查找和替换List中条目内的所有值

Question

Question:

Given a list of the form:

[((x1,y1,z1), (x2,y2,z2), variable_1, variable_2, variable_3),
((x3,y3,z3), (x1,y1,z1), variable_4, variable_5, variable_6),...
((x1,y1,z1), (xN,yN,zN), variable_M, variable_M+1, variable_M+2)]

What is the most efficient (and preferably universal, ie for any list) way to replace all entries in the list matching a value (say (x1,y1,z1) ) with another value (say (xM,yM,zM) )?

Central to my problem is that each entry in the list is composed of multiple elements, and typical methods I use to replace values (eg [new_input if x == current else x for x in pairs] ) do not appear to be useful in this case. Note the form of each entry is: ((x1,y1,z1), (x2,y2,z2), variable_1, variable_2, variable_3) , and the list is: [((x1,y1,z1), (x2,y2,z2), variable_1, variable_2, variable_3), ((x3,y3,z3), (x1,y1,z1), variable_4, variable_5, variable_6)] . I'm looking to change elements within the entry.

Currently I am using a reliable if crude method in which the list is broken into columns and then replaces individual elements in the list, before zipping the columns back together. The code is roughly as follows:

current=(x1,y1,z1)
new_input=(xM,yM,zM)
pairs=#list as above
def column(list, col_num):
    return [column[col_num] for column in list]

p_c_1=column(pairs, 0)
p_c_2=column(pairs, 1)
p_c_3=column(pairs, 2)
p_c_4=column(pairs, 3)
p_c_5=column(pairs, 4)

for i in xrange(len(pairs)):
    if p_c_1[i]==current:
       p_c_1.remove(p_c_1[i])
       p_c_1.insert(i,new_input)
    if p_c_2[i]==current:
       p_c_2.remove(p_c_2[i])
       p_c_2.insert(i,new_input)

pairs=zip(p_c_1,p_c_2,p_c_3,p_c_4,p_c_5)

This method yields an output:

[((xM,yM,zM), (x2,y2,z2), variable_1, variable_2, variable_3),
((x3,y3,z3), (xM,yM,zM), variable_4, variable_5, variable_6),...
((xM,yM,zM), (xN,yN,zN), variable_M, variable_M+1, variable_M+2)]

The method has proved reliable for this particular case, but as there are multiple lists with individual entries composed of several items, I have to construct a new bit of code such as that above for each unique list, which seems crude. I would ideally want to avoid disassembling and reassembling the list every time I change an entry, or automate the process at least.

What is the best way to go about this issue?

Note: A solution would ideally be able to handle the replacement of any element within a list entry, and not just the first two columns. Other lists I have do not necessarily follow the above format.

Answer 1

Create a dict that is a mapping of (things you want to replace) to (things to replace them with).

Then it's straightforward to make a new list comprehension using dict.get to default to the extant value if you haven't specified a replacement in your replacement-dict.

li = [('a','c','b'),('b','b','c'),('c','d'),'5',4]

d = {('b','b','c'):('b','b','b'),('c','d'):('c','c')}

[d.get(x,x) for x in li]
Out[33]: [('a', 'c', 'b'), ('b', 'b', 'b'), ('c', 'c'), '5', 4]

Edit: since it seems you have a peculiar data structure (one giant tuple spanning the entire list ), you can do this:

[tuple(d.get(x,x) for x in li[0])]

Or, if you actually have more of those nested outer-tuples that you're just not showing, try:

[tuple(d.get(x,x) for x in outer_tup) for outer_tup in li]

Both get you back to your original list-of-tuples-of-(tuples and other things) data structure.