We Keep Coding

Compare with another column value

train.loc[:,'nd_mean_2021-04-15':'nd_mean_2021-08-27'] > train['q_5']
I get Automatic reindexing on DataFrame vs Series comparisons is deprecated and will raise ValueError in a future version. Do left, right = left.align(right, axis=1, copy=False)before e.g.left == right` and something strange output with a lot of columns, but I did expect cell values masked with True or False for calculate sum on next step.
Comparing each columns separately works just fine
train['nd_mean_2021-04-15'] > train['q_5']
But works slowly and messy code.

I've tested your original solution, and two additional ways of performing this comparison you want to make.
To cut to the chase, the following option had the smallest execution time:
%%timeit
sliced_df = df.loc[:, 'nd_mean_2021-04-15':'nd_mean_2021-08-27']
comparisson_df = pd.DataFrame({col: df['q_5'] for col in sliced_df.columns})
(sliced_df > comparisson_df)
# 1.46 ms ± 610 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Drawback: it's little bit messy and requires you to create 2 new objects (sliced_df and comparisson_df)
Option 2: Using DataFrame.apply (slower but more readable)
The second option although slower than your original and the above implementations, in my opinion is the cleanest and easiest to read of them all. If you're not trying to process large amounts of data (I assume not, since you're using pandas instead of Dask or Spark that are tools more suitable for processing large volumes of data) then it's worth bringing it to the discussion table:
%%timeit
df.loc[:, 'nd_mean_2021-04-15':'nd_mean_2021-08-27'].apply(lambda col: col > df['q_5'])
# 5.66 ms ± 897 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Original Solution
I've also tested the performance of your original implementation and here's what I got:
%%timeit
df.loc[:, 'nd_mean_2021-04-15':'nd_mean_2021-08-27'] > df['q_5']
# 2.02 ms ± 175 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Side-Note: If the FutureWarning message is bothering you, there's always the option to ignore them, adding the following code after your script imports:
import warnings
warnings.filterwarnings('ignore', category=FutureWarning)
DataFrame Used for Testing
All of the above implementations used the same dataframe, that I created using the following code:
import pandas as pd
import numpy as np
columns = list(
map(
lambda value: f'nd_mean_{value}',
pd.date_range('2021-04-15', '2021-08-27', freq='W').to_series().dt.strftime('%Y-%m-%d').to_list()
)
)
df = pd.DataFrame(
{col: np.random.randint(0, 100, 10) for col in [*columns, 'q_5']}
)
Screenshots

What is the most efficient way to bootstrap the mean of a list of numbers?

I have a list of numbers (floats) and I would like to estimate the mean. I also need to estimate the variation of such mean. My goal is to resample the list 100 times, and my output would be an array with length 100, each element corresponding to the mean of a resampled list.
Here is a simple workable example for what I would like to achieve:
import numpy as np
data = np.linspace(0, 4, 5)
ndata, boot = len(data), 100
output = np.mean(np.array([data[k] for k in np.random.uniform(high=ndata, size=boot*ndata).astype(int)]).reshape((boot, ndata)), axis=1)
This is however quite slow when I have to repeat for many lists with large number of elements. The method also seems very clunky and un-Pythonic. What would be a better way to achieve my goal?
P.S. I am aware of scipy.stats.bootstrap, but I have problem upgrading scipy to 1.7.1 in anaconda to import this.

Use np.random.choice:
import numpy as np
data = np.linspace(0, 4, 5)
ndata, boot = len(data), 100
output = np.mean(
np.random.choice(data, size=(100, ndata)),
axis=1)
If I understood correctly, this expression (in your question's code):
np.array([data[k] for k in np.random.uniform(high=ndata, size=boot*ndata).astype(int)]).reshape((boot, ndata)
is doing a sampling with replacement and that is exactly what np.random.choice does.
Here are some timings for reference:
%timeit np.mean(np.array([data[k] for k in np.random.uniform(high=ndata, size=boot*ndata).astype(int)]).reshape((boot, ndata)), axis=1)
133 µs ± 3.96 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit np.mean(np.random.choice(data, size=(boot, ndata)),axis=1)
41.1 µs ± 538 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
As it can be seen np.random.choice yields 3x improvement.

Speeding up operations on large arrays & datasets (Pandas slow, Numpy better, further improvements?)

