pandas 大熊猫替换（擦除）字符串中的不同字符
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pandas replace (erase) different characters from strings
提问by As3adTintin
I have a list of high schools. I would like to erase certain characters, words, and symbols from the strings.
我有一份高中名单。我想从字符串中删除某些字符、单词和符号。
I currently have:
我目前有：
df['schoolname'] = df['schoolname'].str.replace('high', "")
However, I would like to use a list so I can quickly replace high, school, /, etc.
不过，我想用一个列表，以便我能快捷地更换high，school，/，等。
Any suggestions?
有什么建议？
df['schoolname'] = df['schoolname'].str.replace(['high', 'school'], "")
does not work
不起作用
回答by Andy Hayden
Use regex (seperate the strings by |):
使用正则表达式（用 分隔字符串|）：
df['schoolname'] = df['schoolname'].str.replace('high|school', "")
回答by Amir Imani
You can create a dictionary and then .replace({}, regex=True)method:
您可以创建一个字典，然后.replace({}, regex=True)方法：
replacements = {
'schoolname': {
r'(high|school)': ''}
}
df.replace(replacements, regex=True, inplace=True)
回答by mrHymanlu
My problem: I wanted to find a simple solution in deleting characters / symbols using the replace method with pandas.
我的问题：我想找到一个简单的解决方案，使用Pandas的替换方法删除字符/符号。
I had the following array in a data frame:
我在数据框中有以下数组：
df = array(['2012', '2016', '2011', '2013', '2015', '2017', '2001', '2007',
'[2005], ?2004.', '2005', '2009', '2008', '2009, c2008.', '2006',
'2019', '[2003]', '2018', '2012, c2011.', '[2012]', 'c2012.',
'2014', '2002', 'c2005.', '[2000]', 'c2000.', '2010',
'2008, c2007.', '2011, c2010.', '2011, ?2002.', 'c2011.', '[2017]',
'c1996.', '[2018]', '[2019]', '[2011]', '2000', '2000, c1995.',
'[2004]', '2005, ?2004.', 'c2004.', '[2009]', 'c2009.', '[2014]',
'1999', '[2010]', 'c2010.', '[2006]', '2007, 2006.', '[2013]',
'c2001.', 'C2016.', '2008, c2006.', '2011, ?2010.', '2007, c2005.',
'2009, c2005.', 'c2002.', '[2004], c2003.', '2009, c2007.', '2003',
'?2003.', '[2016]', '[2001]', '2010, c2001.', '[1998]', 'c1998.'],
dtype=object)
As you can see, the years were entered using multiple formats (ugh!) with brackets and copyright symbols and lowercase c and uppercase C.
如您所见，年份是使用多种格式（呃！）输入的，包括括号和版权符号以及小写 c 和大写 C。
Now I wanted to remove those unwanted characters and only have the years in four digits. Since it's an array, you also need to transform it into a string before using replace(). Create a variable of all the characters you want replaced and separate them with ' | '.
现在我想删除那些不需要的字符并且只有四位数的年份。由于它是一个数组，因此您还需要在使用replace() 之前将其转换为字符串。创建一个包含所有要替换的字符的变量，并用 ' | 分隔它们。'。
rep_chars = 'c|C|\]|\[|?|\.'
df[Year] = df['Year'].str.replace(rep_chars,"")
Make sure to use
\.and not just the period. The same with\]and\[.
确保使用
\.而不仅仅是期间。与\]和相同\[。
Output:
输出：
array(['2012', '2016', '2011', '2013', '2015', '2017', '2001', '2007',
'2005, 2004', '2005', '2009', '2008', '2009, 2008', '2006', '2019',
'2003', '2018', '2012, 2011', '2014', '2002', '2000', '2010',
'2008, 2007', '2011, 2010', '2011, 2002', '1996', '2000, 1995',
'2004', '1999', '2007, 2006', '2008, 2006', '2007, 2005',
'2009, 2005', '2004, 2003', '2009, 2007', '2010, 2001', '1998'],
dtype=object)
Happy Data Cleaning!
快乐的数据清理！
