Python print from two dictionaries

Question

I split the dialogue into two dictionaries, each of them contains words which the person say (i have 2 persons). I have to print 4 columns (keyword, number from first directory (how many times use that word first person), number from second directory and count of them) and order by keyword. Can somebody help me ? Output have to look like this:

african   1  0  1
air-speed 1  0  0
an        1  1  2
arthur    1  0  1
...

As you can see I have som text

text = """Bridgekeeper: Hee hee heh. Stop. What... is your name?
King Arthur: It is 'Arthur', King of the Britons.
Bridgekeeper: What... is your quest?
King Arthur: To seek the Holy Grail.
Bridgekeeper: What... is the air-speed velocity of an unladen swallow?
King Arthur: What do you mean? An African or European swallow?"""

Output of bridgekeeper_w and arthur_w:

print (bridgekeeper_w) 

{'hee': 2, 'heh': 1, 'stop': 1, 'what': 3, 'is': 3, 'your': 2, 'name': 1, 'quest': 1, 'the': 1, 'air-speed': 1, 'velocity': 1, 'of': 1, 'an': 1, 'unladen': 1, 'swallow': 1}

print (arthur_w)
{'king': 4, 'it': 1, 'is': 1, 'arthur': 1, 'of': 1, 'the': 2, 'britons': 1, 'to': 1, 'seek': 1, 'holy': 1, 'grail': 1, 'what': 1, 'do': 1, 'you': 1, 'mean': 1, 'an': 1, 'african': 1, 'or': 1, 'european': 1, 'swallow': 1}

Now i need this (keyword, number from first dict, number from second dict, and count):

african   1  0  1
air-speed 1  0  0
an        1  1  2
arthur    1  0  1
...
``

Answer 1

If you already have two dictionaries, the main problem is how to loop over keys which are in either dictionary. But that's not hard;

for key in sorted(set(list(bridgekeeper_w.keys()) + list(arthur_w.keys()))):
    b_count = 0 if key not in bridgekeeper_w else bridgekeeper_w[key]
    a_count = 0 if key not in arthur_w else arthur_w[key]
    print('%-20s %3i %3i %3i' % (key, b_count, a_count, b_count+a_count))

If the integrity of the dictionaries is not important, a more elegant solution might be to add the missing keys to one of the dictionaries, and then simply loop over all its keys.

for key in arthur_w.keys():
    if key not in bridgekeeper_w:
        bridgekeeper_w[key] = 0

for key, b_count in sorted(bridgekeeper_w.items()):
    a_count = 0 if key not in arthur_w else arthur_w[key]
    print('%-20s %3i %3i %3i' % (key, b_count, a_count, b_count+a_count))

This does away with the rather tedious and slightly complex set(list(keys()...)) of the first solution, at the cost of traversing one of the dictionaries twice.

Answer 2

Or a solution without third-party libraries:

bridgekeeper_d = {'hee': 2, 'heh': 1, 'stop': 1, 'what': 3, 'is': 3, 'your': 2, 'name': 1, 'quest': 1, 'the': 1, 'air-speed': 1, 'velocity': 1, 'of': 1, 'an': 1, 'unladen': 1, 'swallow': 1}
arthur_d = {'king': 4, 'it': 1, 'is': 1, 'arthur': 1, 'of': 1, 'the': 2, 'britons': 1, 'to': 1, 'seek': 1, 'holy': 1, 'grail': 1, 'what': 1, 'do': 1, 'you': 1, 'mean': 1, 'an': 1, 'african': 1, 'or': 1, 'european': 1, 'swallow': 1}
joined = dict.fromkeys(list(bridgekeeper_d.keys()) + list(arthur_d.keys()), {})

for key, value in bridgekeeper_d.items():
    joined[key]["bridgekeeper"] = value

for key, value in arthur_d.items():
    joined[key]["arthur"] = value
# At this point, joined looks like this:
# {
#     'hee': {'bridgekeeper': 1, 'arthur': 1},
#     'heh': {'bridgekeeper': 1, 'arthur': 1},
#     'stop': {'bridgekeeper': 1, 'arthur': 1},
#     'what': {'bridgekeeper': 1, 'arthur': 1}
#     ...
# }

for key, dic in joined.items():
    print("%-15s %d %d %d" % (key, dic["bridgekeeper"], dic["arthur"], dic["bridgekeeper"] + dic["arthur"]))

Output:

hee             1 1 2
heh             1 1 2
stop            1 1 2
what            1 1 2
is              1 1 2
your            1 1 2
name            1 1 2
quest           1 1 2
the             1 1 2
air-speed       1 1 2
velocity        1 1 2
of              1 1 2
an              1 1 2
unladen         1 1 2
swallow         1 1 2
king            1 1 2
it              1 1 2
arthur          1 1 2
britons         1 1 2
to              1 1 2
seek            1 1 2
holy            1 1 2
grail           1 1 2
do              1 1 2
you             1 1 2
mean            1 1 2
african         1 1 2
or              1 1 2
european        1 1 2

Answer 3

There are few steps to achieve the below dataframe-

1. Spilt the string based on '\n' new line char.
2. initialize the result as defaultdict(list), then split each row on ':' use value at index 0 as the key and the value at index 1 as value.
3. Convert the value list for each key back to a string via join.
4. Remove puntuations
5. Use Counter to evaluate the value of each word in the string.

Finally, we'll have a JSON like this -

{'Bridgekeeper': Counter({'Hee': 1,
          'hee': 1,
          'heh': 1,
          'Stop': 1,
          'What': 3,
          'is': 3,
          'your': 2,
          'name': 1,
          'quest': 1,
          'the': 1,
          'airspeed': 1,
          'velocity': 1,
          'of': 1,
          'an': 1,
          'unladen': 1,
          'swallow': 1}),

This JSON can be transformed into the required output very easily if we load it into a dataframe.

from collections import defaultdict
import string
from collections import Counter
import pandas as pd

result = defaultdict(list)
for row in text.split('\n'):
    result[row.split(':')[0].strip()].append(row.split(':')[1].strip())

result = {key:(' '.join(value)).translate(str.maketrans('', '', string.punctuation)) for key,value in result.items()}
result = {key:Counter(value.split(' ')) for key,value in result.items()}
df = pd.DataFrame(result).fillna(0).astype(int)
df['sum'] = df['Bridgekeeper'] + df['King Arthur']
df.to_csv('out.csv', sep='\t')

Output Dataframe -

          Bridgekeeper  King Arthur  sum
Hee                  1            0    1
hee                  1            0    1
heh                  1            0    1
Stop                 1            0    1
What                 3            1    4
is                   3            1    4
your                 2            0    2
name                 1            0    1
quest                1            0    1
the                  1            2    3
airspeed             1            0    1
velocity             1            0    1
of                   1            1    2
an                   1            0    1
unladen              1            0    1
swallow              1            1    2
It                   0            1    1
Arthur               0            1    1
King                 0            1    1
Britons              0            1    1
To                   0            1    1
seek                 0            1    1
Holy                 0            1    1
Grail                0            1    1
do                   0            1    1
you                  0            1    1
mean                 0            1    1
An                   0            1    1