使用正则表达式提取不同格式的日期并对它们进行排序 - pandas

Question

我是文本挖掘的新手，我需要从 *.txt 文件中提取日期并对它们进行排序。日期在句子之间（每行），它们的格式可能如下所示：

04/20/2009; 04/20/09; 4/20/09; 4/3/09
Mar-20-2009; Mar 20, 2009; March 20, 2009; Mar. 20, 2009; Mar 20 2009;
20 Mar 2009; 20 March 2009; 20 Mar. 2009; 20 March, 2009
Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009
Feb 2009; Sep 2009; Oct 2010
6/2008; 12/2009
2009; 2010

如果缺少日期，请考虑 1 日，如果缺少月份，请考虑 1 月。

我的想法是提取所有日期并将其转换为 mm/dd/yyyy 格式。但是我对如何查找和替换模式有点怀疑。这就是我所做的：

import pandas as pd

doc = []
with open('dates.txt') as file:
    for line in file:
        doc.append(line)

df = pd.Series(doc)

df2 = pd.DataFrame(df,columns=['text'])

def myfunc(x):
    if len(x)==4:
        x = '01/01/'+x
    else:
        if not re.search('/',x):
            example = re.sub('[-]','/',x)
            terms = re.split('/',x)
            if (len(terms)==2):
                if len(terms[-1])==2:
                    x = '01/'+terms[0]+'/19'+terms[-1]
                else:
                    x = '01/'+terms[0]+'/'+terms[-1] 
            elif len(terms[-1])==2:
                x = terms[0].zfill(2)+'/'+terms[1].zfill(2)+'/19'+terms[-1]
    return x

df2['text'] = df2.text.str.replace(r'(((?:\d+[/-])?\d+[/-]\d+)|\d{4})', lambda x: myfunc(x.groups('Date')[0]))

我只为数字日期格式做过。但是我有点困惑如何使用字母数字日期。

我知道这是一个粗略的代码，但这正是我得到的。

Answer 1

我认为这是 coursera 文本挖掘作业之一。那么你可以使用正则表达式和提取来获得解决方案。 dates.txt 即

doc = []
with open('dates.txt') as file:
    for line in file:
        doc.append(line)

df = pd.Series(doc)

def date_sorter():
    # Get the dates in the form of words
    one = df.str.extract(r'((?:\d{,2}\s)?(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*(?:-|\.|\s|,)\s?\d{,2}[a-z]*(?:-|,|\s)?\s?\d{2,4})')
    # Get the dates in the form of numbers
    two = df.str.extract(r'((?:\d{1,2})(?:(?:\/|-)\d{1,2})(?:(?:\/|-)\d{2,4}))')
    # Get the dates where there is no days i.e only month and year  
    three = df.str.extract(r'((?:\d{1,2}(?:-|\/))?\d{4})')
    #Convert the dates to datatime and by filling the nans in two and three. Replace month name because of spelling mistake in the text file.
    dates = pd.to_datetime(one.fillna(two).fillna(three).replace('Decemeber','December',regex=True).replace('Janaury','January',regex=True))
return pd.Series(dates.sort_values())

date_sorter()

输出：

9     1971-04-10
84    1971-05-18
2     1971-07-08
53    1971-07-11
28    1971-09-12
474   1972-01-01
153   1972-01-13
13    1972-01-26
129   1972-05-06
98    1972-05-13
111   1972-06-10
225   1972-06-15
31    1972-07-20
171   1972-10-04
191   1972-11-30
486   1973-01-01
335   1973-02-01
415   1973-02-01
36    1973-02-14
405   1973-03-01
323   1973-03-01
422   1973-04-01
375   1973-06-01
380   1973-07-01
345   1973-10-01
57    1973-12-01
481   1974-01-01
436   1974-02-01
104   1974-02-24
299   1974-03-01

如果你只想 return 索引那么 return pd.Series(dates.sort_values().index)

第一个正则表达式的解析

 #?: Non-capturing group 

((?:\d{,2}\s)? # The two digits group. `?` refers to preceding token or group. Here the digits of 2 or 1 and space occurring once or less.  

 (?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* # The words in group ending with any letters `[]` occuring any number of times (`*`). 

 (?:-|\.|\s|,) # Pattern matching -,.,space 

 \s? #(`?` here it implies only to space i.e the preceding token)

 \d{,2}[a-z]* # less than or equal to two digits having any number of letters at the end (`*`). (Eg: may be 1st, 13th , 22nd , Jan , December etc ) . 

 (?:-|,|\s)?# The characters -/,/space may occur once and may not occur because of `?` at the end

 \s? # space may occur or may not occur at all (maximum is 1) (`?` here it refers only to space)

 \d{2,4}) # Match digit which is 2 or 4

希望对您有所帮助。

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