问题描述
c1. c2. c3. l
1. 2. 3 [1,2,3,4,5,6,7]
3. 4. 8. [8,9,0]
我想分解它,以便 l 列中每个列表中的每 3 个元素将是一个新行,以及原始列表中三元组索引的列。所以我会得到:
c1. c2. c3. l idx
1. 2. 3 [1,3]. 0
1. 2. 3. [4,6]. 1
3. 4. 8. [8,0]. 0
最好的方法是什么?
解决方法
先将列表元素分成块然后explode
:
df.l = df.l.apply(lambda lst: [lst[3*i:3*(i+1)] for i in range(len(lst) // 3)])
df
# c1 c2 c3 l
#0 1 2 3 [[1,2,3],[4,5,6]]
#1 3 4 8 [[8,9,0]]
df.explode('l')
# c1 c2 c3 l
#0 1 2 3 [1,3]
#0 1 2 3 [4,6]
#1 3 4 8 [8,0]
如果需要索引列:
# store index as second element of the tuple
df.l = df.l.apply(lambda lst: [(lst[3*i:3*(i+1)],i) for i in range(len(lst) // 3)])
df
# c1 c2 c3 l
#0 1 2 3 [([1,0),([4,6],1)]
#1 3 4 8 [([8,0],0)]
df = df.explode('l')
df
# c1 c2 c3 l
#0 1 2 3 ([1,0)
#0 1 2 3 ([4,1)
#1 3 4 8 ([8,0)
# extract list and index from the tuple column
df['l'],df['idx'] = df.l.str[0],df.l.str[1]
df
# c1 c2 c3 l idx
#0 1 2 3 [1,3] 0
#0 1 2 3 [4,6] 1
#1 3 4 8 [8,0] 0