Python，而不是一个包含4000万个连续数字的列表，我如何制作40个100万个连续数字列表的2-d列表？

Question

I am working on a MapReduce algorithm currently and I need to build my data source a little better. This program is to give a list of nonces to use in a hash algorithm to find "good" (low value) hashes; very similar to bitcoin. Right now I make a single list of 40 million consecutive numbers (nonces). But the overhead in IO (using mincemeat.py) is making the program very slow.

Currently I am using this to create my list

#Build the data source
nonces = [i for i in range(0, 400000)]
#Create a dict with a single entry
datasource = dict(enumerate(nonces))

How could I alter the first line of code to create a list of size 40, containing lists of size 1 million; so the first list would be 1-1mil, then 1mil to 2mil, etc? Do I need to break down and make a for loop, or is there a simple one liner I could implement to achieve this?

Here is how I would implement the for loop to do it, can it be condensed? (I know I have repeating numbers..)

nonceList = []
for j in range(0, 40):
    nonceList.append([i for i in range(j*1000000, (j+1)*1000000)])
datasource = dict(enumerate(nonceList))

Answer 1

Don't produce consecutive numbers up front; just have your mapreduce tasks produce them from a starter number.

Eg for 40 tasks, number those 0-40 and use a multiplier to generate numbers in the task; in Python 2, use xrange() to generate numbers, as range() will produce a list, materializing a million integer objects for no gain.