遍历行和列以在Pandas中添加计数

Question

I am trying to iterate over columns AND rows in Pandas to cross-reference a list I have and count the cooccurrences.

My dataframe looks like:

+-------+-----+-----+----+----+-------+-------+------+
| Lemma | Dog | Cat | Sg | Pl |  Good |  Okay |  Bad |
+-------+-----+-----+----+----+-------+-------+------+
| Dog   |   0 |   0 |  0 |  0 |   0   |   0   |  0   |
| Cat   |   0 |   0 |  0 |  0 |   0   |   0   |  0   |
+-------+-----+-----+----+----+-------+-------+------+

I have a list like:

c=[[dog, Sg, Good], [cat, Pl, Okay], [dog, Pl, Bad]

I want to go through every item in Lemma , find it in c and then for that list item look for any of the column names. If those column names are seen, I was to add +1. I also want to add a count if the Lemma items occur in a 3 word window of each other.

I've tried something like the following (ignoring the word window issue):

for idx, row in df.iterrows():
    for columns in df:
        for i in c:
            if i[0]==row:
                if columns in c[1]:
                    df.ix['columns','row'] +=1

But I get the error: "ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()."

My ideal results look like:

+-------+-----+-----+----+----+-------+-------+------+
| Lemma | Dog | Cat | Sg | Pl |  Good |  Okay |  Bad |
+-------+-----+-----+----+----+-------+-------+------+
| Dog   |   1 |   1 |  1 |  1 |   1   |   0   |  1   |
| Cat   |   2 |   0 |  0 |  1 |   0   |   1   |  0   |
+-------+-----+-----+----+----+-------+-------+------+

Thanks!

Answer 1

You have several things that need to be changed.

1) Your list probably needs to have Dog instead of dog , Cat instead of cat

2) You probably want: for column in df.columns instead of for columns in df

3) You probably want: if i[0] == row['Lemma'] instead of if i[0]==row: (this is where it was breaking

4) You probably want if column in i instead of if columns in c[1]

5) You probably want df.ix[idx, column] += 1 instead of df.ix['columns','row'] +=1

Answer 2

1. The ideal result shown in the question is not accurate. There should never be a cat in the dog column and vise versa.
2. I wouldn't iterate through the DataFrame , I'd unpack the list of lists into a dict then load the dict into a DataFrame , as shown below.

Code:

import pandas as pd

c=[['dog', 'Sg', 'Good'], ['cat', 'Pl', 'Okay'], ['dog', 'Pl', 'Bad'],
   ['dog', 'Sg', 'Good'], ['cat', 'Pl', 'Okay'], ['dog', 'Pl', 'Okay'],
   ['dog', 'Sg', 'Good'], ['cat', 'Sg', 'Good'], ['dog', 'Pl', 'Bad'],
   ['dog', 'Sg', 'Good'],['cat', 'Pl', 'Okay'], ['dog', 'Pl', 'Bad']]

Lemma = {'dog': {'dog': 0, 'Sg': 0, 'Pl': 0, 'Good': 0, 'Okay': 0, 'Bad': 0},
         'cat': {'cat': 0, 'Sg': 0, 'Pl': 0, 'Good': 0, 'Okay': 0, 'Bad': 0}}

Note: Each value in a list from c is a key in Lemma . Reference python dictionaries . eg With x = ['dog', 'Sg', 'Good'] , Lemma[x[0]][x[2]] is the same as Lemma['dog']['Good'] . The initial value of Lemma['dog']['Good'] = 0, therefore Lemma['dog']['Good'] = 0 + 1, then next time it would be 1 + 1, etc.

for x in c:
    Lemma[x[0]][x[0]] = Lemma[x[0]][x[0]] + 1
    Lemma[x[0]][x[1]] = Lemma[x[0]][x[1]] + 1
    Lemma[x[0]][x[2]] = Lemma[x[0]][x[2]] + 1

df = pd.DataFrame.from_dict(Lemma, orient='index')

Output:

Plot

df.plot(kind='bar', figsize=(6, 6))

Create the dict programmatically:

create sets of words for the dict keys from the list of lists :

outer_keys = set()
inner_keys = set()
for x in c:
    outer_keys.add(x[0])  # first word is outer key
    inner_keys |= set(x[1:])  # all other words

create dict of dicts :

Lemma = {j: dict.fromkeys(inner_keys | {j}, 0) for j in outer_keys}

final dict :

{'dog': {'Okay': 0, 'Pl': 0, 'Good': 0, 'Bad': 0, 'Sg': 0, 'dog': 0},
 'cat': {'Okay': 0, 'Pl': 0, 'Good': 0, 'Bad': 0, 'Sg': 0, 'cat': 0}}