I'm working with a very large data set (about 75 million entries) and I'm trying to shorten the length of time that running my code takes by a significant margin (with a loop right now it will take a couple days) and keep memory usage extremely low.
I have two numpy arrays (clients and units) of the same length. My goal is to get a list of every index that a value occurs in my first list (clients) and then find a sum of the entries in my second list at each of those indices.
This is what I've tried (np is the previously imported numpy library)
# create a list of each value that appears in clients
unq = np.unique(clients)
arr = np.zeros(len(unq))
tmp = np.arange(len(clients))
# for each unique value i in clients
for i in range(len(unq)) :
#create a list inds of all the indices that i occurs in clients
inds = tmp[clients==unq[i]]
# add the sum of all the elements in units at the indices inds to a list
arr[i] = sum(units[inds])
Does anyone know a method that will allow me to find these sums without looping through each element in unq?
With Pandas, this can easily be done using the grouby() function:
import pandas as pd
# some fake data
df = pd.DataFrame({'clients': ['a', 'b', 'a', 'a'], 'units': [1, 1, 1, 1]})
print df.groupby(['clients'], sort=False).sum()
which gives you the desired output:
units
clients
a 3
b 1
I use the sort=False option since that might lead to a speed-up (by default the entries will be sorted which can take some time for huge datsets).
This is a typical group-by type operation, which can be performed elegantly and efficiently using the numpy-indexed package (disclaimer: I am its author):
import numpy_indexed as npi
unique_clients, units_per_client = npi.group_by(clients).sum(units)
Note that unlike the pandas approach, there is no need to create a temporary datastructure just to perform this kind of elementary operation.
Related
I have a dictionary that is filled with multiple dataframes. Now I am searching for an efficient way for changing the key structure, but the solution I have found is rather slow when more dataframes / bigger dataframes are involved. Thats why I wanted to ask if anyone might know a more convenient / efficient / faster way or approach than mine. So first, I created this example to show where I initially started:
import pandas as pd
import numpy as np
# assign keys to dic
teams = ["Arsenal", "Chelsea", "Manchester United"]
dic_teams = {}
# fill dic with random entries
for t1 in teams:
dic_teams[t1] = pd.DataFrame({'date': pd.date_range("20180101", periods=30),
'Goals': pd.Series(np.random.randint(0,5, size = 30)),
'Chances': pd.Series(np.random.randint(0,15, size = 30)),
'Fouls': pd.Series(np.random.randint(0, 20, size = 30)),
'Offside': pd.Series(np.random.randint(0, 10, size = 30))})
dic_teams[t1] = dic_teams[t1].set_index('date')
dic_teams[t1].index.name = None
Now I basically have a dictionary where every key is a team, which means I have a dataframe for every team with information on their game performance over time. Now I would prefer to change this particular dictionary so I get a structure where the key is the date, instead of a team. This would mean that I have a dataframe for every date, which is filled with the performance of each team on that date. I managed to do that using the following code, which works but is really slow once I add more teams and performance factors:
# prepare lists for looping
dates = dic_teams["Arsenal"].index.to_list()
perf = dic_teams["Arsenal"].columns.to_list()
dic_dates = {}
# new structure where key = date
for d in dates:
dic_dates[d] = pd.DataFrame(index = teams, columns = perf)
for t2 in teams:
dic_dates[d].loc[t2] = dic_teams[t2].loc[d]
Because I am using a nested loop, the restructuring of my dictionary is slow. Does anyone have an idea how I could improve the second piece of code? I'm not necessarily searching just for a solution, also for a logic or idea how to do better.
Thanks in advance, any help is highly appreciated
Creating a Pandas dataframes the way you do is (strangely) awfully slow, as well as direct indexing.
Copying a dataframe is surprisingly quite fast. Thus you can use an empty reference dataframe copied multiple times. Here is the code:
dates = dic_teams["Arsenal"].index.to_list()
perf = dic_teams["Arsenal"].columns.to_list()
zygote = pd.DataFrame(index = teams, columns = perf)
dic_dates = {}
# new structure where key = date
for d in dates:
dic_dates[d] = zygote.copy()
for t2 in teams:
dic_dates[d].loc[t2] = dic_teams[t2].loc[d]
This is about 2 times faster than the reference on my machine.
