python - 创建一个 one-hot pandas 数据框

Question

我有一组从 0 到 9 的标签，例如:

2 7 5 3

我想将其转换为单热编码，如下所示:

0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 0 0

所以我做了这个方法:

def make_one_hot(m):
    result = pd.DataFrame([])
    for i in range(0, len(m)):
        x = [0] * 10
        x[m[i]] = 1
        result = result.append(x)
        print("result: " + result)
    return result

当打印结果时，我得到这个错误:

Traceback (most recent call last):
  File "../src/script.py", line 23, in <module>
    train_labels = make_one_hot(train_data.ix[:,0])
  File "../src/script.py", line 18, in make_one_hot
    print("result: " + result)
  File "/opt/conda/lib/python3.6/site-packages/pandas/core/ops.py", line 1241, in f
8.8s
2
    return self._combine_const(other, na_op)
  File "/opt/conda/lib/python3.6/site-packages/pandas/core/frame.py", line 3641, in _combine_const
    raise_on_error=raise_on_error)
  File "/opt/conda/lib/python3.6/site-packages/pandas/core/internals.py", line 3197, in eval
    return self.apply('eval', **kwargs)
  File "/opt/conda/lib/python3.6/site-packages/pandas/core/internals.py", line 3091, in apply
    applied = getattr(b, f)(**kwargs)
  File "/opt/conda/lib/python3.6/site-packages/pandas/core/internals.py", line 1205, in eval
8.8s
3
    repr(other))
TypeError: Could not compare ['result: '] with block values

由于我不擅长 Python，所以我不确定是只有 print 语句有问题还是我计算数组的方式也有问题。

那么简单而正确的方法是什么？

Answer 1

方法 #1:这是使用 NumPy 广播 的一种方法 -

In [143]: a = [2 ,7 ,5 ,3]

In [144]: pd.DataFrame((np.asarray(a)[:,None] == np.arange(10)).astype(int))
Out[144]: 
   0  1  2  3  4  5  6  7  8  9
0  0  0  1  0  0  0  0  0  0  0
1  0  0  0  0  0  0  0  1  0  0
2  0  0  0  0  0  1  0  0  0  0
3  0  0  0  1  0  0  0  0  0  0

方法 #2: 另一种使用 zeros-initialization -

In [145]: out = np.zeros((len(a), 10),dtype=int)

In [146]: out[np.arange(len(a)), a] = 1

In [147]: pd.DataFrame(out)
Out[147]: 
   0  1  2  3  4  5  6  7  8  9
0  0  0  1  0  0  0  0  0  0  0
1  0  0  0  0  0  0  0  1  0  0
2  0  0  0  0  0  1  0  0  0  0
3  0  0  0  1  0  0  0  0  0  0

方法 #3: 使用 Scipy 的稀疏矩阵 -

In [166]: from scipy.sparse import csr_matrix

In [167]: n = len(a)

In [169]: pd.DataFrame(csr_matrix(([1]*n, (range(n), a)), shape=(n, 10)).toarray())
Out[169]: 
   0  1  2  3  4  5  6  7  8  9
0  0  0  1  0  0  0  0  0  0  0
1  0  0  0  0  0  0  0  1  0  0
2  0  0  0  0  0  1  0  0  0  0
3  0  0  0  1  0  0  0  0  0  0