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python - 基于两列标记数据

问题描述

我有一个包含几列的数据框，如下所示：

id | x1 | text | x2 | num | x3 | class
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | 
2nn| TT | word | QQ |  1  | TT | # They don't all have the same vals
2nn| TT | word | QQ |  1  | TT | # This is just to illustrate it

我也有以下lists字符串：

class1 = ["",...]
class2 = ["",...]
class3 = ["",...]
class4 = ["",...]
class5 = ["",...] # Multiple strings, I just used '...' for simplicity

我正在尝试在列中分配类，class以便如果事务（行）text列中的单词包含在任何 中找到的任何单词lists，则将列表名称分配为类。

我这样做是为了标记一些我最终将用于分类的数据。

我只想对从第 10,000 行开始的数据执行此标记。我正在使用的是：

# last 6000 rows
for index, row in df.tail(6000).iterrows():
    if df[df['text'].str.contains(class1)==True]:
        df.loc[row, 'class'] = "class1"

    if df[df['text'].str.contains(class2)==True]:
        df.loc[row, 'class'] = "class2"

    if df[df['text'].str.contains(class3)==True]:
        df.loc[row, 'class'] = "class3"

    if df[df['text'].str.contains(class4)==True]:
        df.loc[row, 'class'] = "class4"

    if df[df['text'].str.contains(class5)==True]:
        df.loc[row, 'class'] = "class5"

我收到打印以下错误的响应：

TypeError: unhashable type: 'list'

以下是我在尝试 Chris A 的回复时的代码：

# Word bins for the various labels
complaint = ["sucks", "worst", "doesn't", "didn't", "won't", "bad", "horrible", "unusable", "cannot", "can't", "not", "did not", "waste", "hate", "hated", "awful", "useless", "sucked", "freezing", "freezes", "froze", "does not", "crap", "stupid"]

compliment = ["awesome", "great", "amazing", "cool", "good", "nice", "nicest", "successful", "thanks", ":)", "successfully"]

neutral = ["Eh", "meh", "works"]

bug = ["please", "fix", "won't", "cannot", "can't", "not", "freezing", "freezes", "froze", "does not", "did not", "help", "plz"]

feature = ["it would be", "id like", "i'd like", "could", "can you", "implement", "feature", "lacks", "wish"]
def label_data(df):

    d = {'Compliment': compliment,
         'Complaint': complaint,
         'Neutral': neutral,
         'Bug': bug,
         'Feature': feature}

    for name, values in d.items():
        df.loc[df['review'].isin(values), 'label'] = name

我的主类从文本文件中调用数据，然后使用以下方法调用此方法：

df_orig = pd.read_table("PRIVATEPATH/data.txt", delimiter=",")
label_data(df_labelled)

标签： pythonpandasdataframeanaconda

解决方案

为此使用列表中的 python 可能会有所帮助dictionary。

使用时，str.contains您还必须通过使用|正则表达式“OR”运算符连接每个值来“构建”正则表达式字符串。

笔记

正如您所发现的那样，这里的一个问题是，以这种方式构建正则表达式模式将需要您转义列表中的任何特殊正则表达式字符。示例 - 您:)的赞美列表中有“ ”。这将需要成为 ' \:\)'

d = {'class1': class1,
     'class2': class2,
     'class3': class3,
     'class4': class4}

for name, values in d.items():
    # Create a regex string joining all the values in the list with the regex OR '|'
    pat = '|'.join(values)
    df.loc[df['text'].str.contains(pat), 'class'] = name

简化示例

df = pd.DataFrame({'id': {0: '2nn',1: '2nn',2: '2nn',3: '2nn',4: '2nn',5: '2nn',6: '2nn',7: '2nn',8: '2nn',9: '2nn',10: '2nn',11: '2nn',12: '2nn'},
                    'x1': {0: 'TT',1: 'TT',2: 'TT',3: 'TT',4: 'TT',5: 'TT',6: 'TT',7: 'TT',8: 'TT',9: 'TT',10: 'TT',11: 'TT',12: 'TT'},
                    'text': {0: 'abc',1: 'abc',2: 'e',3: 'h',4: 'm',5: 'p',6: 'q',7: 'd',8: 's',9: 'j',10: 'h',11: 'o',12: 'z'},
                    'x2': {0: 'QQ',1: 'QQ',2: 'QQ',3: 'QQ',4: 'QQ',5: 'QQ',6: 'QQ',7: 'QQ',8: 'QQ',9: 'QQ',10: 'QQ',11: 'QQ',12: 'QQ'},
                    'num': {0: 1,1: 1,2: 1,3: 1,4: 1,5: 1,6: 1,7: 1,8: 1,9: 1,10: 1,11: 1,12: 1},
                    'x3': {0: 'TT',1: 'TT',2: 'TT',3: 'TT',4: 'TT',5: 'TT',6: 'TT',7: 'TT',8: 'TT',9: 'TT',10: 'TT',11: 'TT',12: 'TT'},
                    'class': {0: np.nan,1: np.nan,2: np.nan,3: np.nan,4: np.nan,5: np.nan,6: np.nan,7: np.nan,8: np.nan,9: np.nan,10: np.nan,11: np.nan,12: np.nan}})

class1 = list('abcde')
class2 = list('fghi')
class3 = list('jklmn')
class4 = list('opqrs')

d = {'class1': class1,
     'class2': class2,
     'class3': class3,
     'class4': class4}

for name, values in d.items():
    pat = '|'.join(values)
    df.loc[df['text'].str.contains(pat), 'class'] = name

print(df)

[出去]

     id  x1 text  x2  num  x3   class
0   2nn  TT    a  QQ    1  TT  class1
1   2nn  TT    b  QQ    1  TT  class1
2   2nn  TT    e  QQ    1  TT  class1
3   2nn  TT    h  QQ    1  TT  class2
4   2nn  TT    m  QQ    1  TT  class3
5   2nn  TT    p  QQ    1  TT  class4
6   2nn  TT    q  QQ    1  TT  class4
7   2nn  TT    d  QQ    1  TT  class1
8   2nn  TT    s  QQ    1  TT  class4
9   2nn  TT    j  QQ    1  TT  class3
10  2nn  TT    h  QQ    1  TT  class2
11  2nn  TT    o  QQ    1  TT  class4
12  2nn  TT    z  QQ    1  TT     NaN

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