
27                                               #     return sub_df
28                                           
29         1            1      1.0      0.0      bins = int(max_bins) # bin using highest resolution that the data supports, subject to constraints above (max 100 bins, min 5 bins)
30                                           
31                                               #should track bin resolution for all data.  To add.
32                                           
33                                               #if get here, then neither nans_ranked, nor sub_df are empty
34                                               # sub_df_ranked = pd.qcut(sub_df, bins, labels=False)
35         1       160530 160530.0     45.0      sub_df_ranked = (sub_df.rank(ascending = True) - 1)*100/len(x)
36                                           
37         1         5777   5777.0      1.6      ranked_df = pd.concat([sub_df_ranked, nans_ranked])
38                                               
39         1            1      1.0      0.0      return ranked_df

Answer 1

I'd build a function using numpy
I plan on using this within each group defined within a pandas groupby

def rnk(df):
    a = df.values.argsort(0)
    n, m = a.shape
    r = np.arange(a.shape[1])
    b = np.empty_like(a)
    b[a, np.arange(m)[None, :]] = np.arange(n)[:, None]
    return pd.DataFrame(b / n, df.index, df.columns)

gcols = ['date_id', 'category']
rcols = ['var_1', 'var_2', 'var_3']
df.groupby(gcols)[rcols].apply(rnk).add_suffix('_ranked')

   var_1_ranked  var_2_ranked  var_3_ranked
0      0.333333      0.809524      0.428571
1      0.160000      0.360000      0.240000
2      0.153846      0.384615      0.461538
3      0.000000      0.315789      0.105263
4      0.560000      0.200000      0.160000
...

---

How It Works

• Because I know that ranking is related to sorting, I want to use some clever sorting to do this quicker.

• numpy's argsort will produce a permutation that can be used to slice the array into a sorted array.

  a = np.array([25, 300, 7])
  b = a.argsort()
  print(b)
  
  [2 0 1]
  
  print(a[b])
  
  [  7  25 300]
  

• So, instead, I'm going to use the argsort to tell me where the first, second, and third ranked elements are.

  # create an empty array that is the same size as b or a
  # but these will be ranks, so I want them to be integers
  # so I use empty_like(b) because b is the result of 
  # argsort and is already integers.
  u = np.empty_like(b)
  
  # now just like when I sliced a above with a[b]
  # I slice u the same way but instead I assign to
  # those positions, the ranks I want.
  # In this case, I defined the ranks as np.arange(b.size) + 1
  u[b] = np.arange(b.size) + 1
  
  print(u)
  
  [2 3 1]
  

• And that was exactly correct. The 7 was in the last position but was our first rank. 300 was in the second position and was our third rank. 25 was in the first position and was our second rank.

• Finally, I divide by the number in the rank to get the percentiles. It so happens that because I used zero based ranking np.arange(n), as opposed to one based np.arange(1, n+1) or np.arange(n) + 1 as in our example, I can do the simple division to get the percentiles.
• What's left to do is apply this logic to each group. We can do this in pandas with groupby
• Some of the missing details include how I use argsort(0) to get independent sorts per column` and that I do some fancy slicing to rearrange each column independently.

---

Can we avoid the groupby and have numpy do the whole thing?
I'll also take advantage of numba's just in time compiling to speed up some things with njit

from numba import njit

@njit
def count_factor(f):
    c = np.arange(f.max() + 2) * 0
    for i in f:
        c[i + 1] += 1
    return c

@njit
def factor_fun(f):
    c = count_factor(f)
    cc = c[:-1].cumsum()
    return c[1:][f], cc[f]

def lexsort(a, f):
    n, m = a.shape
    f = f * (a.max() - a.min() + 1)
    return (f.reshape(-1, 1) + a).argsort(0)


def rnk_numba(df, gcols, rcols):
    tups = list(zip(*[df[c].values.tolist() for c in gcols]))
    f = pd.Series(tups).factorize()[0]
    a = lexsort(np.column_stack([df[c].values for c in rcols]), f)
    c, cc = factor_fun(f)
    c = c[:, None]
    cc = cc[:, None]
    n, m = a.shape
    r = np.arange(a.shape[1])
    b = np.empty_like(a)
    b[a, np.arange(m)[None, :]] = np.arange(n)[:, None]
    return pd.DataFrame((b - cc) / c, df.index, rcols).add_suffix('_ranked')

How it works

• Honestly, this is difficult to process mentally. I'll stick with expanding on what I explained above.
• I want to use argsort again to drop rankings into the correct positions. However, I have to contend with the grouping columns. So what I do is compile a list of tuples and factorize them as was addressed in this question here
• Now that I have a factorized set of tuples I can perform a modified lexsort that sorts within my factorized tuple groups. This question addresses the lexsort.
• A tricky bit remains to be addressed where I must off set the new found ranks by the size of each group so that I get fresh ranks for every group. This is taken care of in the tiny snippet b - cc in the code below. But calculating cc is a necessary component.

So that's some of the high level philosophy. What about @njit?

• Note that when I factorize, I am mapping to the integers 0 to n - 1 where n is the number of unique grouping tuples. I can use an array of length n as a convenient way to track the counts.
• In order to accomplish the groupby offset, I needed to track the counts and cumulative counts in the positions of those groups as they are represented in the list of tuples or the factorized version of those tuples. I decided to do a linear scan through the factorized array f and count the observations in a numba loop. While I had this information, I'd also produce the necessary information to produce the cumulative offsets I also needed.
• numba provides an interface to produce highly efficient compiled functions. It is finicky and you have to acquire some experience to know what is possible and what isn't possible. I decided to numbafy two functions that are preceded with a numba decorator @njit. This coded works just as well without those decorators, but is sped up with them.

---

Timing

%%timeit 
ranked_cols = [col + '_ranked' for col in to_rank]
​
ranked = df[['date_id', 'category'] + to_rank].groupby(['date_id', 'category'], as_index = False).apply(lambda x: rank_fun(x, to_rank)) 
ranked.columns = ranked_cols        
ranked.reset_index(inplace = True)
ranked.set_index('level_1', inplace = True)  
1 loop, best of 3: 481 ms per loop

gcols = ['date_id', 'category']
rcols = ['var_1', 'var_2', 'var_3']

%timeit df.groupby(gcols)[rcols].apply(rnk_numpy).add_suffix('_ranked')
100 loops, best of 3: 16.4 ms per loop

%timeit rnk_numba(df, gcols, rcols).head()
1000 loops, best of 3: 1.03 ms per loop

Answer 2

I suggest you try this code. It's 3 times faster than yours, and more clear.

rank function:

def rank(x):
    counts = x.value_counts()
    bins = int(0 if len(counts) == 0 else x.count() / counts.iloc[0])
    bins = 100 if bins > 100 else bins
    if bins < 5:
        return x.apply(lambda x: 50)
    else:
        return (pd.qcut(x, bins, labels=False) * (100 / bins)).fillna(50).astype(int)

single thread apply:

for col in to_rank:
    df[col + '_ranked'] = df.groupby(['date_id', 'category'])[col].apply(rank)

mulple thread apply:

import sys
from multiprocessing import Pool

def tfunc(col):
    return df.groupby(['date_id', 'category'])[col].apply(rank)

pool = Pool(len(to_rank))
result = pool.map_async(tfunc, to_rank).get(sys.maxint)
for (col, val) in zip(to_rank, result):
    df[col + '_ranked'] = val