使用Python从简历中提取多个字段

时间:2013-08-14 15:05:29

标签: python python-2.7

我正在尝试用Python处理很多简历。简历的示例可能如下所示。不幸的是,每个简历可能不会使用相同的格式。除了使用正则表达式从简历中提取某些字段(假设我将所有字段都转换为纯文本)并使用python之外,有没有一种方法可以做到这一点?

Name: Someone
Tel: xxx-xxxxxxx
Add: 123 Some Street
Email: Someone@gmail.com

Objective/Goal
To obtain a position in...

Education
2004 - 2006: University of XYZ


Work Experience
2006 - 2008: Programmer

Skills
Programming skills: Python, ..

假设我只想在那里提取一些字段,如何获得字段名称和下一个字段之间的所有文本?例如,我只想获得名称和工作经验字段,它应返回以下内容。

NameField = 'Someone'
WorkExpField = '2006 - 2008: Programmer...'

2 个答案:

答案 0 :(得分:3)

我的“我要尝试这个,但是懒得做漂亮”的方法用于不同格式的简历。我愿意用不同的简历格式来测试它。 欢迎提供其他建议/意见!

import string

class Resume():
    def __init__(self,filename):
        self.filepath = filename
        self.load()
        self.parse()

    def load(self):
        with open(self.filepath,'rb') as f:
            self.content = f.read().splitlines()

    def checkLine(self,word,value, content, line):
        if word in content.lower():
            value = self.addValue(value,line)
        return value

    def addValue(self,value,line):
        value[line] = value.get(line,0) + 1
        return value

    def dict_List(self,dict_, content):
        new = [(key,value) for key,value in dict_.items() if dict_[key] == max(dict_.values())]
        return [(x[0],content[x[0]]) for x in sorted(new)]

    def get_name(self):
        names = []
        for each in self.name:
            if each[0] not in self.headings:
                each = each[1].replace('Name',"")
                if each[0] not in string.letters:
                    each = each[1:]
                names.append(each.strip())
            else:
                index = self.headings[self.headings.index(each[0])+1]
                names.append("\n".join(self.content[each[0]+1:index]))
        if len(names) == 1:
            return names[0]
        else:
            return names

    def get_work(self):
        experience = []
        for each in self.work:
            index = self.headings[self.headings.index(each[0])+1]
            experience.append("\n".join(self.content[each[0]+1:index]))
        if len(experience) == 1:
            return experience[0]
        else:
            return epxerience

    def parse(self):
        name = dict()
        work_experience = dict()
        isHeading = dict()
        for line_num in range(len(self.content)):
            for checkName in ["name",":"]:
                name.update(self.checkLine(checkName,name,self.content[line_num], line_num))
            for checkWork in ["work","experience"]:
                work_experience.update(self.checkLine(checkWork,work_experience, self.content[line_num],line_num))
            if line_num != len(self.content) - 1:
                if len(self.content[line_num + 1]) > len(self.content[line_num]):
                    isHeading.update(self.addValue(isHeading,line_num))
            if line_num > 0:
                if self.content[line_num - 1] == "":
                    isHeading.update(self.addValue(isHeading,line_num))
            if len(self.content[line_num]) == len(self.content[line_num].lstrip()):
                isHeading.update(self.addValue(isHeading,line_num))
            if self.content[line_num] == "":
                isHeading[line_num] = isHeading.get(line_num,0) - 1

        self.name = self.dict_List(name, self.content)
        self.work = self.dict_List(work_experience, self.content)
        self.headings = self.dict_List(isHeading, self.content)
        self.headings = [x[0] for x in self.headings]



if __name__ == "__main__":
    resume = Resume(filename = 'sampleresume.txt')
    print resume.get_name()
    print resume.get_work()

收率:

Someone
2006 - 2008: Programmer

答案 1 :(得分:0)

您应该查看regexp s。

它们允许您解析文本。

一个例子:

#!/usr/local/bin/python2.7
import re

prog = re.compile("\s*(Name|name|nick).*")
result = prog.match("Name: Bob Exampleson")

if result:
    print result.group(0)

result = prog.match("University: MIT")

if result:
    print result.group(0)

根据候选人使用的不同文本,您必须优化搜索。

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