I have a large dataset comprising millions of rows and around 6 columns. The data is currently in a Pandas dataframe and I'm looking for the fastest way to operate on it. For example, let's say I want to drop all the rows where the value in one column is "1".
Here's my minimal working example:
# Create dummy data arrays and pandas dataframe
array_size = int(5e6)
array1 = np.random.rand(array_size)
array2 = np.random.rand(array_size)
array3 = np.random.rand(array_size)
array_condition = np.random.randint(0, 3, size=array_size)
df = pd.DataFrame({'array_condition': array_condition, 'array1': array1, 'array2': array2, 'array3': array3})
def method1():
df_new = df.drop(df[df.array_condition == 1].index)
EDIT: As Henry Yik pointed out in the comments, a faster Pandas approach is this:
def method1b():
df_new = df[df.array_condition != 1]
I believe that Pandas can be quite slow at this sort of thing, so I also implemented a method using numpy, processing each column as a separate array:
def method2():
masking = array_condition != 1
array1_new = array1[masking]
array2_new = array2[masking]
array3_new = array3[masking]
array_condition_new = array_condition[masking]
And the results:
%timeit method1()
625 ms ± 7.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit methodb()
158 ms ± 7.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit method2()
138 ms ± 3.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
So we do see a slight significant performance boost using numpy. However, this is at the cost of much less readable code (i.e. having to create a mask and apply it to each array). This method doesn't seem as scalable either as if I have, say, 30 columns of data, I'll need a lot of lines of code that apply the mask to every array! Additionally, it would be useful to allow optional columns, so this method may fail trying to operate on arrays which are empty.
Therefore, I have 2 questions:
1) Is there a cleaner / more flexible way to implement this in numpy?
2) Or better, is there any higher performance method I could use here? e.g. JIT (numba?), Cython or something else?
PS, in practice, in-place operations can be used, replacing the old array with the new one once data is dropped

Part 1: Pandas and (maybe) Numpy
Compare your method1b and method2:
method1b generates a DataFrame, which is probably what you want,
method2 generates a Numpy array, so to get fully comparable result,
you should subsequently generate a DataFrame from it.
So I changed your method2 to:
def method2():
masking = array_condition != 1
array1_new = array1[masking]
array2_new = array2[masking]
array3_new = array3[masking]
array_condition_new = array_condition[masking]
df_new = pd.DataFrame({ 'array_condition': array_condition[masking],
'array1': array1_new, 'array2': array2_new, 'array3': array3_new})
and then compared execution times (using %timeit).
The result was that my (expanded) version of method2 executed about 5% longer
than method1b (check on your own).
So my opinion is that as long as a single operation is concerned,
it is probably better to stay with Pandas.
But if you want to perform on your source DataFrame a couple of operations
in sequence and / or you are satisfied with the result as a Numpy array,
it is worth to:
Call arr = df.values to get the underlying Numpy array.
Perform all required operations on it using Numpy methods.
(Optionally) create a DataFrame from the final reslut.
I tried Numpy version of method1b:
def method3():
a = df.values
arr = a[a[:,0] != 1]
but the execution time was about 40 % longer.
The reason is probably that Numpy array has all elements of the
same type, so array_condition column is coerced to float and then
the whole Numpy array is created, what takes some time.
Part 2: Numpy and Numba
An alternative to consider is to use Numba package - a Just-In-Time
Python compiler.
I made such test:
Created a Numpy array (as a preliminary step):
a = df.values
The reason is that JIT compiled methods are able to use Numpy methods and types,
but not those of Pandas.
To perform the test, I used almost the same method as above,
but with #njit annotation (requires from numba import njit):
#njit
def method4():
arr = a[a[:,0] != 1]
This time:
The execution time was about 45 % of the time for method1b.
But since a = df.values has been executed before the test loop,
there are doubts whether this result is comparable with earlier tests.
Anyway, try Numba on your own, maybe it will be an interesting option for you.

