pandas AttributeError: 'str' 对象在 Seaborn 和 Scatterplot 中没有属性 'view'
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AttributeError: 'str' object has no attribute 'view' in Seaborn , Scatterplot
提问by user8270077
I am getting a bizarre exception in Seaborn.
我在 Seaborn 遇到了一个奇怪的例外。
For a reproducible example:
对于可重现的示例：
toy_data.to_json()
'{"X":{"0":0.12765045,"1":0.0244816152,"2":0.1263715245,"3":0.0246376768,"4":0.1108581319,"5":0.1406719382,"6":0.1358105564,"7":0.1245863432,"8":0.1175445352,"9":0.1188479018,"10":0.1113148159,"11":0.117455495,"12":0.110555662,"13":0.1328567106,"14":0.103064284,"15":0.1119474442,"16":0.119390455,"17":0.1246727756,"18":0.1117827155,"19":0.1169972547},"Y":{"0":0.1241083714,"1":0.1394242378,"2":0.1225010796,"3":0.0077080173,"4":0.1198371354,"5":0.0029026989,"6":0.1259473297,"7":0.0,"8":0.0,"9":0.1214620231,"10":0.1204110739,"11":0.0,"12":0.1194570059,"13":0.0014971676,"14":0.1184584731,"15":0.1212061305,"16":0.1221438778,"17":0.0,"18":0.1209991075,"19":0.0},"Label":{"0":"17","1":"3","2":"17","3":"0","4":"14","5":"21","6":"16","7":"23","8":"20","9":"15","10":"14","11":"20","12":"14","13":"22","14":"13","15":"14","16":"15","17":"23","18":"14","19":"20"},"Probability":{"0":1.0,"1":1.0,"2":1.0,"3":1.0,"4":1.0,"5":1.0,"6":1.0,"7":1.0,"8":1.0,"9":1.0,"10":1.0,"11":1.0,"12":1.0,"13":1.0,"14":1.0,"15":1.0,"16":1.0,"17":1.0,"18":0.9101796407,"19":1.0}}'
toy_data.head()
X Y Label Probability
0 0.127650 0.124108 17 1.0
1 0.024482 0.139424 3 1.0
2 0.126372 0.122501 17 1.0
3 0.024638 0.007708 0 1.0
4 0.110858 0.119837 14 1.0
sns.scatterplot(x = toy_data.X, y = toy_data.Y, hue = toy_data.Label.values, alpha = 0.5)
AttributeError: 'str' object has no attribute 'view'
Similar exception with this syntax:
此语法的类似异常：
sns.scatterplot(x = 'X', y = 'Y', data = toy_data, hue = 'Label', alpha = 0.5)
回答by Jorge
As presented, your column Label is an object. It need to be a number (int or float).
如图所示，您的列 Label 是一个对象。它必须是一个数字（整数或浮点数）。
toy_data.info()
<class 'pandas.core.frame.DataFrame'>
Index: 20 entries, 0 to 9
Data columns (total 4 columns):
X 20 non-null float64
Y 20 non-null float64
Label 20 non-null object
Probability 20 non-null float64
dtypes: float64(3), object(1)
memory usage: 800.0+ bytes
Do this:
做这个：
toy_data.loc[:,'Label'] = toy_data.Label.astype(np.float)
to change it to float.
将其更改为浮动。
Then your command:
然后你的命令：
sns.scatterplot(x = toy_data.X, y = toy_data.Y, hue = toy_data.label.values, alpha = 0.5)
should work.
应该管用。
I am using this command to generate the data frame
我正在使用此命令生成数据框
dict = {"X":{"0":0.12765045,"1":0.0244816152,"2":0.1263715245,"3":0.0246376768,"4":0.1108581319,"5":0.1406719382,"6":0.1358105564,"7":0.1245863432,"8":0.1175445352,"9":0.1188479018,"10":0.1113148159,"11":0.117455495,"12":0.110555662,"13":0.1328567106,"14":0.103064284,"15":0.1119474442,"16":0.119390455,"17":0.1246727756,"18":0.1117827155,"19":0.1169972547},"Y":{"0":0.1241083714,"1":0.1394242378,"2":0.1225010796,"3":0.0077080173,"4":0.1198371354,"5":0.0029026989,"6":0.1259473297,"7":0.0,"8":0.0,"9":0.1214620231,"10":0.1204110739,"11":0.0,"12":0.1194570059,"13":0.0014971676,"14":0.1184584731,"15":0.1212061305,"16":0.1221438778,"17":0.0,"18":0.1209991075,"19":0.0},"Label":{"0":"17","1":"3","2":"17","3":"0","4":"14","5":"21","6":"16","7":"23","8":"20","9":"15","10":"14","11":"20","12":"14","13":"22","14":"13","15":"14","16":"15","17":"23","18":"14","19":"20"},"Probability":{"0":1.0,"1":1.0,"2":1.0,"3":1.0,"4":1.0,"5":1.0,"6":1.0,"7":1.0,"8":1.0,"9":1.0,"10":1.0,"11":1.0,"12":1.0,"13":1.0,"14":1.0,"15":1.0,"16":1.0,"17":1.0,"18":0.9101796407,"19":1.0}}
toy_data = pd.DataFrame(dict)
You will need numpy import numpy as npto convert the values to float
您将需要 numpyimport numpy as np将值转换为浮点数
回答by Prageeth Jayathissa
This is a peculiar issue with seaborn where the color paletteneeds to equal the number of labels
这是 seaborn 的一个特殊问题，其中颜色palette需要等于标签的数量
sns.scatterplot(x = toy_data.X, y = toy_data.Y, hue = toy_data.Label, alpha = 0.5,
palette=sns.color_palette("Set1", toy_data.Label.nunique()) )
If not, seaborn will apply numerical functions to map the number of huecategories to the default paletteof fourcolours. This is why you get an error when the values are a string
如果没有，seaborn 将应用数值函数将hue类别数量映射到默认palette的四种颜色。这就是为什么当值是字符串时会出现错误的原因
