使用pandas.map更改值

您是对的,我认为您执行了同一语句两次(1比1)。在python交互终端上执行的以下语句澄清了这一点。

注: 如果传递字典,map()将序列中的所有值替换为 NaN 如果它与字典的键不匹配(我认为,您也做了同样的操作,即执行语句两次)。检查 pandas map(), apply() .

熊猫文件注释 :当 精氨酸 是字典,值在 系列 不在字典中的(作为键)将转换为 南 .

>>> import pandas as pd
>>>
>>> d = {
...     "label": ["ham", "ham", "spam", "ham", "ham"],
...     "sms_messsage": [
...     "Go until jurong point, crazy.. Available only ...",
...     "Ok lar... Joking wif u oni...",
...     "Free entry in 2 a wkly comp to win FA Cup fina...",
...     "U dun say so early hor... U c already then say...",
...     "Nah I don't think he goes to usf, he lives aro..."
...    ]
... }
>>>
>>> df = pd.DataFrame(d)
>>> df
  label                                       sms_messsage
0   ham  Go until jurong point, crazy.. Available only ...
1   ham                      Ok lar... Joking wif u oni...
2  spam  Free entry in 2 a wkly comp to win FA Cup fina...
3   ham  U dun say so early hor... U c already then say...
4   ham  Nah I don't think he goes to usf, he lives aro...
>>>
>>> df['label'] = df.label.map({'ham':0, 'spam':1})
>>> df
   label                                       sms_messsage
0      0  Go until jurong point, crazy.. Available only ...
1      0                      Ok lar... Joking wif u oni...
2      1  Free entry in 2 a wkly comp to win FA Cup fina...
3      0  U dun say so early hor... U c already then say...
4      0  Nah I don't think he goes to usf, he lives aro...
>>>
>>> df['label'] = df.label.map({'ham':0, 'spam':1})
>>> df
   label                                       sms_messsage
0    NaN  Go until jurong point, crazy.. Available only ...
1    NaN                      Ok lar... Joking wif u oni...
2    NaN  Free entry in 2 a wkly comp to win FA Cup fina...
3    NaN  U dun say so early hor... U c already then say...
4    NaN  Nah I don't think he goes to usf, he lives aro...
>>>

获得相同结果的其他方法

>>> import pandas as pd
>>>
>>> d = {
...     "label": ['spam', 'ham', 'ham', 'ham', 'spam'],
...     "sms_message": ["M1", "M2", "M3", "M4", "M5"]
... }
>>>
>>> df = pd.DataFrame(d)
>>> df
  label sms_message
0  spam          M1
1   ham          M2
2   ham          M3
3   ham          M4
4  spam          M5
>>>

第一路使用 map() 具有 dictionary 参数

>>> new_values = {'spam': 1, 'ham': 0}
>>>
>>> df
  label sms_message
0  spam          M1
1   ham          M2
2   ham          M3
3   ham          M4
4  spam          M5
>>>
>>> df.label = df.label.map(new_values)
>>> df
   label sms_message
0      1          M1
1      0          M2
2      0          M3
3      0          M4
4      1          M5
>>>

第二路使用 MAP() 具有 function 参数

>>> df.label = df.label.map(lambda v: 0 if v == 'ham' else 1)
>>> df
   label sms_message
0      1          M1
1      0          M2
2      0          M3
3      0          M4
4      1          M5
>>>

第三路使用 apply() 具有 功能 参数

>>> df.label = df.label.apply(lambda v: 0 if v == "ham" else 1)
>>>
>>> df
   label sms_message
0      1          M1
1      0          M2
2      0          M3
3      0          M4
4      1          M5
>>>

谢谢您。

也许你的问题是读表功能。

试着去做:

df = pd.read_table('smsspamcollection/SMSSpamCollection',
                   sep='\t', 
                   header=None,
                   names=['label', 'sms_message'])