jfive jfive - 1 year ago 164
Python Question

Why does concatenation of DataFrames get exponentially slower?

I have a function which processes a DataFrame, largely to process data into buckets create a binary matrix of features in a particular column using


To avoid processing all of my data using this function at once (which goes out of memory and causes iPython to crash), I have broken the large DataFrame into chunks using:

chunks = (len(df) / 10000) + 1
df_list = np.array_split(df, chunks)

will automatically create new columns based on the contents of
and these are likely to differ for each

After processing, I am concatenating the DataFrames back together using:

for i, df_chunk in enumerate(df_list):
print "chunk", i
[x, y] = preprocess_data(df_chunk)
super_x = pd.concat([super_x, x], axis=0)
super_y = pd.concat([super_y, y], axis=0)
print datetime.datetime.utcnow()

The processing time of the first chunk is perfectly acceptable, however, it grows per chunk! This is not to do with the
as there is no reason for it to increase. Is this increase in time occurring as a result of the call to

Please see log below:

chunks 6
chunk 0
2016-04-08 00:22:17.728849
chunk 1
2016-04-08 00:22:42.387693
chunk 2
2016-04-08 00:23:43.124381
chunk 3
2016-04-08 00:25:30.249369
chunk 4
2016-04-08 00:28:11.922305
chunk 5
2016-04-08 00:32:00.357365

Is there a workaround to speed this up? I have 2900 chunks to process so any help is appreciated!

Open to any other suggestions in Python!

Answer Source

Never call DataFrame.append or pd.concat inside a for-loop. It leads to quadratic copying.

pd.concat returns a new DataFrame. Space has to be allocated for the new DataFrame, and data from the old DataFrames have to be copied into the new DataFrame. Consider the amount of copying required by this line inside the for-loop (assuming each x has size 1):

super_x = pd.concat([super_x, x], axis=0)

| iteration | size of old super_x | size of x | copying required |
|         0 |                   0 |         1 |                1 |
|         1 |                   1 |         1 |                2 |
|         2 |                   2 |         1 |                3 |
|       ... |                     |           |                  |
|       N-1 |                 N-1 |         1 |                N |

1 + 2 + 3 + ... + N = N(N-1)/2. So there is O(N**2) copies required to complete the loop.

Now consider

super_x = []
for i, df_chunk in enumerate(df_list):
    [x, y] = preprocess_data(df_chunk)
super_x = pd.concat(super_x, axis=0)

Appending to a list is an O(1) operation and does not require copying. Now there is a single call to pd.concat after the loop is done. This call to pd.concat requires N copies to be made, since super_x contains N DataFrames of size 1. So when constructed this way, super_x requires O(N) copies.

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