Pandas .loc taking a very long time - python

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'

Related

What is quicker, (=.at), (=.loc), (.drop), or (.append) to filter a large dataframe?

I want to sort through a Dataframe of about 400k rows, with 4 columns, taking out roughly half of them with an if statement:
for a in range (0, howmanytimestorunthrough):
if ('Primary' not in DataFrameexample[a]):
#take out row
So far I've been testing either one of the 4 below:
newdf.append(emptyline,)
nefdf.at[b,'column1'] = DataFrameexample.at[a,'column1']
nefdf.at[b,'column2'] = DataFrameexample.at[a,'column2']
nefdf.at[b,'column3'] = DataFrameexample.at[a,'column3']
nefdf.at[b,'column4'] = DataFrameexample.at[a,'column4']
b = b + 1
or the same with .loc
newdf.append(emptyline,)
nefdf.loc[b,:] = DataFrameexample.loc[a,:]
b = b + 1
or changing the if (not in) to an if (in) and using:
DataFrameexample = DataFrameexample.drop([k])
or trying to set emptyline to have values, and then append it:
notemptyline = pd.Series(DataFrameexample.loc[a,:].values, index = ['column1', 'column2', ...)
newdf.append(notemptyline, ignore_index=True)
So from what I've managed to test so far, they all seem to work ok on a small number of rows (2000), but once I start getting a lot more rows they take exponentially longer. .at seems slighly faster than .loc even if I need it to run 4 times, but still gets slow (10 times the rows, takes longer than 10 times). .drop I think tries to copy the dataframe each time, so really doesn't work? I can't seem to get .append(notemptyline) to work properly, it just replaces index 0 over and over again.
I know there must be an efficient way of doing this, I just can't seem to quite get there. Any help?
Your speed problem has nothing to do with .loc vs .at vs ... (for a comparisson between .loc and .at look have a look at this question) but comes from explicitly looping over every row of your dataframe. Pandas is all about vectorising your operations.
You want to filter your dataframe based on a comparison. You can transform that to a boolean indexer.
indexer = df!='Primary'
This will give you a 4 by n rows dataframe with boolean values. Now you want to reduce the dimension to 1 x n rows such that the value is true if all values in the row (axis 1) are true.
indexer = indexer.all(axis=1)
Now we can use .loc to to get only the rows were indexer is True
df = df.loc[indexer]
This will be much faster then iterating over the rows.
EDIT:
To check if the df entry contains a string you can replace the first row:
indexer = df.apply(lambda x: x.str.contains('Primary'))
Note that you normally don't want to use an apply statement (internally it uses a for loop for custom functions) to iterate over a lot of elements. In this case we are looping over the columns which is fine if you just have a couple of those.

fastest way to generate column with random elements based on another column

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

All indices of each unique element in a list python

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.

Pandas - expanding inverse quantile function

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'
)

Should pandas dataframes be nested?

I am creating a python script that drives an old fortran code to locate earthquakes. I want to vary the input parameters to the fortran code in the python script and record the results, as well as the values that produced them, in a dataframe. The results from each run are also convenient to put in a dataframe, leading me to a situation where I have a nested dataframe (IE a dataframe assigned to an element of a data frame). So for example:
import pandas as pd
import numpy as np
def some_operation(row):
results = np.random.rand(50, 3) * row['p1'] / row['p2']
res = pd.DataFrame(results, columns=['foo', 'bar', 'rms'])
return res
# Init master df
df_master = pd.DataFrame(columns=['p1', 'p2', 'results'], index=range(3))
df_master['p1'] = np.random.rand(len(df_master))
df_master['p2'] = np.random.rand(len(df_master))
df_master = df_master.astype(object) # make sure generic types can be used
# loop over each row, call some_operation and store results DataFrame
for ind, row in df_master.iterrows():
df_master.loc[ind, "results"] = some_operation(row)
Which raises this exception:
ValueError: Incompatible indexer with DataFrame
It works as expected, however, if I change the last line to this:
df_master["results"][ind] = some_operation(row)
I have a few questions:
Why does .loc (and .ix) fail when the slice assignment succeeds? If the some_operation function returned a list, dictionary, etc., it seems to work fine.
Should the DataFrame be used in this way? I know that dtype object can be ultra slow for sorting and whatnot, but I am really just using the dataframe a convenient container because the column/index notation is quite slick. If DataFrames should not be used in this way is there similar alternative? I was looking at the Panel class but I am not sure if it is the proper solution for my application. I would hate forge ahead and apply the hack shown above to some code and then have it not supported in future releases of pandas.
Why does .loc (and .ix) fail when the slice assignment succeeds? If the some_operation function returned a list, dictionary, etc. it seems to work fine.
This is a strange little corner case of the code. It stems from the fact that if the item being assigned is a DataFrame, loc and ix assume that you want to fill the given indices with the content of the DataFrame. For example:
>>> df1 = pd.DataFrame({'a':[1, 2, 3], 'b':[4, 5, 6]})
>>> df2 = pd.DataFrame({'a':[100], 'b':[200]})
>>> df1.loc[[0], ['a', 'b']] = df2
>>> df1
a b
0 100 200
1 2 5
2 3 6
If this syntax also allowed storing a DataFrame as an object, it's not hard to imagine a situation where the user's intent would be ambiguous, and ambiguity does not make a good API.
Should the DataFrame be used in this way?
As long as you know the performance drawbacks of the method (and it sounds like you do) I think this is a perfectly suitable way to use a DataFrame. For example, I've seen a similar strategy used to store the trained scikit-learn estimators in cross-validation across a large grid of parameters (though I can't recall the exact context of this at the moment...)

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