Huge memory usage of loading large dictionaries in memory
Lots of ideas. However, if you want practical help, edit your question to show ALL of your code. Also tell us what is the "it" that shows memory used, what it shows when you load a file with zero entries, and what platform you are on, and what version of Python.
You say that "the word can be 1-5 words long". What is the average length of the key field in BYTES? Are the ids all integer? If so what are the min and max integer? If not, what is the average length if ID in bytes? To enable cross-achecking of all of above, how many bytes are there in your 6.5M-line file?
Looking at your code, a 1-line file word1,1 will create a dict d['1'] = 'word1' ... isn't that bassackwards?
Update 3: More questions: How is the "word" encoded? Are you sure you are not carrying a load of trailing spaces on any of the two fields?
Update 4 ... You asked "how to most efficiently store key/value pairs in memory with python" and nobody's answered that yet with any accuracy.
You have a 168 Mb file with 6.5 million lines. That's 168 * 1.024 ** 2 / 6.5 = 27.1 bytes per line. Knock off 1 byte for the comma and 1 byte for the newline (assuming it's a *x platform) and we're left with 25 bytes per line. Assuming the "id" is intended to be unique, and as it appears to be an integer, let's assume the "id" is 7 bytes long; that leaves us with an average size of 18 bytes for the "word". Does that match your expectation?
So, we want to store an 18-byte key and a 7-byte value in an in-memory look-up table.
Let's assume a 32-bit CPython 2.6 platform.
>>> K = sys.getsizeof('123456789012345678')
>>> V = sys.getsizeof('1234567')
>>> K, V
(42, 31)
Note that sys.getsizeof(str_object) => 24 + len(str_object)
Tuples were mentioned by one answerer. Note carefully the following:
>>> sys.getsizeof(())
28
>>> sys.getsizeof((1,))
32
>>> sys.getsizeof((1,2))
36
>>> sys.getsizeof((1,2,3))
40
>>> sys.getsizeof(("foo", "bar"))
36
>>> sys.getsizeof(("fooooooooooooooooooooooo", "bar"))
36
>>>
Conclusion: sys.getsizeof(tuple_object) => 28 + 4 * len(tuple_object) ... it only allows for a pointer to each item, it doesn't allow for the sizes of the items.
A similar analysis of lists shows that sys.getsizeof(list_object) => 36 + 4 * len(list_object) ... again it is necessary to add the sizes of the items. There is a further consideration: CPython overallocates lists so that it doesn't have to call the system realloc() on every list.append() call. For sufficiently large size (like 6.5 million!) the overallocation is 12.5 percent -- see the source (Objects/listobject.c). This overallocation is not done with tuples (their size doesn't change).
Here are the costs of various alternatives to dict for a memory-based look-up table:
List of tuples:
Each tuple will take 36 bytes for the 2-tuple itself, plus K and V for the contents. So N of them will take N * (36 + K + V); then you need a list to hold them, so we need 36 + 1.125 * 4 * N for that.
Total for list of tuples: 36 + N * (40.5 + K + v)
That's 26 + 113.5 * N (about 709 MB when is 6.5 million)
Two parallel lists:
(36 + 1.125 * 4 * N + K * N) + (36 + 1.125 * 4 * N + V * N) i.e. 72 + N * (9 + K + V)
Note that the difference between 40.5 * N and 9 * N is about 200MB when N is 6.5 million.
Value stored as int not str:
But that's not all. If the IDs are actually integers, we can store them as such.
>>> sys.getsizeof(1234567)
12
That's 12 bytes instead of 31 bytes for each value object. That difference of 19 * N is a further saving of about 118MB when N is 6.5 million.
Use array.array('l') instead of list for the (integer) value:
We can store those 7-digit integers in an array.array('l'). No int objects, and no pointers to them -- just a 4-byte signed integer value. Bonus: arrays are overallocated by only 6.25% (for large N). So that's 1.0625 * 4 * N instead of the previous (1.125 * 4 + 12) * N, a further saving of 12.25 * N i.e. 76 MB.
So we're down to 709 - 200 - 118 - 76 = about 315 MB.
N.B. Errors and omissions excepted -- it's 0127 in my TZ :-(
convert your data into a dbm (import anydbm, or use berkerley db by import bsddb ...), and then use dbm API to access it.
the reason to explode is that python has extra meta information for any objects, and the dict needs to construct a hash table (which would require more memory). you just created so many objects (6.5M) so the metadata becomes too huge.
import bsddb
a = bsddb.btopen('a.bdb') # you can also try bsddb.hashopen
for x in xrange(10500) :
a['word%d' %x] = '%d' %x
a.close()
This code takes only 1 second to run, so I think the speed is OK (since you said 10500 lines per second). btopen creates a db file with 499,712 bytes in length, and hashopen creates 319,488 bytes.
With xrange input as 6.5M and using btopen, I got 417,080KB in ouput file size and around 1 or 2 minute to complete insertion. So I think it's totally suitable for you.
Take a look (Python 2.6, 32-bit version)...:
>>> sys.getsizeof('word,1')
30
>>> sys.getsizeof(('word', '1'))
36
>>> sys.getsizeof(dict(word='1'))
140
The string (taking 6 bytes on disk, clearly) gets an overhead of 24 bytes (no matter how long it is, add 24 to its length to find how much memory it takes). When you split it into a tuple, that's a little bit more. But the dict is what really blows things up: even an empty dict takes 140 bytes -- pure overhead of maintaining a blazingly-fast hash-based lookup take. To be fast, a hash table must have low density -- and Python ensures a dict is always low density (by taking up a lot of extra memory for it).
The most memory-efficient way to store key / value pairs is as a list of tuples, but lookup of course will be very slow (even if you sort the list and use bisect for the lookup, it's still going to be extremely slower than a dict).
Consider using shelve instead -- that will use little memory (since the data reside on disk) and still offer pretty spiffy lookup performance (not as fast as an in-memory dict, of course, but for a large amount of data it will be much faster than lookup on a list of tuples, even a sorted one, can ever be!-).