Creating A Separate Counter() Object And Pandas Dataframe For Each List Within A List Of Lists

All the other answers I could find specifically referred to aggregating across all of the nested lists within a list of lists, where as I'm looking to aggregate separately for each

Solution 1:

IMO, this question can show the real pandas's power. Let's do the following - instead of counting boring [a,a,b,b,b,c,c,c], [d,d,d,a,a,a,c,c,c], [c,c,c,a,a,f,f,f] we will count the frequency of words in real books. I've chosen the following three: 'Faust', 'Hamlet', 'Macbeth'.

Code:

#!/usr/bin/env python# -*- coding: utf-8 -*-from collections import defaultdict
import string
import requests
import pandas as pd

books = {
  'Faust': 'http://www.gutenberg.org/cache/epub/2229/pg2229.txt',
  'Hamlet': 'http://www.gutenberg.org/cache/epub/2265/pg2265.txt',
  'Macbeth': 'http://www.gutenberg.org/cache/epub/2264/pg2264.txt',
}

# prepare translate table, which will remove all punctuations and digits
chars2remove = list(string.punctuation + string.digits)
transl_tab = str.maketrans(dict(zip(chars2remove, list(' ' * len(chars2remove)))))
# replace 'carriage return' and 'new line' characters with spaces
transl_tab[10] = ' '
transl_tab[13] = ' 'deftokenize(s):
    return s.translate(transl_tab).lower().split()

defget_data(url):
    r = requests.get(url)
    if r.status_code == requests.codes.ok:
        return r.text
    else:
        r.raise_for_status()

# generate DF containing words from books
d = defaultdict(list)
for name, url in books.items():
    d[name] = tokenize(get_data(url))

df = pd.concat([pd.DataFrame({'book': name, 'word': tokenize(get_data(url))})
                for name, url in books.items()], ignore_index=True)

# let's count the frequency
frequency = df.groupby(['book','word']) \
              .size() \
              .sort_values(ascending=False)

# outputprint(frequency.head(30))
print('[Macbeth]: macbeth\t', frequency.loc['Macbeth', 'macbeth'])
print('[Hamlet]: nay\t', frequency.loc['Hamlet', 'nay'])
print('[Faust]: faust\t', frequency.loc['Faust', 'faust'])

Output:

book     word
Hamlet   the      1105
         and       919
Faust    und       918
Hamlet   to        760
Macbeth  the       759
Hamlet   of        698
Faust    ich       691
         die       668
         der       610
Macbeth  and       602
Hamlet   you       588
         i         560
         a         542
         my        506
Macbeth  to        460
Hamlet   it        439
Macbeth  of        426
Faust    nicht     426
Hamlet   in        409
Faust    das       403
         ein       399
         zu        380
Hamlet   that      379
Faust    in        365
         ist       363
Hamlet   is        346
Macbeth  i         344
Hamlet   ham       337
         this      328
         not       316
dtype: int64

[Macbeth]: macbeth      67
[Hamlet]: nay    27
[Faust]: faust   272

Solution 2:

You can create a list and append the counters to it. (Also, you are using Counter, but still doing the counts yourself, which is unnecessary.)

master_list =[[a,a,b,b,b,c,c,c],[d,d,d,a,a,a,c,c,c],[c,c,c,a,a,f,f,f]]
counters =[]for list_ in master_list:
    counters.append(Counter(list_))

Now you can address each separate list with counters[i].