如何在Python中将多个项目从一个列表移到另一个

Question

I would like to move more than one item from one list to another.

list1 = ['2D','  ','  ','  ','  ','  ','  ','  ','  ']
list2 = ['XX','XX','5D','4S','3D','  ','  ','  ','  ']
list3 = ['XX','XX','XX','8C','7H','6C','  ','  ','  ']

In the code above ' ' is a double space

I would like to be able to move '5D','4S','3D' from list2 onto '8C','7H','6C' in list3 .

I've tried the code below but it doesn't work.

list1 = ['2D','  ','  ','  ','  ','  ','  ','  ','  ']
list2 = ['XX','XX','5D','4S','3D','  ','  ','  ','  ']
list3 = ['XX','XX','XX','8C','7H','6C','  ','  ','  ']


items_to_be_moved = list2[list2.index('XX')+2 : list2.index('  ')]

list3[list3.index('  ')] = items_to_be_moved
del list2[list2.index('XX')+2 : list2.index('  ')]

print('list2',list2)
print('list3',list3)

and this returns

list2 ['XX', 'XX', '  ', '  ', '  ', '  ']
list3 ['XX', 'XX', 'XX', '8C', '7H', '6C', ['5D', '4S', '3D'], '  ', '  ']

However, i dont want to use list2.index('XX')+2 , i would like to use a code that gives the last index of 'XX' just like list2.index(' ') gives the first index of ' ' .

Also, i dont want the moved items to be in a separate list of their own within another list. For example: instead of returning

"list3 ['XX', 'XX', 'XX', '8C', '7H', '6C', ['5D', '4S', '3D'], '  ', '  ']"

the list

"list3 ['XX', 'XX', 'XX', '8C', '7H', '6C','5D', '4S', '3D', '  ', '  ']"

should be returned.

Answer 1

To replace the correct items, use slice assignment.

list1 = ['2D','  ','  ','  ','  ','  ','  ','  ','  ']
list2 = ['XX','XX','5D','4S','3D','  ','  ','  ','  ']
list3 = ['XX','XX','XX','8C','7H','6C','  ','  ','  ']

list3[3:6] = list2[2:5]

print list3
# ['XX', 'XX', 'XX', '5D', '4S', '3D', '  ', '  ', '  ']

If you want to start from after the final consecutive 'XX' , you can use:

from itertools import takewhile
i = sum(1 for _ in takewhile(lambda elem: elem == 'XX', list3))

list3[i:i+3] = list2[i-1:i+2]

to find the index of the last one.

Answer 2

well if all of the 'xx' are always before the needed there is another way (beside the one suggested by @turek) using 'list.count('xx')' and the slicing :

list[count-1:count+2]

you can use the slice operator to do the insert also I would probably iterate like suggested by @intra-c

Answer 3

First, here's how to find the last index . But beware, if you get the index number of the last copy of XX , what if the list reads, eg, ['XX', '1D', 'XX', '4S'] so that there's a valid item in the middle?

Anyway, once you have a span that you want to move from one list to another, you need only use the slice operator to do the insert and delete:

>>> list2[2:5]
['5D', '4S', '3D']
>>> list3[6:6] = list2[2:5]
>>> list3
['XX', 'XX', 'XX', '8C', '7H', '6C', '5D', '4S', '3D', '  ', '  ', '  ']

You can use del to remove from list2, or you can just assign to the slice:

>>> list2[2:5] = []
>>> list2
['XX', 'XX', '  ', '  ', '  ', '  ']

Answer 4

Try looping through with

for item in items_to_be_moved:
    # Add items one at a time

Answer 5

On the second topic:

Don't think there's a function for finding last occurance in lists, but you could either do a loop

pos = 0
while 1:
    try:
        pos = list3.index('XX',pos+1)
    except ValueError:
        break
if pos == 0 and list3[0] != 'XX':
    #Do something to pos if there was no XX in list...

or reverse the list in-place do the index-thing reverse it back and recalc the position.

list3.reverse()
pos = list3.index('XX')
list3.reverse()
pos = len(list3) - pos - 1

Answer 6

Don't forget about negative indices for slicing from the end:

i2_begin = -list2[::-1].index('XX')
i2_end = list2.index('  ')
i3 = list3.index('  ')

list3[i3:i3] = list2[i2_begin:i2_end]
del list2[i2_begin:i2_end]