概述

假设我有一个看起来像这样的文本文件：

Item,Date,Time,Location
1,01/01/2016,13:41,[45.2344:-78.25453]
2,01/03/2016,19:11,[43.3423:-79.23423,41.2342:-81242]
3,01/10/2016,01:27,[51.2344:-86.24432]

我希望能够做的是用pandas.read_csv读取,但第二行将抛出错误.这是我目前使用的代码：

import pandas as pd
df = pd.read_csv("path/to/file.txt",sep=",",dtype=str)

我试图将quotechar设置为“[”,但是这显然只是占用了行,直到下一个打开括号并添加一个右括号会导致“找到长度为2的字符串”错误.任何见解将不胜感激.谢谢！

更新

提供了三种主要解决方案：1)为数据框提供大量名称,以允许读入所有数据,然后对数据进行后处理,2)在方括号中查找值并在其周围加上引号,或者3)用分号替换前n个逗号.

总的来说,我认为选项3通常不是一个可行的解决方案(虽然对我的数据来说很好),因为a)如果我在一个包含逗号的列中引用了值,b)如果我的方括号列是不是最后一栏？这留下了解决方案1和2.我认为解决方案2更具可读性,但解决方案1更有效,仅运行1.38秒,而解决方案2则运行3.02秒.测试在包含18列和超过208,000行的文本文件上运行.

import pandas as pd
import io

with open('file2.csv','r') as f:
    lines = f.readlines()
    fo = io.StringIO()
    fo.writelines(u"" + line.replace(',',';',3) for line in lines)
    fo.seek(0)    

df = pd.read_csv(fo,sep=';')
print df
   Item        Date   Time                            Location
0     1  01/01/2016  13:41                 [45.2344:-78.25453]
1     2  01/03/2016  19:11  [43.3423:-79.23423,41.2342:-81242]
2     3  01/10/2016  01:27                 [51.2344:-86.24432]

或者可以尝试这种复杂的方法,因为主要问题是,分隔符,列表中的值与其他列值的分隔符相同.

所以你需要后期处理：

import pandas as pd
import io

temp=u"""Item,41.2342:-81242,[51.2344:-86.24432]"""
#after testing replace io.StringIO(temp) to filename
#estimated max number of columns
df = pd.read_csv(io.StringIO(temp),names=range(10))
print df
      0           1      2                    3               4  \
0  Item        Date   Time             Location             NaN   
1     1  01/01/2016  13:41  [45.2344:-78.25453]             NaN   
2     2  01/03/2016  19:11   [43.3423:-79.23423  41.2342:-81242   
3     3  01/10/2016  01:27  [51.2344:-86.24432]             NaN   

                 5   6   7   8   9  
0              NaN NaN NaN NaN NaN  
1              NaN NaN NaN NaN NaN  
2  41.2342:-81242] NaN NaN NaN NaN  
3              NaN NaN NaN NaN NaN  

#remove column with all NaN
df = df.dropna(how='all',axis=1)
#first row get as columns names
df.columns = df.iloc[0,:]
#remove first row
df = df[1:]
#remove columns name
df.columns.name = None

#get position of column Location
print df.columns.get_loc('Location')
3
#df1 with Location values
df1 = df.iloc[:,df.columns.get_loc('Location'): ]
print df1
              Location             NaN              NaN
1  [45.2344:-78.25453]             NaN              NaN
2   [43.3423:-79.23423  41.2342:-81242  41.2342:-81242]
3  [51.2344:-86.24432]             NaN              NaN

#combine values to one column
df['Location'] = df1.apply( lambda x : ','.join([e for e in x if isinstance(e,basestring)]),axis=1)

#subset of desired columns
print df[['Item','Date','Time','Location']]
  Item        Date   Time                                           Location
1    1  01/01/2016  13:41                                [45.2344:-78.25453]
2    2  01/03/2016  19:11  [43.3423:-79.23423,41.2342:-8...
3    3  01/10/2016  01:27                                [51.2344:-86.24432]

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