Python 字典，多个键指向 memory 有效方式中的同一列表

Question

I have this unique requirement which can be explained by this code. This is working code but not memory efficient.

data = [[
        "A 5408599",
        "B 8126880",
        "A 2003529",
    ],
    [
        "C 9925336",
        "C 3705674",
        "A 823678571",
        "C 3205170186",
    ],
    [
        "C 9772980",
        "B 8960327",
        "C 4185139021",
        "D 1226285245",
        "C 2523866271",
        "D 2940954504",
        "D 5083193",
    ]]

temp_dict = {
    item: index for index, sublist in enumerate(data)
        for item in sublist
}

print(data[temp_dict["A 2003529"]])

out: ['A 5408599', 'B 8126880', 'A 2003529']

In short, I want each item of sub-list to be indexable and should return the sublist.

The above method works but It takes a lot of memory when data is large. Is there any better, memory and CPU friendly way? The data is stored as a JSON file.

Edit I tried the answers for the largest possible use case scenario (1000 sublist, 100 items in each sublist, 1 million queries) and here are results (mean of 10 runs):

Method,    Time (seconds),    Extra Memory used
my,        0.637              40 Mb
deceze,    0.63               40 Mb
James,     0.78               200 kb
Pant,      > 300              0 kb
mcsoini,   forever            0 kb

Answer 1

You can try something like this:

list(filter(lambda x: any(["C 9772980" in x]),data))

No need to make a mapping structure.

Answer 2

You are really in a trade-off space between the time/memory it takes to generate the dictionary versus the time it takes to scan the entire data for an on-the-fly method.

If you want a low memory method, you can use a function that searches each sublist for the value. Using a generator will get initial results faster to the user, but for large data sets, this will be slow between returns.

data = [[
        "A 5408599",
        "B 8126880",
        "A 2003529",
    ],
    [
        "C 9925336",
        "C 3705674",
        "A 823678571",
        "C 3205170186",
    ],
    [
        "C 9772980",
        "B 8960327",
        "C 4185139021",
        "D 1226285245",
        "C 2523866271",
        "D 2940954504",
        "D 5083193",
    ]]


def find_list_by_value(v, data):
    for sublist in data:
        if v in sublist:
            yield sublist

for s in find_list_by_value("C 9772980", data):
    print(s)

As mentioned in the comments, building a hash table based just on the first letter or first 2 or 3 character might be a good place to start. This will allow you to build a candidate list of sublists, then scan those to see if the value is in the sublist.

from collections import defaultdict

def get_key(v, size=3):
    return v[:size]

def get_keys(sublist, size=3):
    return set(get_key(v, size) for v in sublist)

def find_list_by_hash(v, data, hash_table, size=3):
    key = get_key(v, size)
    candidate_indices = hash_table.get(key, set())
    for ix in candidates:
        if v in data[ix]:
            yield data[ix]

# generate the small hash table
quick_hash = defaultdict(set)
for i, sublist in enumerate(data):
    for k in get_keys(sublist, 3):
        quick_hash[k].add(i)

# lookup a value by the small hash
for s in find_list_by_hash("C 9772980", data, quick_hash, 3):
    print(s)

In this code quick_hash will take some time to build, because you are scanning your entire data structure. However, the memory foot print will be much smaller. You main parameter for tuning performance is size . Smaller size will have a smaller memory footprint, but will take longer when running find_list_by_hash because your candidate pool will be larger. You can do some testing to see what the right size should be for your data. Just be mindful that all of your values are at least as long as size .

Answer 3

try this, using pandas

import pandas as pd
df=pd.DataFrame(data)
rows = df.shape[0]
for row in range(rows):
    print[[row]]    #Do something with your data

this looks simple solution, even if your data grows big, this will handle that efficiently

Answer 4

I'm not entirely sure how this would behave for larger amounts data, but you could try something along the lines of:

import pandas as pd
df = pd.DataFrame(data).T
df.loc[:, (df == 'A 2003529').any(axis=0)]
Out[39]: 
           0
0  A 5408599
1  B 8126880
2  A 2003529
3       None
4       None
5       None
6       None

Edit: Does not seem to be beneficial in terms of time, based on a quick test with some fake larger scale data.