我有一张看起来像:
Groupe Id MotherName FatherName Field
Advanced 56 Laure James English-107,Economics, Management, History, Philosophy
Middle 11 Ann Nicolas Web-development, Java-2
Advanced 6 Helen Franc Literature, English-2
Beginner 43 Laure James Mathematics, History, Philosophy, Literature
Middle 14 Naomi Franc Java-2, Management, English-107为了进一步处理数据,我需要拆分Field列,并将其替换为多个列,如下所示:
Id English-107 Economics Management History Web-development Java-2 Literature English-2 Mathematics Philosophy
56 1 1 1 1 0 0 0 0 0 1
11 0 0 0 0 1 1 0 0 0 0因此,这些列可以附加到初始的dataframe中。我不知道怎么做,因为基本的分裂就像
pd.DataFrame(df.Field.str.split(',',1).tolist())不能解决我的问题,因为我需要列不仅基于列表中的位置,而且基于列表中的每个唯一值。你知道我怎样才能接近它吗?
发布于 2016-02-29 23:15:17
您可以使用concat和str.get_dummies
print pd.concat([df['Id'], df['Field'].str.get_dummies(sep=",")], axis=1)
Id Economics English-107 English-2 History Java-2 Literature \
0 56 1 1 0 1 0 0
1 11 0 0 0 0 1 0
2 6 0 0 1 0 0 1
3 43 0 0 0 1 0 1
4 14 0 1 0 0 1 0
Management Mathematics Philosophy Web-development
0 1 0 1 0
1 0 0 0 1
2 0 0 0 0
3 0 1 1 0
4 1 0 0 0 如果需要计数值,可以使用pivot_table (我添加了一个字符串Economics进行测试):
df1 = df['Field'].str.split(',',expand=True).stack()
.groupby(level=0)
.value_counts()
.reset_index()
df1.columns=['a','b','c']
print df1.pivot_table(index='a',columns='b',values='c').fillna(0)
b Economics English-107 English-2 History Java-2 Literature Management \
a
0 2 1 0 1 0 0 1
1 0 0 0 0 1 0 0
2 0 0 1 0 0 1 0
3 0 0 0 1 0 1 0
4 0 1 0 0 1 0 1
b Mathematics Philosophy Web-development
a
0 0 1 0
1 0 0 1
2 0 0 0
3 1 1 0
4 0 0 0 https://stackoverflow.com/questions/35711580
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