问题
I currently have a .h5 file, with a table in it consisting of three columns: a text columns of 64 chars, an UInt32 column relating to the source of the text and a UInt32 column which is the xxhash of the text. The table consists of ~ 2.5e9 rows
I am trying to find and count the duplicates of each text entry in the table - essentially merge them into one entry, while counting the instances. I have tried doing so by indexing on the hash column and then looping through table.itersorted(hash)
, while keeping track of the hash value and checking for collisions - very similar to finding a duplicate in a hdf5 pytable with 500e6 rows. I did not modify the table as I was looping through it but rather wrote the merged entries to a new table - I am putting the code at the bottom.
Basically the problem I have is that the whole process takes significantly too long - it took me about 20 hours to get to iteration #5 4e5. I am working on a HDD however, so it is entirely possible the bottleneck is there. Do you see any way I can improve my code, or can you suggest another approach? Thank you in advance for any help.
P.S. I promise I am not doing anything illegal, it is simply a large scale leaked password analysis for my Bachelor Thesis.
ref = 3 #manually checked first occuring hash, to simplify the below code
gen_cnt = 0
locs = {}
print("STARTING")
for row in table.itersorted('xhashx'):
gen_cnt += 1 #so as not to flush after every iteration
ps = row['password'].decode(encoding = 'utf-8', errors = 'ignore')
if row['xhashx'] == ref:
if ps in locs:
locs[ps][0] += 1
locs[ps][1] |= row['src']
else:
locs[ps] = [1, row['src']]
else:
for p in locs:
fill_password(new_password, locs[ps]) #simply fills in the columns, with some fairly cheap statistics procedures
new_password.append()
if (gen_cnt > 100):
gen_cnt = 0
new_table.flush()
ref = row['xhashx']```
回答1:
Your dataset is 10x larger than the referenced solution (2.5e9 vs 500e6 rows). Have you done any testing to identify where the time is spent? The table.itersorted()
method may not be linear - and might be resource intensive. (I don't have any experience with itersorted.)
Here is a process that might be faster:
- Extract a NumPy array of the hash field (column
xhashx
) - Find the unique hash values
- Loop thru the unique hash values and extract a NumPy array of rows that match each value
- Do your uniqueness tests against the rows in this extracted array
- Write the unique rows to your new file
Code for this process below:
Note: This has been not tested, so may have small syntax or logic gaps
# Step 1: Get a Numpy array of the 'xhashx' field/colmu only:
hash_arr = table.read(field='xhashx')
# Step 2: Get new array with unique values only:
hash_arr_u = np.unique(hash_arr)
# Alternately, combine first 2 steps in a single step
hash_arr_u = np.unique(table.read(field='xhashx'))
# Step 3a: Loop on rows unique hash values
for hash_test in hash_arr_u :
# Step 3b: Get an array with all rows that match this unique hash value
match_row_arr = table.read_where('xhashx==hash_test')
# Step 4: Check for rows with unique values
# Check the hash row count.
# If there is only 1 row, uniqueness tested not required
if match_row_arr.shape[0] == 1 :
# only one row, so write it to new.table
else :
# check for unique rows
# then write unique rows to new.table
##################################################
# np.unique has an option to save the hash counts
# these can be used as a test in the loop
(hash_arr_u, hash_cnts) = np.unique(table.read(field='xhashx'), return_counts=True)
# Loop on rows in the array of unique hash values
for cnt in range(hash_arr_u.shape[0]) :
# Get an array with all rows that match this unique hash value
match_row_arr = table.read_where('xhashx==hash_arr_u(cnt)')
# Check the hash row count.
# If there is only 1 row, uniqueness tested not required
if hash_cnts[cnt] == 1 :
# only one row, so write it to new.table
else :
# check for unique rows
# then write unique rows to new.table
来源:https://stackoverflow.com/questions/60783486/pytables-duplicates-2-5-giga-rows