0

我正在编写一个应该删除重复条目的脚本。数据中有些人输入了两次他们的名字，因为他们有两个电话号码，并且因为电话号码字段不是数组，要输入多个，他们输入了多个条目。

我的脚本将条目更改为具有与列名对应的键的字典，然后遍历每一行。有一个遍历每一行的主 for 循环，然后是一个嵌套的 for 循环，遍历每个元素的所有元素，比较它们以检测重复项。当我点击重复时，我的代码应该比较电话、电子邮件和网站，然后如果它们是唯一的/不匹配，则将它们附加到一个区域。

这是我的代码：

import csv

# This function takes a tab-delim csv and merges the ones with the same name but different phone / email / websites.
def merge_duplicates(sheet):

    myjson = [] # myjson = list of dictionaries where each dictionary

    with(open("ieca_first_col_fake_text.txt", "rU")) as f:

        sheet = csv.DictReader(f,delimiter="\t")
        for row in sheet:
            myjson.append(row)

        write_file = csv.DictWriter(open('duplicates_deleted.csv','w'), ['name','phone','email','website'], restval='', delimiter = '\t')

        for row in myjson:

            # convert phone, email, and web to lists so that extra can be appended
            row['phone'] = row['phone'].split()
            row['email'] = row['email'].split()
            row['website'] = row['website'].split()
            print row

        for i in len(myjson):

            # if the names match, check to see if phone, em, web match. If any match, append to first row.
            try:
                if myjson[i]['name'] == myjson[i+1]['name']:
                    if myjson[i]['phone'] != myjson[i+1]['phone']:
                        myjson[i]['phone'].append(myjson[i+1]['phone'])
#                     if row['email'] != myjson[rowvalue+1]['email']:
#                         row['email'].append(myjson[rowvalue+1]['email'])
#                     if row['website'] != myjson[rowvalue+1]['website']:
#                         row['website'].append(myjson[rowvalue+1]['website'])
            except IndexError:
                print("We're at the end now") 

            write_file.writerow(row)

merge_duplicates('ieca_first_col_fake_text.txt')

所以我的代码中的一切都很花哨，然后它遇到了第一个重复项，我收到了这个错误：

{'website': [], 'phone': [], 'name': 'Diane Grant Albrecht M.S.', 'email': []}
{'website': ['www.got.com'], 'phone': ['111-222-3333'], 'name': 'Lannister G. Cersei M.A.T., CEP', 'email': ['cersei@got.com']}
{'website': [], 'phone': [], 'name': 'Argle D. Bargle Ed.M.', 'email': []}
{'website': ['www.daManWithThePlan.com'], 'phone': ['000-000-1111'], 'name': 'Sam D. Man Ed.M.', 'email': ['dman123@gmail.com']}
Traceback (most recent call last):
  File "/Users/samuelfinegold/Documents/noodle/delete_duplicates.py", line 40, in <module>
    merge_duplicates('ieca_first_col_fake_text.txt')
  File "/Users/samuelfinegold/Documents/noodle/delete_duplicates.py", line 20, in merge_duplicates
    row['email'] = row['email'].split()
AttributeError: 'NoneType' object has no attribute 'split'
logout

非常感谢你的帮忙！

前数据，如果它有帮助：

name    phone   email   website
Diane Grant Albrecht M.S.           
"Lannister G. Cersei M.A.T., CEP"   111-222-3333    cersei@got.com  www.got.com
Argle D. Bargle Ed.M.           
Sam D. Man Ed.M.    000-000-1111    dman123@gmail.com   www.daManWithThePlan.com
Sam D. Man Ed.M.    
Sam D. Man Ed.M.    111-222-333     dman123@gmail.com   www.daManWithThePlan.com
D G Bamf M.S.           
Amy Tramy Lamy Ph.D.
4

2 回答 2

3

错误是，如果row['phone']是None，则不能拆分它。

你可以这样做

row['phone'] = row['phone'].split() if row['phone'] else []
row['email'] = row['email'].split() if row['email'] else []
row['website'] = row['website'].split() if row['website'] else []

[]可以替换为您要指定的任何默认值（例如：None或""）。

一种更清洁的方法是

row['phone'] = row['phone'].split() if row.get('phone') else []
row['email'] = row['email'].split() if row.get('email') else []
row['website'] = row['website'].split() if row.get('website') else []
于 2013-07-03T19:54:26.527 回答
1

Personally, I would use and to do this:

row['email'] = row.get('email',[]) and row['email'].split()

The logic is the same as:

if row.get('email'):
    row['email'] = row['email'].split()

Although, strictly speaking, this does reassign if the key is missing (or email has already been made into a list), so you might want to do this:

# you could also use hasattr(row['email'],'split')
if 'email' in row and isinstance(row['email'],str):
    row['email'] = row['email'].split()
于 2013-07-03T19:55:06.060 回答