Source code for nlp_primitives.punctuation_count

# -*- coding: utf-8 -*-

import re
import string

import numpy as np
from featuretools.primitives.base import TransformPrimitive
from featuretools.variable_types import NaturalLanguage, Numeric

[docs]class PunctuationCount(TransformPrimitive): """Determines number of punctuation characters in a string. Description: Given list of strings, determine the number of punctuation characters in each string. Looks for any of the following: !"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~ If a string is missing, return `NaN`. Examples: >>> x = ['This is a test file.', 'This is second line', 'third line: $1,000'] >>> punctuation_count = PunctuationCount() >>> punctuation_count(x).tolist() [1.0, 0.0, 3.0] """ name = "punctuation_count" input_types = [NaturalLanguage] return_type = Numeric default_value = 0 def get_function(self): pattern = "(%s)" % '|'.join([re.escape(x) for x in string.punctuation]) def punctuation_count(x): x = x.reset_index(drop=True) counts = x.str.extractall(pattern).groupby(level=0).count()[0] counts = counts.reindex_like(x).fillna(0) counts[x.isnull()] = np.nan return counts.astype(float) return punctuation_count