pandas Python：如何从熊猫系列的字典中获取值
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Python: how to get values from a dictionary from pandas series
提问by Kexin Xu
I am very new to python and trying to get value from dictionary where keys are defined in a dataframe column (pandas). I searched quite a bit and the closest thing is a question in the link below, but it doesnt come with an answer.
我对 python 非常陌生，并试图从字典中获取值，其中键在数据框列（pandas）中定义。我搜索了很多，最接近的是下面链接中的一个问题，但它没有给出答案。
So, here I am trying to find answer for the same type of question.
所以，在这里我试图找到相同类型问题的答案。
Select from dictionary using pandas series
I have a dictionary
我有字典
type_dict = {3: 'foo', 4:'bar',5:'foobar', 6:'foobarbar'}
and a data frame with the following column:
以及包含以下列的数据框：
>>> df.type
0 3
1 4
2 5
3 6
4 3
5 4
6 5
7 6
8 3
I want to create a new column containing the corresponding type_dict value, but the following was the only thing I could come up and was not working:
我想创建一个包含相应 type_dict 值的新列，但以下是我唯一能想到的并且不起作用：
type_dict[df.type]
TypeError: 'Series' objects are mutable, thus they cannot be hashed
类型错误：“系列”对象是可变的，因此它们不能被散列
type_dict[df.type.values]
TypeError: unhashable type: 'numpy.ndarray'
类型错误：不可散列的类型：'numpy.ndarray'
Updated question:
更新的问题：
for pandas DataFrame, say 'df', how can i plot speed over meters with type as the key of marker dictionary.
对于pandas DataFrame，比如说'df'，我如何用类型作为标记字典的键来绘制米上的速度。
mkr_dict = {'gps': 'x', 'phone': '+', 'car': 'o'}
x = {'speed': [10, 15, 20, 18, 19], 'meters' : [122, 150, 190, 230, 300], 'type': ['phone', 'phone', 'gps', 'gps', 'car']}
df = pd.DataFrame(x)
meters speed type
0 122 10 phone
1 150 15 phone
2 190 20 gps
3 230 18 gps
4 300 19 car
plt.scatter(df.meters, df.Speed, marker = df.type.map(mkr_dict))
the scatter plot doesn't work for me...
散点图对我不起作用...
回答by EdChum
Pass the dict as an arg to map:
将 dict 作为 arg 传递给map：
In [79]:
df['type'].map(type_dict)
Out[79]:
0 foo
1 bar
2 foobar
3 foobarbar
4 foo
5 bar
6 foobar
7 foobarbar
8 foo
Name: type, dtype: object
This will lookup the key value in the dict and return the associated value from the dict.
这将在字典中查找键值并从字典中返回关联的值。
回答by Julien Spronck
In pandas, this should work
在Pandas中，这应该有效
df['val'] = df.apply(lambda x: type_dict[x['type']], axis=1)
