我正在嘗試采用具有匹配 ID 的藝術(shù)家，使音樂(lè)跨越各種單一到流派的組合。這就是我想要做的Artist | Id | Genre | Jazz | Blues | Rock | Trap | Rap | Hip-Hop | Pop | Rb |----------------------------------------------------------------------------------------------------Bob | 1 | [Jazz, Blues] | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0----------------------------------------------------------------------------------------------------Fred | 2 | [Rock,Jazz] | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0----------------------------------------------------------------------------------------------------Jeff | 3 | [Trap, Rap, Hip-Hop] | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0----------------------------------------------------------------------------------------------------Amy | 4 | [Pop, Rock, Jazz] | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0----------------------------------------------------------------------------------------------------Mary | 5 | [Hip-Hop, Jazz, Rb] | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1----------------------------------------------------------------------------------------------------這是我得到的錯(cuò)誤---------------------------------------------------------------------------TypeError Traceback (most recent call last)<ipython-input-50-7a4ed81e14d7> in <module> 11 for index, row in artist_df.iterrows(): 12 x.append(index)---> 13 for i in row['genre']: 14 artists_with_genres.at[index, genre] = 1 15 TypeError: 'float' object is not iterable這些（藝術(shù)家）流派是我將在結(jié)合其他因素（如年份、歌曲或人口統(tǒng)計(jì)數(shù)據(jù)）時(shí)用來(lái)幫助確定相似藝術(shù)家的屬性。我正在創(chuàng)建和迭代的新專欄將指定藝術(shù)家是否屬于某個(gè)流派。用 1/0 來(lái)簡(jiǎn)單地表示藝術(shù)家是否是搖滾/嘻哈/陷阱等。使用屬性的二進(jìn)制表示。這是當(dāng)前的數(shù)據(jù)框獲取我的數(shù)據(jù)框并將流派拆分為單獨(dú)的類型，以便我可以轉(zhuǎn)換為 1/0 二進(jìn)制表示。我需要將流派設(shè)置為索引嗎？
1 回答
aluckdog
TA貢獻(xiàn)1847條經(jīng)驗(yàn) 獲得超7個(gè)贊
嘗試使用get_dummies：
df['Genre'] = df['Genre'].str.split('|')
dfx = pd.get_dummies(pd.DataFrame(df['Genre'].tolist()).stack()).sum(level=0)
df = pd.concat([df, dfx], axis=1).drop(columns=['Genre'])
print(df)
Artist Id Blues Hip-Hop Jazz Pop Rap Rb Rock Trap
0 Bob 1 1 0 1 0 0 0 0 0
1 Fred 2 0 0 1 0 0 0 1 0
2 Jeff 3 0 1 0 0 1 0 0 1
3 Amy 4 0 0 1 1 0 0 1 0
4 Mary 5 0 1 1 0 0 1 0 0
詳細(xì)解釋看這里 -> Pandas column of lists to separate columns
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