在python中将1个列表中的项目替换为另一个列表中的项目

Question

I'm trying to replace items in 1 list of lists with items from another list of lists.

    alist = [[1,2,3], [4,5,6], [7,8,9]]
    blist = [[a,b,c], [d,e,f], [g,h,i]]

Basically, I want to replace the items in alist with it's blist counterpart. For example, if someone chose 1 in alist, I want to replace it with the same index from blist which would be a.

    alist = [[a,2,3], [4,5,6], [7,8,9]]

Any suggestions?

Answer 1

One way is to loop over both lists together, then loop over each subgroup to replace the matched choice.

For lockstep iteration, the zip function is your best friend. For transforming lists a list comprehension is a great tool. Here's how to use them:

>>> alist = [[1,2,3], [4,5,6], [7,8,9]]
>>> blist = [[10,20,30], [40,50,60], [70,80,90]]
>>> choice = 1
>>> alist = [[(bx if ax==choice else ax) for ax, bx in zip(aseq, bseq)] for aseq, bseq in zip(alist, blist)]
>>> alist
[[10, 2, 3], [4, 5, 6], [7, 8, 9]]

What is nice about this approach is that it can do multiple replacements (for example, if alist contains several 1 entries, they all get replaced with their corresponding entries in the blist ).

Another nice feature is that the ax==choice decision can easily be replaced with other predicates (for example, ax in {2, 5, 8} will search for multiple substitution targets in a single pass). Another example is ax % 2 == 0 which would make blist replacements for any alist value that is even).

Answer 2

>>> z = [(i, x.index(item)) for i, x in enumerate(alist) if item in x][0]
>>> alist[z[0]][z[1]] = blist[z[0]][z[1]]
>>> alist
[['a', 2, 3], [4, 5, 6], [7, 8, 9]]

See it working online: ideone

Answer 3

Crude but easily explained:

blist = [['a','b','c'], ['d','e','f'], ['g','h','i']]
alist = [[1,2,3], [4,5,2], [7,8,9]]
x = 2

for y in xrange(len(alist)):
    if x in alist[y]:
        i = alist[y].index(x)
        alist[y][i] = blist[y][i]

print alist

Output:

[[1, 'b', 3], [4, 5, 'f'], [7, 8, 9]]