I have a table like below:
URN Firm_Name
0 104472 R.X. Yah & Co
1 104873 Big Building Society
2 109986 St James's Society
3 114058 The Kensington Society Ltd
4 113438 MMV Oil Associates Ltd
And I want to count the frequency of all the words within the Firm_Name column, to get an output like below:
I have tried the following code:
import pandas as pd
import nltk
data = pd.read_csv("X:\Firm_Data.csv")
top_N = 20
word_dist = nltk.FreqDist(data['Firm_Name'])
print('All frequencies')
print('='*60)
rslt=pd.DataFrame(word_dist.most_common(top_N),columns=['Word','Frequency'])
print(rslt)
print ('='*60)
However the following code does not produce a unique word count.
IIUIC, use value_counts()
In [3361]: df.Firm_Name.str.split(expand=True).stack().value_counts()
Out[3361]:
Society 3
Ltd 2
James's 1
R.X. 1
Yah 1
Associates 1
St 1
Kensington 1
MMV 1
Big 1
& 1
The 1
Co 1
Oil 1
Building 1
dtype: int64
Or,
pd.Series(np.concatenate([x.split() for x in df.Firm_Name])).value_counts()
Or,
pd.Series(' '.join(df.Firm_Name).split()).value_counts()
For top N, for example 3
In [3379]: pd.Series(' '.join(df.Firm_Name).split()).value_counts()[:3]
Out[3379]:
Society 3
Ltd 2
James's 1
dtype: int64
Details
In [3380]: df
Out[3380]:
URN Firm_Name
0 104472 R.X. Yah & Co
1 104873 Big Building Society
2 109986 St James's Society
3 114058 The Kensington Society Ltd
4 113438 MMV Oil Associates Ltd
You need str.cat
with lower
first for concanecate all values to one string
, then need word_tokenize
and last use your solution:
top_N = 4
#if not necessary all lower
a = data['Firm_Name'].str.lower().str.cat(sep=' ')
words = nltk.tokenize.word_tokenize(a)
word_dist = nltk.FreqDist(words)
print (word_dist)
<FreqDist with 17 samples and 20 outcomes>
rslt = pd.DataFrame(word_dist.most_common(top_N),
columns=['Word', 'Frequency'])
print(rslt)
Word Frequency
0 society 3
1 ltd 2
2 the 1
3 co 1
Also is possible remove lower
if necessary:
top_N = 4
a = data['Firm_Name'].str.cat(sep=' ')
words = nltk.tokenize.word_tokenize(a)
word_dist = nltk.FreqDist(words)
rslt = pd.DataFrame(word_dist.most_common(top_N),
columns=['Word', 'Frequency'])
print(rslt)
Word Frequency
0 Society 3
1 Ltd 2
2 MMV 1
3 Kensington 1
This answer can also be used - Count distinct words from a Pandas Data Frame . It utilizes the Counter method and applies it to each row.
from collections import Counter
c = Counter()
df = pd.DataFrame(
[[104472,"R.X. Yah & Co"],
[104873,"Big Building Society"],
[109986,"St James's Society"],
[114058,"The Kensington Society Ltd"],
[113438,"MMV Oil Associates Ltd"]
], columns=["URN","Firm_Name"])
df.Firm_Name.str.split().apply(c.update)
Counter({'R.X.': 1,
'Yah': 1,
'&': 1,
'Co': 1,
'Big': 1,
'Building': 1,
'Society': 3,
'St': 1,
"James's": 1,
'The': 1,
'Kensington': 1,
'Ltd': 2,
'MMV': 1,
'Oil': 1,
'Associates': 1})
This is faster:
df.Firm_Name.str.split().explode().value_counts()
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