Question

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

我还有以下带有字符串的lists：

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

我正在尝试分配列class中的类，以便如果事务（行）的text列中的单词包含在任何lists中找到的单词，请分配列表名称作为类。

我这样做是为了标记一些最终将用于分类的数据。

我只想对从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"

我得到的响应显示以下错误：

TypeError: unhashable type: 'list'

下面是我尝试Chris A响应时的代码：

# 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)

Answer 1

为此使用python dictionary列表可能会有所帮助。

使用str.contains时，还必须通过将每个值与正则表达式“或”运算符|结合起来来“构建”正则表达式字符串。

注释

这里您发现的一个陷阱是，以这种方式构建正则表达式模式将要求您转义列表中的所有特殊正则表达式字符。示例-您的称赞列表中有“ :)”。这需要变成'\:\)'

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]

     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

基于两列标记数据

1 个答案:

简化示例