我有一个列表
:
L = ['a', 'b']
我需要通过连接从 1
到 k
的原始 list
来创建一个新的 list
.示例:
I need to create a new list
by concatenating an original list
which range goes from 1
to k
. Example:
k = 4
L1 = ['a1','b1', 'a2','b2','a3','b3','a4','b4']
我试试:
l1 = L * k
print l1
#['a', 'b', 'a', 'b', 'a', 'b', 'a', 'b']
l = [ [x] * 2 for x in range(1, k + 1) ]
print l
#[[1, 1], [2, 2], [3, 3], [4, 4]]
l2 = [item for sublist in l for item in sublist]
print l2
#[1, 1, 2, 2, 3, 3, 4, 4]
print zip(l1,l2)
#[('a', 1), ('b', 1), ('a', 2), ('b', 2), ('a', 3), ('b', 3), ('a', 4), ('b', 4)]
print [x+ str(y) for x,y in zip(l1,l2)]
#['a1', 'b1', 'a2', 'b2', 'a3', 'b3', 'a4', 'b4']
但我认为这很复杂.什么是最快和最通用的解决方案?
But I think it is very complicated. What is the fastest and most generic solution?
你可以使用列表推导:
L = ['a', 'b']
k = 4
L1 = ['{}{}'.format(x, y) for y in range(1, k+1) for x in L]
print(L1)
输出
['a1', 'b1', 'a2', 'b2', 'a3', 'b3', 'a4', 'b4']
这篇关于Python:将列表与范围列表合并的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!