最好的是根据元组的第一个元素重新排列一个元组列表以匹配字符串？

Question

I have a list of tuples that looks like:

[(['A', 'E', 'J', 'M', 'S'],), (['E', 'C', 'D', 'A', 'A'],), (['F', 'B', 'F', 'C', 'C'],)]

or if it's easier, before I used zip() it looked like this:

[['A', 'E', 'J', 'M', 'S'],
['E', 'C', 'D', 'A', 'A'],
['F', 'B', 'F', 'C', 'C']]

I'm trying to re-order the top row so it reads JAMES, and have the bottom rows follow. What is the best way to go about this. Thanks for any help.

The output i'm aiming for is

[['J', 'A', 'M', 'E', 'S'],
['D', 'E', 'A', 'C', 'A'],
['F', 'F', 'C', 'B', 'C']]

Answer 1

assuming you arrange the first row manually, make a a numpy.ndarray , and slice it with the manually chosen index array:

In [117]: a=np.array([['A', 'E', 'J', 'M', 'S'],
     ...: ['E', 'C', 'D', 'A', 'A'],
     ...: ['F', 'B', 'F', 'C', 'C']])

In [118]: idx=[2,0,3,1,4]  #index array

In [119]: a[:,idx]
Out[119]: 
array([['J', 'A', 'M', 'E', 'S'],
       ['D', 'E', 'A', 'C', 'A'],
       ['F', 'F', 'C', 'B', 'C']], 
      dtype='|S1')

Answer 2

Create a dictionary that contains keys like 'J' , 'A' , 'M' , .. pointing to some numbers to get the desired order and then use this dictionary to sort the unzipped version( zip(*..) ) of your list of lists.

>>> lis = [['A', 'E', 'J', 'M', 'S'],
['E', 'C', 'D', 'A', 'A'],
['F', 'B', 'F', 'C', 'C']]
>>> d = {'J':0, 'A':1, 'M':2, 'E':3, 'S':4}
>>> zip(*sorted(zip(*lis), key= lambda x: (d[x[0]])))
[('J', 'A', 'M', 'E', 'S'), ('D', 'E', 'A', 'C', 'A'), ('F', 'F', 'C', 'B', 'C')]

Answer 3

You want to associate the data with your sort keys, sort, then extract the data in order. More or less DSU. In this case, zip is a quick way to 'cross' the matrix, so zip the tuples, do your sorting (however you want to do it... in the below, i used a fixed mapping since i don't know your sort criteria), zip the other way:

input = [['A', 'E', 'J', 'M', 'S'],
    ['E', 'C', 'D', 'A', 'A'],
    ['F', 'B', 'F', 'C', 'C']]

input_zip = zip(*input)
print input_zip
[('A', 'E', 'F'),
 ('E', 'C', 'B'),
 ('J', 'D', 'F'),
 ('M', 'A', 'C'),
 ('S', 'A', 'C')]

indices = [2,0,3,1,4]
output = [input_zip[x] for x in indices]
[('J', 'D', 'F'),
 ('A', 'E', 'F'),
 ('M', 'A', 'C'),
 ('E', 'C', 'B'),
 ('S', 'A', 'C')]

# Zip it the other way now.
output_zip = zip(*output)
[('J', 'A', 'M', 'E', 'S'),
 ('D', 'E', 'A', 'C', 'A'),
 ('F', 'F', 'C', 'B', 'C')]

hopefully that doesn't just answer the question, but explain the thinking behind it.

Answer 4

So it sounds like what you want is to first order the data so that you have the corresponding element from each of your lists in a list of its owns so:

    input = [['A', 'E', 'J', 'M', 'S'],
            ['E', 'C', 'D', 'A', 'A'],
            ['F', 'B', 'F', 'C', 'C']]
    rearrangedlist = [[input[x][y] for x in range(len(input))] for y in range(len(input[0]))]

    # Gives rearrangedlist = [['A', 'E', 'F'],
                             ['E', 'C', 'B'],
                             ['J', 'D', 'F'],
                             ['M', 'A', 'C'],
                             ['S', 'A', 'C']]

then:

    correctorder = ['J','A','M','E','S']
    rearrangeddict = [dict((x[0],x[1:]) for x in rearrangedlist)]
    for lettertofind in correctorder:
        print rearrangeddict[lettertofind]

Gives:

    ['J', 'D', 'F']
    ['A', 'E', 'F']
    ['M', 'A', 'C']
    ['E', 'C', 'B']
    ['S', 'A', 'C']