You may find using numpy.where useful here. It converts a Boolean mask to array indices, making life much cheaper. Combining this with numpy.vstack allows for some memory-cheap operations:
def method3():
wh = np.where(array_condition == 1)
return np.vstack(tuple(col[wh] for col in (array1, array2, array3)))
This gives the following timeits:
>>> %timeit method2()
180 ms ± 6.66 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit method3()
96.9 ms ± 2.5 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Tuple unpacking allows the operation to be fairly light on memory, as when the object is vstack-ed back together, it is smaller. If you need to get your columns out of a DataFrame directly, the following code snippet may be useful:
def method3b():
wh = np.where(array_condition == 1)
col_names = ['array1','array2','array3']
return np.vstack(tuple(col[wh] for col in tuple(df[col_name].to_numpy()
for col_name in col_names)))
This allows one to grab columns by name from the DataFrame, which are then tuple unpacked on the fly. The speed is about the same:
>>> %timeit method3b()
96.6 ms ± 3.09 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Enjoy!

How to optimize a numpy loop that sums values from an array which is indexed by another array where values equal the loop index

I have this piece of code that is called multiple times during the run of the application.
It takes an array of numbers which represent values (value_array).
These should be summed up in zones, which are defined in the zone_array.
zone_ids represents a list of all the possible zones in zone_array.
Its basically something in the lines of: i got a population raster map and i want to know how many people live in each zone of the zone map.
the code:
values = np.zeros(len(zone_ids))
for i in zone_ids:
values[i] = round(np.nansum(value_array[zone_array == i]), 2)
return values
The culprit seems to be the for loop, but i have not found a way to eliminate it and have the same results.
I tried it with bincount but i did not succeed.
Using numba jit also has no effect.
I would like to stay away from cython as this code will be used in a Qgis plugin which has no cython support.
test code:
import numpy as np
def fill_values(zone_array, value_array, zone_ids):
values = np.zeros(len(zone_ids))
for i in zone_ids:
values[i] = round(np.nansum(value_array[zone_array == i]), 2)
return values
def run():
# 300 different zones
zone_ids = range(300)
# zone map with 300 zones
zone_array = (np.random.rand(2000, 2000) * 300).astype(int)
# value map from which we want the sum of values per zone (real map can have NaN values)
value_array = (np.random.rand(2000, 2000) * 10.)
value_array[5, 5] = np.NAN
fill_values(zone_array, value_array, zone_ids)
if __name__ == '__main__':
run()
1.92 s ± 17.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
With the implementation of bincount as suggested by Divakar :
203 ms ± 15.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

With a direct usage of bincount, you would have NaNs in the summations. So, you can simply replace the NaNs with zeros and use bincount. This should be much faster, being a vectorized solution.
Hence, the implementation would be -
val_nonan = np.where(np.isnan(value_array), 0, value_array)
out = np.round(np.bincount(zone_array.ravel(), val_nonan.ravel()),2)

Pandas DataFrame performance

Pandas is really great, but I am really surprised by how inefficient it is to retrieve values from a Pandas.DataFrame. In the following toy example, even the DataFrame.iloc method is more than 100 times slower than a dictionary.
The question: Is the lesson here just that dictionaries are the better way to look up values? Yes, I get that that is precisely what they were made for. But I just wonder if there is something I am missing about DataFrame lookup performance.
I realize this question is more "musing" than "asking" but I will accept an answer that provides insight or perspective on this. Thanks.
import timeit
setup = '''
import numpy, pandas
df = pandas.DataFrame(numpy.zeros(shape=[10, 10]))
dictionary = df.to_dict()
'''
f = ['value = dictionary[5][5]', 'value = df.loc[5, 5]', 'value = df.iloc[5, 5]']
for func in f:
print func
print min(timeit.Timer(func, setup).repeat(3, 100000))
value = dictionary[5][5]
0.130625009537
value = df.loc[5, 5]
19.4681699276
value = df.iloc[5, 5]
17.2575249672

