如何解析熊猫数据框对象

Question

我在 pandas Dataframe 中读取 csv 文件，然后获取它的 dummy 并将它们连接起来，但例如我有名为“Genre”的列，它包含“comedy, Drama”和“action, comedy”所以当我得到 dummy 和 concat它们为每个句子创建一个对象，但我想解析它们。例如，我想创建对象 'Genre.comedy'、'Genre.Drama'、 'Genre.action' 而不是 'Genre.comedy,drama'和'Genre.action,comedy' 这是我的代码：

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import csv
from sklearn import preprocessing
trainset = pd.read_csv("/Users/yada/Downloads/IMDBMovieData.csv", encoding='latin-1')
X = trainset.drop(['Description', 'Runtime'], axis=1)
features = ['Genre','Actors']
for f in features:
    X_dummy = pd.get_dummies(X[f], prefix = f)
    X = X.drop([f], axis = 1)
    X = pd.concat((X, X_dummy), axis = 1)

这是我的 csv 文件的某一行： csv

Answer 1

我认为需要str.get_dummies 和add_prefix：

features = ['Genre','Actors']
for f in features:
    X_dummy = X[f].str.get_dummies(', ').add_prefix(f + '.')
    X = X.drop([f], axis = 1)
    X = pd.concat((X, X_dummy), axis = 1)

或者：

trainset = pd.DataFrame({'Description':list('abc'),
                   'Genre':['comedy, drama','action, comedy','action'],
                   'Actors':['a, b','a, c','d, a'],
                   'Runtime':[1,3,5],
                   'E':[5,3,6],
                   'F':list('aaa')})

print (trainset)
  Description           Genre Actors  Runtime  E  F
0           a   comedy, drama   a, b        1  5  a
1           b  action, comedy   a, c        3  3  a
2           c          action   d, a        5  6  a

X = trainset.drop(['Description', 'Runtime'], axis=1)
features = ['Genre','Actors']
X_dummy_list = [X.pop(f).str.get_dummies(', ').add_prefix(f + '.') for f in features]
X = pd.concat([X] + X_dummy_list , axis = 1)
print (X)

   E  F  Genre.action  Genre.comedy  Genre.drama  Actors.a  Actors.b  \
0  5  a             0             1            1         1         1   
1  3  a             1             1            0         1         0   
2  6  a             1             0            0         1         0   

   Actors.c  Actors.d  
0         0         0  
1         1         0  
2         0         1