Overcoming the slow dataframe direct indexing is tricky. We can use numpy to do that. Indeed, we can convert the dataframe to a 3D numpy array, use numpy to perform the transposition, and finally convert the slices into dataframes again. Note that this approach assumes that all values are integers and that the input dataframe are well structured.
Here is the final implementation:
dates = dic_teams["Arsenal"].index.to_list()
perf = dic_teams["Arsenal"].columns.to_list()
dic_dates = {}
# Create a numpy array from Pandas dataframes
# Assume the order of the `dates` and `perf` indices are the same in all dataframe (and their order)
full = np.empty(shape=(len(teams), len(dates), len(perf)), dtype=int)
for tId,tName in enumerate(teams):
full[tId,:,:] = dic_teams[tName].to_numpy()
# New structure where key = date, created from the numpy array
for dId,dName in enumerate(dates):
dic_dates[dName] = pd.DataFrame({pName: full[:,dId,pId] for pId,pName in enumerate(perf)}, index = teams)
This implementation is 6.4 times faster than the reference on my machine. Note that about 75% of the time is sadly spent in the pd.DataFrame calls. Thus, if you want a faster code, use a basic 3D numpy array!
I have a 10 GB csv file with 170,000,000 rows and 23 columns that I read in to a dataframe as follows:
import pandas as pd
d = pd.read_csv(f, dtype = {'tax_id': str})
I also have a list of strings with nearly 20,000 unique elements:
h = ['1123787', '3345634442', '2342345234', .... ]
I want to create a new column called class in the dataframe d. I want to assign d['class'] = 'A' whenever d['tax_id'] has a value that is found in the list of strings h. Otherwise, I want d['class'] = 'B'.
The following code works very quickly on a 1% sample of my dataframe d:
d['class'] = 'B'
d.loc[d['tax_num'].isin(h), 'class'] = 'A'
However, on the complete dataframe d, this code takes over 48 hours (and counting) to run on a 32 core server in batch mode. I suspect that indexing with loc is slowing down the code, but I'm not sure what it could really be.
In sum: Is there a more efficient way of creating the class column?
If your tax numbers are unique, I would recommend setting tax_num to the index and then indexing on that. As it stands, you call isin which is a linear operation. However fast your machine is, it can't do a linear search on 170 million records in a reasonable amount of time.
df.set_index('tax_num', inplace=True) # df = df.set_index('tax_num')
df['class'] = 'B'
df.loc[h, 'class'] = 'A'
If you're still suffering from performance issues, I'd recommend switching to distributed processing with dask.
"I also have a list of strings with nearly 20,000 unique elements"
Well, for starters, you should make that list a set if you are going to be using it for membership testing. list objects have linear time membership testing, set objects have very optimized constant-time performance for membership testing. That is the lowest hanging fruit here. So use
h = set(h) # convert list to set
d['class'] = 'B'
d.loc[d['tax_num'].isin(h), 'class'] = 'A'
I have a dataframe of ~20M lines
I have a column called A that gives me an id (there are ~10K ids in total).
The value of this id defines a random distribution's parameters.
Now I want to generate a column B, that is randomly drawn from the distribution that is defined by the value in the column A
What is the fastest way to do this? Doing something with iterrows or apply is extremely slow. Another possiblity is to group by A, and generate all my data for each value of A (so I only draw from one distribution). But then I don't end up with a Dataframe but with a "groupBy" object, and I don't know how to go back to having the initial dataframe, plus my new column.
I think this approach is similar to what you were describing, where you generate the samples for each id. On my machine, it appears this would take around 5 minutes to run. I assume you can trivially get the ids.
import numpy as np
num_ids = 10000
num_rows = 20000000
ids = np.arange(num_ids)
loc_params = np.random.random(num_ids)
A = np.random.randint(0, num_ids, num_rows)
B = np.zeros(A.shape)
for idx in ids:
A_idxs = A == idx
B[A_idxs] = np.random.normal(np.sum(A_idxs), loc_params[idx])
This question is pretty vague, but how would this work for you?
df['B'] = df.apply(lambda row: distribution(row.A), axis=1)
Editing from question edits (apply is too slow):
You could create a mapping dictionary for the 10k ids to their generated value, then do something like
df['B'] = df['A'].map(dictionary)
I'm unsure if this will be faster than apply, but it will require fewer calls to your random distribution generator
I have a dataframe of values:
df = pd.DataFrame(np.random.uniform(0,1,(500,2)), columns = ['a', 'b'])
>>> print df
a b
1 0.277438 0.042671
.. ... ...
