Changing a single element of a list while iterating in python

Question

So I'm given a list of tuples. Each tuple contains an integer, a string, and a boolean. I want to go through the list and find the first tuple that has a False for that boolean, do some operation with the tuple, then change the boolean to true. Here's some sample code:

def sendnext(packets):
    for (number, data, sent) in packets:
        if not sent:
            sendsegment(number, data)
            sent = True
            break

Obviously, I can't just change sent to True. The list can also grow arbitrarily large, so I don't want to make a copy of it or use a list comprehension. I know I could use indexing, but that will probably reduce readability. I feel like there should be a pythonic way to do this. Any suggestions?

Thanks for your help!

Do you need to preserve the packets later on? Or you just need to do the sendsegment if sent = False? — shad0w_wa1k3r
– shad0w_wa1k3r, Commented Nov 14, 2013 at 3:28
Yes, they'll be used later. Packets is actually a global list. — Championcake
– Championcake, Commented Nov 14, 2013 at 3:32

MxLDevs · Accepted Answer · 2013-11-14 03:34:06Z

2

I would define a Packet class and then simply perform operations on each Packet object

# assuming a Packet class has been defined and you have a list of Packet objects
for packet in packets:
  if not packet.sent:
    sendsegment(packet.number, packet.data)
    packet.sent = true
    break

Especially since you'll be using the packets elsewhere, and presumably is a major component in your program, it might be easier and more manageable to avoid working with arbitrary lists of data.

edited Nov 14, 2013 at 3:34

answered Nov 14, 2013 at 3:28

MxLDevs

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1 Comment

Championcake Over a year ago

This is a great idea, but unfortunately I'm not able to change how the packets are defined.

Pi Marillion · Accepted Answer · 2013-11-14 14:09:49Z

1

Just use an index variable so that you're changing the actual values, instead of a copy:

def sendnext(packets):
    for i in range(len(packets)):
        if not packets[i][2]:
            sendsegment(packets[i][0], packets[i][1])
            packets[i] = (packets[i][0], packets[i][1], True)
            break

For simplicity, iteration can be done with enumerate:

def sendnext(packets):
    for i, (number, data, sent) in enumerate(packets):
        if not sent:
            sendsegment(number, data)
            packets[i] = (number, data, True)
            break

edited Nov 14, 2013 at 14:09

answered Nov 14, 2013 at 3:29

Pi Marillion

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6 Comments

Championcake Over a year ago

The second to last line would be packets[i] = (packets[i][0], packets[i][1], True). I think I'll end up using this, but that line just makes me uneasy.

dstromberg Over a year ago

This looks good, except it's briefer and less error-prone to use "for index, packet in enumerate(packets)" instead of range(len).

Pi Marillion Over a year ago

@Championcake Whoops, I always forget that tuple elements can't be assigned to (since tuples are immutable). I fixed the answer.

Pi Marillion Over a year ago

@dstromberg My preference is usually for indexing (perhapse due to being a C-style transplant), but that is, of course, briefer. I've included a more readable option including enumerate.

Tim Peters Over a year ago

Note that tuple unpacking can be done in the for line, like so: for i, (number, data, sent) in enumerate(packets):. It's clearer that way, provided you're used to it ;-)

|

dyoo · Accepted Answer · 2013-11-14 03:33:18Z

0

Can you use a generator approach here to map some computation on your iteration of packets? Something like:

def sendnext(packets):
    for (number, data, sent) in packets:
        if not sent:
            sendsegment(number, data)
            yield (number, data, True)
        else:
            yield (number, data, True)

where we relinquish control of packets to sendnext, and then just use the packets that get yielded by sendnext?

answered Nov 14, 2013 at 3:33

dyoo

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