A dict is to a DataFrame as a bicycle is to a car.
You can pedal 10 feet on a bicycle faster than you can start a car, get it in gear, etc, etc. But if you need to go a mile, the car wins.
For certain small, targeted purposes, a dict may be faster.
And if that is all you need, then use a dict, for sure! But if you need/want the power and luxury of a DataFrame, then a dict is no substitute. It is meaningless to compare speed if the data structure does not first satisfy your needs.
Now for example -- to be more concrete -- a dict is good for accessing columns, but it is not so convenient for accessing rows.
import timeit
setup = '''
import numpy, pandas
df = pandas.DataFrame(numpy.zeros(shape=[10, 1000]))
dictionary = df.to_dict()
'''
# f = ['value = dictionary[5][5]', 'value = df.loc[5, 5]', 'value = df.iloc[5, 5]']
f = ['value = [val[5] for col,val in dictionary.items()]', 'value = df.loc[5]', 'value = df.iloc[5]']
for func in f:
print(func)
print(min(timeit.Timer(func, setup).repeat(3, 100000)))
yields
value = [val[5] for col,val in dictionary.iteritems()]
25.5416321754
value = df.loc[5]
5.68071913719
value = df.iloc[5]
4.56006002426
So the dict of lists is 5 times slower at retrieving rows than df.iloc. The speed deficit becomes greater as the number of columns grows. (The number of columns is like the number of feet in the bicycle analogy. The longer the distance, the more convenient the car becomes...)
This is just one example of when a dict of lists would be less convenient/slower than a DataFrame.
Another example would be when you have a DatetimeIndex for the rows and wish to select all rows between certain dates. With a DataFrame you can use
df.loc['2000-1-1':'2000-3-31']
There is no easy analogue for that if you were to use a dict of lists. And the Python loops you would need to use to select the right rows would again be terribly slow compared to the DataFrame.

It seems the performance difference is much smaller now (0.21.1 -- I forgot what was the version of Pandas in the original example). Not only the performance gap between dictionary access and .loc reduced (from about 335 times to 126 times slower), loc (iloc) is less than two times slower than at (iat) now.
In [1]: import numpy, pandas
...: ...: df = pandas.DataFrame(numpy.zeros(shape=[10, 10]))
...: ...: dictionary = df.to_dict()
...:
In [2]: %timeit value = dictionary[5][5]
85.5 ns ± 0.336 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
In [3]: %timeit value = df.loc[5, 5]
10.8 µs ± 137 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [4]: %timeit value = df.at[5, 5]
6.87 µs ± 64.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [5]: %timeit value = df.iloc[5, 5]
14.9 µs ± 114 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [6]: %timeit value = df.iat[5, 5]
9.89 µs ± 54.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [7]: print(pandas.__version__)
0.21.1
---- Original answer below ----
+1 for using at or iat for scalar operations. Example benchmark:
In [1]: import numpy, pandas
...: df = pandas.DataFrame(numpy.zeros(shape=[10, 10]))
...: dictionary = df.to_dict()
In [2]: %timeit value = dictionary[5][5]
The slowest run took 34.06 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 310 ns per loop
In [4]: %timeit value = df.loc[5, 5]
10000 loops, best of 3: 104 µs per loop
In [5]: %timeit value = df.at[5, 5]
The slowest run took 6.59 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 9.26 µs per loop
In [6]: %timeit value = df.iloc[5, 5]
10000 loops, best of 3: 98.8 µs per loop
In [7]: %timeit value = df.iat[5, 5]
The slowest run took 6.67 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 9.58 µs per loop
It seems using at (iat) is about 10 times faster than loc (iloc).

I encountered the same problem. you can use at to improve.
"Since indexing with [] must handle a lot of cases (single-label access, slicing, boolean indexing, etc.), it has a bit of overhead in order to figure out what you’re asking for. If you only want to access a scalar value, the fastest way is to use the at and iat methods, which are implemented on all of the data structures."
see official reference http://pandas.pydata.org/pandas-docs/stable/indexing.html chapter "Fast scalar value getting and setting"

I experienced different phenomenon about accessing the dataframe row.
test this simple example on dataframe about 10,000,000 rows.
dictionary rocks.
def testRow(go):
go_dict = go.to_dict()
times = 100000
ot= time.time()
for i in range(times):
go.iloc[100,:]
nt = time.time()
print('for iloc {}'.format(nt-ot))
ot= time.time()
for i in range(times):
go.loc[100,2]
nt = time.time()
print('for loc {}'.format(nt-ot))
ot= time.time()
for i in range(times):
[val[100] for col,val in go_dict.iteritems()]
nt = time.time()
print('for dict {}'.format(nt-ot))

I think the fastest way of accessing a cell, is
df.get_value(row,column)
df.set_value(row,column,value)
Both are faster than (I think)
df.iat(...)
df.at(...)