python pandas/numpy 根据映射方案替换所有值的快速方法

Question

假设我有一个巨大的 Pandas 数据框/numpy 数组，其中每个元素都是一个有序值列表:

sequences = np.array([12431253, 123412531, 12341234,12431253, 145345],
                     [5463456, 1244562, 23452],
                     [243524, 141234,12431253, 456367], 
                     [456345, 253451], 
                     [75635, 14145, 12346,12431253])

或者，

sequences = pd.DataFrame({'sequence': [[12431253, 123412531, 12341234,12431253, 145345],
                                      [5463456, 1244562, 23452],
                                      [243524, 141234, 456367,12431253],
                                      [456345, 253451],
                                      [75635, 14145, 12346,12431253]]})

我想用另一组从 0 开始的标识符替换它们，所以我设计了一个这样的映射:

from compiler.ast import flatten
from sets import Set
mapping = pd.DataFrame({'v0': list(Set(flatten(sequences['sequence']))), 'v1': range(len(Set(flatten(sequences['sequence'])))})

……

所以我正在寻找的结果:

sequences = np.array([1, 2, 3,1, 4], [5, 6, 7], [8, 9, 10,1], [11, 12], [13, 14, 15,1])

我怎样才能将其扩展到一个巨大的数据框/numpy 序列？

非常感谢您的指导!非常感谢!

Answer 1

这是一种将1D 数组扁平化的方法，使用np.unique 为每个元素分配唯一的 ID，然后拆分回数组列表 -

lens = np.array(map(len,sequences))
seq_arr = np.concatenate(sequences)
ids = np.unique(seq_arr,return_inverse=1)[1]
out = np.split(ids,lens[:-1].cumsum())

sample 运行-

In [391]: sequences = np.array([[12431253, 123412531, 12341234,12431253, 145345],
     ...:                      [5463456, 1244562, 23452],
     ...:                      [243524, 141234,12431253, 456367], 
     ...:                      [456345, 12431253], 
     ...:                      [75635, 14145, 12346,12431253]])

In [392]: out
Out[392]: 
[array([12, 13, 11, 12,  5]),
 array([10,  9,  2]),
 array([ 6,  4, 12,  8]),
 array([ 7, 12]),
 array([ 3,  1,  0, 12])]

In [393]: np.array(map(list,out)) # If you need NumPy array as final o/p
Out[393]: 
array([[12, 13, 11, 12, 5], [10, 9, 2], [6, 4, 12, 8], [7, 12],
       [3, 1, 0, 12]], dtype=object)