499 0.570952 0.865869
[500 rows x 2 columns]
I want to transform this by replacing the values with their percentile, where the percentile is taken over the distribution of all values in prior rows. i.e., if you do df.T.unstack(), it would be a pure expanding sample. This might be more intuitive if you think of the index as a DatetimeIndex, and I'm asking to take the expanding percentile over the entire cross-sectional history.
So the goal is this guy:
a b
0 99 99
.. .. ..
499 58 84
(Ideally I'd like to take the distribution of a value over the set of all values in all rows before and including that row, so not exactly an expanding percentile; but if we can't get that, that's fine.)
I have one really ugly way of doing this, where I transpose and unstack the dataframe, generate a percentile mask, and overlay that mask on the dataframe using a for loop to get the percentiles:
percentile_boundaries_over_time = pd.DataFrame({integer:
pd.expanding_quantile(df.T.unstack(), integer/100.0)
for integer in range(0,101,1)})
percentile_mask = pd.Series(index = df.unstack().unstack().unstack().index)
for integer in range(0,100,1):
percentile_mask[(df.unstack().unstack().unstack() >= percentile_boundaries_over_time[integer]) &
(df.unstack().unstack().unstack() <= percentile_boundaries_over_time[integer+1])] = integer
I've been trying to get something faster to work, using scipy.stats.percentileofscore() and pd.expanding_apply(), but it's not giving the correct output and I'm driving myself insane trying to figure out why. This is what I've been playing with:
perc = pd.expanding_apply(df, lambda x: stats.percentileofscore(x, x[-1], kind='weak'))
Does anyone have any thoughts on why this gives incorrect output? Or a faster way to do this whole exercise? Any and all help much appreciated!
As several other commenters have pointed out, computing percentiles for each row likely involves sorting the data each time. This will probably be the case for any current pre-packaged solution, including pd.DataFrame.rank or scipy.stats.percentileofscore. Repeatedly sorting is wasteful and computationally intensive, so we want a solution that minimizes that.
Taking a step back, finding the inverse-quantile of a value relative to an existing data set is analagous to finding the position we would insert that value into the data set if it were sorted. The issue is that we also have an expanding set of data. Thankfully, some sorting algorithms are extremely fast with dealing with mostly sorted data (and inserting a small number of unsorted elements). Hence our strategy is to maintain our own array of sorted data, and with each row iteration, add it to our existing list and query their positions in the newly expanded sorted set. The latter operation is also fast given that the data is sorted.
I think insertion sort would be the fastest sort for this, but its performance will probably be slower in Python than any native NumPy sort. Merge sort seems to be the best of the available options in NumPy. An ideal solution would involve writing some Cython, but using our above strategy with NumPy gets us most of the way.
This is a hand-rolled solution:
def quantiles_by_row(df):
""" Reconstruct a DataFrame of expanding quantiles by row """
# Construct skeleton of DataFrame what we'll fill with quantile values
quantile_df = pd.DataFrame(np.NaN, index=df.index, columns=df.columns)
# Pre-allocate numpy array. We only want to keep the non-NaN values from our DataFrame
num_valid = np.sum(~np.isnan(df.values))
sorted_array = np.empty(num_valid)
# We want to maintain that sorted_array[:length] has data and is sorted
length = 0
# Iterates over ndarray rows
for i, row_array in enumerate(df.values):
# Extract non-NaN numpy array from row
row_is_nan = np.isnan(row_array)
add_array = row_array[~row_is_nan]
# Add new data to our sorted_array and sort.
new_length = length + len(add_array)
sorted_array[length:new_length] = add_array
length = new_length
sorted_array[:length].sort(kind="mergesort")
# Query the relative positions, divide by length to get quantiles
quantile_row = np.searchsorted(sorted_array[:length], add_array, side="left").astype(np.float) / length
# Insert values into quantile_df
quantile_df.iloc[i][~row_is_nan] = quantile_row
return quantile_df
Based on the data that bhalperin provided (offline), this solution is up to 10x faster.
