pd.get_dummies（）在很大程度上很慢

Question

我不确定这是否已经是最快的方法，或者如果我这样做效率低下。

我想热编码一个特定的具有27k +级别的分类列。列有2点不同的数据集不同的值，所以我第一次使用前get_dummies（）

def hot_encode_column_in_both_datasets(column_name,df,df2,sparse=True): 
    col1b = set(df2[column_name].unique()) 
    col1a = set(df[column_name].unique()) 
    combined_cats = list(col1a.union(col1b)) 
    df[column_name] = df[column_name].astype('category', categories=combined_cats) 
    df2[column_name] = df2[column_name].astype('category', categories=combined_cats) 

    df = pd.get_dummies(df, columns=[column_name],sparse=sparse) 
    df2 = pd.get_dummies(df2, columns=[column_name],sparse=sparse) 
    try: 
     del df[column_name] 
     del df2[column_name] 
    except: 
     pass 
    return df,df2 

但是，它已经运行了2个多小时的联合水平，它仍然停留热码。

我可以在这里做错吗？还是仅仅是在大型数据集上运行它的本质？

Df有6.8m行和27列，Df2有19990行和27列，然后热编码我想要的列。

建议感激，谢谢！ :)

Answer 1

我简要回顾了get_dummies source code，我认为它可能没有充分利用您的用例的稀疏性。下面的方法可以更快，但我并没有试图一路扩展它你有19M记录：

import numpy as np 
import pandas as pd 
import scipy.sparse as ssp 

np.random.seed(1) 
N = 10000 

dfa = pd.DataFrame.from_dict({ 
    'col1': np.random.randint(0, 27000, N) 
    , 'col2b': np.random.choice([1, 2, 3], N) 
    , 'target': np.random.choice([1, 2, 3], N) 
    }) 

# construct an array of the unique values of the column to be encoded 
vals = np.array(dfa.col1.unique()) 
# extract an array of values to be encoded from the dataframe 
col1 = dfa.col1.values 
# construct a sparse matrix of the appropriate size and an appropriate, 
# memory-efficient dtype 
spmtx = ssp.dok_matrix((N, len(vals)), dtype=np.uint8) 
# do the encoding. NB: This is only vectorized in one of the two dimensions. 
# Finding a way to vectorize the second dimension may yield a large speed up 
for idx, val in enumerate(vals): 
    spmtx[np.argwhere(col1 == val), idx] = 1 

# Construct a SparseDataFrame from the sparse matrix and apply the index 
# from the original dataframe and column names. 
dfnew = pd.SparseDataFrame(spmtx, index=dfa.index, 
          columns=['col1_' + str(el) for el in vals]) 
dfnew.fillna(0, inplace=True) 

UPDATE

借用其他答案见解here和here ，我能够在两个维度上矢量化解决方案。在我有限的测试中，我注意到构建SparseDataFrame似乎将执行时间增加了几倍。因此，如果您不需要返回类似DataFrame的对象，则可以节省大量时间。此解决方案还处理您需要将2+ DataFrames编码为具有相同列数的2-d数组的情况。

import numpy as np 
import pandas as pd 
import scipy.sparse as ssp 

np.random.seed(1) 
N1 = 10000 
N2 = 100000 

dfa = pd.DataFrame.from_dict({ 
    'col1': np.random.randint(0, 27000, N1) 
    , 'col2a': np.random.choice([1, 2, 3], N1) 
    , 'target': np.random.choice([1, 2, 3], N1) 
    }) 

dfb = pd.DataFrame.from_dict({ 
    'col1': np.random.randint(0, 27000, N2) 
    , 'col2b': np.random.choice(['foo', 'bar', 'baz'], N2) 
    , 'target': np.random.choice([1, 2, 3], N2) 
    }) 

# construct an array of the unique values of the column to be encoded 
# taking the union of the values from both dataframes. 
valsa = set(dfa.col1.unique()) 
valsb = set(dfb.col1.unique()) 
vals = np.array(list(valsa.union(valsb)), dtype=np.uint16) 


def sparse_ohe(df, col, vals): 
    """One-hot encoder using a sparse ndarray.""" 
    colaray = df[col].values 
    # construct a sparse matrix of the appropriate size and an appropriate, 
    # memory-efficient dtype 
    spmtx = ssp.dok_matrix((df.shape[0], vals.shape[0]), dtype=np.uint8) 
    # do the encoding 
    spmtx[np.where(colaray.reshape(-1, 1) == vals.reshape(1, -1))] = 1 

    # Construct a SparseDataFrame from the sparse matrix 
    dfnew = pd.SparseDataFrame(spmtx, dtype=np.uint8, index=df.index, 
           columns=[col + '_' + str(el) for el in vals]) 
    dfnew.fillna(0, inplace=True) 
    return dfnew 

dfanew = sparse_ohe(dfa, 'col1', vals) 
dfbnew = sparse_ohe(dfb, 'col1', vals)