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我正在使用fuzzywuzzy 在公司名称的csv 中查找近似匹配项。我正在将手动匹配的字符串与不匹配的字符串进行比较，希望能找到一些有用的邻近匹配，但是，我在fuzzywuzzy 中遇到了字符串或缓冲区错误。我的代码是：

from fuzzywuzzy import process
from pandas import read_csv

if __name__ == '__main__':
    df = read_csv("usm_clean.csv", encoding = "ISO-8859-1")
    df_false = df[df['match_manual'].isnull()]  
    df_true = df[df['match_manual'].notnull()]
    sss_false = df_false['sss'].values.tolist()
    sss_true = df_true['sss'].values.tolist()


    for sssf in sss_false:
        mmm = process.extractOne(sssf, sss_true) # find best choice
        print sssf + str(tuple(mmm))

这会产生以下错误：

Traceback (most recent call last):
File "fuzzywuzzy_usm2_csv_test.py", line 21, in <module>
mmm = process.extractOne(sssf, sss_true) # find best choice
File "/usr/local/lib/python2.7/site-packages/fuzzywuzzy/process.py", line 123, in extractOne
best_list = extract(query, choices, processor, scorer, limit=1)
File "/usr/local/lib/python2.7/site-packages/fuzzywuzzy/process.py", line 84, in extract
processed = processor(choice)
File "/usr/local/lib/python2.7/site-packages/fuzzywuzzy/utils.py", line 63, in full_process
string_out = StringProcessor.replace_non_letters_non_numbers_with_whitespace(s)
File "/usr/local/lib/python2.7/site-packages/fuzzywuzzy/string_processing.py", line 25, in replace_non_letters_non_numbers_with_whitespace
return cls.regex.sub(u" ", a_string)
TypeError: expected string or buffer

这与导入指定编码的熊猫的效果有关，我添加了它以防止UnicodeDecodeErrors但具有导致此错误的连锁反应。我试图强制使用对象，str(sssf)但这不起作用。

因此，我在这里隔离了导致错误的一行：（#N/A,,,,,,下面粘贴的代码中的第 29 行）。我认为#是导致错误的原因，但奇怪的是，它不是A导致问题的字符，因为文件在被删除时有效。对我来说奇怪的是，下面两行的字符串N/A可以很好地解析，但是，当我删除符号时，第 29 行不会解析#，即使该字段看起来与下面的字段相同。

sss,sid,match_manual,notes,match_date,source,match_by
N20 KIDS,1095543_cha,,,2014-10-12,,20140429_fuzzy_match.ktr (stream 3)
N21 FESTIVAL,08190588_com,,,2014-10-12,,20140429_fuzzy_match.ktr (stream 3)
N21 LTD,,,,,,
N21 LTD.,04615294_com,true,,2014-12-02,,OpenCorps
N2 CHECK,08105000_com,,,2014-10-12,,20140429_fuzzy_match.ktr (stream 3)
N2 CHECK LIMITED,06139690_com,true,,2014-12-02,,OpenCorps
N2CHECK LIMITED,08184223_com,,,2014-05-05,,20140429_fuzzy_match.ktr (stream 3)
N 2 CHECK LTD,05729595_com,,,2014-05-05,,20140429_fuzzy_match.ktr (stream 2)
N2 CHECK LTD,06139690_com,true,,2014-12-02,,OpenCorps
N2CHECK LTD,05729595_com,,,2014-05-05,,20140429_fuzzy_match.ktr (stream 2)
N2E & BACK LTD,05218805_com,,,2014-05-05,,20140429_fuzzy_match.ktr (stream 2)
N2 GROUP LLC,04627044_com,,,2014-10-12,,20140429_fuzzy_match.ktr (stream 3)
N2 GROUP LTD,04475764_com,true,,2014-05-05,data taken from u_supplier_match,20140429_fuzzy_match.ktr (stream 2)
N2R PRODUCTIONS,SC266951_com,,,2014-10-12,,20140429_fuzzy_match.ktr (stream 3)
N2 VISUAL COMMUNICATIONS LIMITED,,,,,,
N2 VISUAL COMMUNICATIONS LTD,03144224_com,true,,2014-12-02,data taken from u_supplier_match,OpenCorps
N2WEB,07636689_com,,,2014-10-12,,20140429_fuzzy_match.ktr (stream 3)
N3 DISPLAY GRAPHICS LTD,04008480_com,true,,2014-12-02,data taken from u_supplier_match,OpenCorps
N3O LIMITED,06561158_com,true,,2014-12-02,,OpenCorps
N3O LTD,,,,,,
N400138,,,,,,
N400360,,,,,,
N4K LTD,07054740_com,,,2014-05-05,,20140429_fuzzy_match.ktr (stream 2)
N51 LTD,,,,,,
N68 LTD,,,,,,
N8 LTD,,,,,,
N9 DESIGN,07342091_com,true,,2015-02-07,openrefine/opencorporates,IM
#N/A,,,,,,
N A,,,,,,
N/A,red_general_xtr,true,Matches done manually,2015-04-16,manual matching,IM
(N) A & A BUILDERS LTD,,,,,,
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2 回答 2

2

默认情况下，pandas.read_csv将字符串解析为非'N/A'数字 ( NaN)

在您的情况下，这意味着您最终得到一个nan值而不是字符串。在您的样本数据集中，这发生在两个地方

底部的第三行（您在问题中突出显示的行）导致sss_false[-3] == nan

最后一行导致sss_true[-1] == nan.

选项1

如果要将字符串解析'N/A'为字符串而不是nan，则执行此操作的方法是替换

df = read_csv("usm_clean.csv", encoding = "ISO-8859-1")

和

df = read_csv("usm_clean.csv", encoding = "ISO-8859-1", keep_default_na=False, na_values='')

pandas 文档中描述了这些额外选项的含义。

na_values :类列表或字典，默认无

要识别为 NA/NaN 的附加字符串。如果 dict 通过，特定的每列 NA 值

keep_default_na :布尔值，默认 True

如果指定了 na_values 并且 keep_default_na 为 False，则默认 NaN 值将被覆盖，否则将附加到

所以，上面的修改告诉 pandas 将空字符串识别为 NA 并丢弃默认值'N/A'

选项 2

如果要丢弃'N/A'第一列中的行，则需要从和中删除nan成员。一种方法是：sss_truesss_false

sss_true = [x for x in sss_true if type(x) != str]
sss_false = [x for x in sss_false if type(x) != str]
于 2015-06-15T10:40:00.273 回答
0

您的sss_true变量包含：

[
    u'N21 LTD.',
    u'N2 CHECK LIMITED',
    u'N2 CHECK LTD',
    u'N2 GROUP LTD',
    u'N2 VISUAL COMMUNICATIONS LTD',
    u'N3 DISPLAY GRAPHICS LTD',
    u'N3O LIMITED',
    u'N9 DESIGN',
    nan              # <---- note this
]

一旦你摆脱了那个非数字值，一切都开始按预期工作。

于 2015-06-15T06:29:45.267 回答