One final comment: np.searchsorted has options for 'left' and 'right' which determines whether you want your prospective inserted position to be the first or last suitable position possible. This matters if you have a lot of duplicates in your data. A more accurate version of the above solution will take the average of 'left' and 'right':
# Query the relative positions, divide to get quantiles
left_rank_row = np.searchsorted(sorted_array[:length], add_array, side="left")
right_rank_row = np.searchsorted(sorted_array[:length], add_array, side="right")
quantile_row = (left_rank_row + right_rank_row).astype(np.float) / (length * 2)
Yet not quite clear, but do you want a cumulative sum divided by total?
norm = 100.0/df.a.sum()
df['cum_a'] = df.a.cumsum()
df['cum_a'] = df.cum_a * norm
ditto for b
Here's an attempt to implement your 'percentile over the set of all values in all rows before and including that row' requirement. stats.percentileofscore seems to act up when given 2D data, so squeezeing seems to help in getting correct results:
a_percentile = pd.Series(np.nan, index=df.index)
b_percentile = pd.Series(np.nan, index=df.index)
for current_index in df.index:
preceding_rows = df.loc[:current_index, :]
# Combine values from all columns into a single 1D array
# * 2 should be * N if you have N columns
combined = preceding_rows.values.reshape((1, len(preceding_rows) *2)).squeeze()
a_percentile[current_index] = stats.percentileofscore(
combined,
df.loc[current_index, 'a'],
kind='weak'
)
b_percentile[current_index] = stats.percentileofscore(
combined,
df.loc[current_index, 'b'],
kind='weak'
)
This operation needs to be applied as fast as possible as the actual arrays which contain millions of elements. This is a simple version of the problem.
So, I have a random array of unique integers (normally millions of elements).
totalIDs = [5,4,3,1,2,9,7,6,8 ...]
I have another array (normally a tens of thousands) of unique integers which I can create a mask.
subsampleIDs1 = [5,1,9]
subsampleIDs2 = [3,7,8]
subsampleIDs3 = [2,6,9]
...
I can use numpy to do
mask = np.in1d(totalIDs,subsampleIDs,assume_unique=True)
I can then extract the information I want of another array using the mask (say column 0 contains the one I want).
variable = allvariables[mask][:,0]
Now given that the IDs are unique in both arrays, is there any way to speed this up significantly. It takes a long time to construct the mask for a few thousand points (subsampleIDs) matching against millions of IDs (totalIDs).
I thought of going through it once and writing out a binary file of an index (to speed up future searches).
for i in range(0,3):
mask = np.in1d(totalIDs,subsampleIDs,assume_unique=True)
index[mask] = i
where X is in subsampleIDsX. Then I can just do:
for i in range(0,3):
if index[i] == i:
rowmatch = i
break
variable = allvariables[rowmatch:len(subsampleIDs),0]
right? But this is also slow because there is a conditional in the loop to find when it first matches. Is there a faster way to find when a number first appears in an ordered array so the conditional doesn't slow the loop?
I suggest you use DataFrame in Pandas. the index of the DataFrame is the totalIDs, and you can select subsampleIDs by: df.ix[subsampleIDs].
Create some test data first:
import numpy as np
N = 2000000
M = 5000
totalIDs = np.random.randint(0, 10000000, N)
totalIDs = np.unique(totalIDs)
np.random.shuffle(totalIDs)
v1 = np.random.rand(len(totalIDs))
v2 = np.random.rand(len(totalIDs))
subsampleIDs = np.random.choice(totalIDs, M)
subsampleIDs = np.unique(subsampleIDs)
np.random.shuffle(subsampleIDs)
Then convert you data in to a DataFrame:
import pandas as pd
df = pd.DataFrame(data = {"v1":v1, "v2":v2}, index=totalIDs)
df.ix[subsampleIDs]
DataFrame use a hashtable to map the index to it's location, it's very fast.
Often this kind of indexing is best performed using a DB (with proper column-indexing).
Another idea is to sort totalIDs once, as a preprocessing stage, and implement your own version of in1d, which avoids sorting everything. The numpy implementation of in1d (at least in the version that I have installed) is fairly simple, and should be easy to copy and modify.
EDIT:
Or, even better, use bucket sort (or radix sort). That should give you O(N+M), N being the size of totalIDs, and M the size of sampleIDs (times a constant you can play with by changing the number of buckets). Here too, you can split totalIDs to buckets only once, which gives you a nifty O(N+M1+M2+...).
Unfortunately, I'm not aware of a numpy implementation, but I did find this: http://en.wikipedia.org/wiki/Radix_sort#Example_in_Python