Pandas Dataframe：如何将一列拆分为多个单热编码列

Question

我有一个这样的文本文件：

444537110                         3 11112111022002200022022111121222002...

输入文件中的最后一个字段的长度为 50k 个字符，并且只有 0,1 或 2。我想要这个最后一个字段的一个热编码版本。所以我的预期结果是这样的数据框：

id          chip   g1_0 g1_1 g1_2 g2_0 g2_1 g2_2 g3_0 g3_1 g3_2 g4_0 ... 
444537110   3      0    1    0    0    1    0    0    1    0    0

我通过读取输入文件创建了一个初始数据框：

df = pd.read_csv('test.txt', index_col=0, sep='\s+', header=None, names = ['chip', 'genos'])

这将创建一个包含 3 列的数据框：

id        chip  genos
444537110    3  1111211102200220000022022111121222000200022002...

我想我可以使用下面的方法创建初始的单独的列，然后使用 pandas get_dummies 函数进行一次热编码，但我一直无法创建单独的列.我试过了

[c for c in df['genos'].str]

但这不是分隔字符

我在这里看过一个类似的问题和答案：How can I one hot encode in Python?

但这只处理一种热编码，不处理拆分非常大的列所增加的复杂性

Answer 1

首先创建 DataFrame 并将字符串转换为列表并调用 get_dummies:

df1 = pd.DataFrame([list(x) for x in df['genos']], index=df.index).add_prefix('g')
df2 = pd.get_dummies(df1)

如果需要在原始列中添加新列（如果可能缺少某些组合）使用 DataFrame.reindex by splitted columns with _ and by all combination created by MultiIndex.from_product:

df1 = pd.DataFrame([list(x) for x in df.pop('genos')], index=df.index).add_prefix('g')
df2 = pd.get_dummies(df1)

splitted = df2.columns.str.split('_')
df2.columns = [splitted.str[0].astype(int) + 1, splitted.str[1].astype(int)]
#
mux = pd.MultiIndex.from_product([df2.columns.get_level_values(0), [0,1,2]])
df2 = df2.reindex(mux, axis=1, fill_value=0)
df2.columns = [f'g{a}_{b}' for a, b in df2.columns]
print (df2)
   g1_0  g1_1  g1_2  g2_0  g2_1  g2_2  g3_0  g3_1  g3_2  g4_0  ...  g32_2  \
0     0     1     0     0     1     0     0     1     0     0  ...      1   

   g33_0  g33_1  g33_2  g34_0  g34_1  g34_2  g35_0  g35_1  g35_2  
0      1      0      0      1      0      0      0      0      1  

[1 rows x 105 columns]

Answer 2

考虑到@Dan 对您将以 50k*3 列结尾的问题的评论，您可以通过这样做获得所需的输出：

string ="444537110 3 11112111022002200022022111121222002"
df = pd.DataFrame([string.split(" ")],columns=['id','chip','genos'])
max_number_of_genes = int(df.genos.apply(lambda x : len([y for y in x])).max())

#Create columns 
for gene in range(1,max_number_of_genes+1):
    for y in range(4):
        df['g{}_{}'.format(gene, y)] = 0

#Iterating over genos values 
for row_number, row in df.iterrows():
    genos = [int(x) for x in df.iloc[row_number, 2]]
    for gene_number, gene in enumerate(genos):     
        df.loc[row_number, 'g{}_{}'.format(gene_number+1, gene)] = 1 

print(df)

输出

+----+------------+-------+--------------------------------------+-------+-------+-------+-------+-------+-------+-------+------+--------+--------+--------+--------+--------+--------+--------+--------+--------+-------+
|    |    id      | chip  |                genos                 | g1_0  | g1_1  | g1_2  | g1_3  | g2_0  | g2_1  | g2_2  | ...  | g33_2  | g33_3  | g34_0  | g34_1  | g34_2  | g34_3  | g35_0  | g35_1  | g35_2  | g35_3 |
+----+------------+-------+--------------------------------------+-------+-------+-------+-------+-------+-------+-------+------+--------+--------+--------+--------+--------+--------+--------+--------+--------+-------+
| 0  | 444537110  |    3  | 11112111022002200022022111121222002  |    0  |    1  |    0  |    0  |    0  |    1  |    0  | ...  |     0  |     0  |     1  |     0  |     0  |     0  |     0  |     0  |     1  |     0 |
+----+------------+-------+--------------------------------------+-------+-------+-------+-------+-------+-------+-------+------+--------+--------+--------+--------+--------+--------+--------+--------+--------+-------+

Answer 3

如果您只拆分 50k 个字符，您可以使用原始字符 Python（为了便于阅读）：

>>> a,b,c = zip(*[{0:(1,0,0),1:(0,1,0),2:(0,0,1)}[int(c)] for c in df['genos'][0]])
>>> a,b,c
((0, 0, 0, 0, 0, 0, ...), (1, 1, 1, 1, 0, 1, ...), (0, 0, 0, 0, 1, 0, ...))

Pandas Dataframe：如何将一列拆分为多个单热编码列

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