So lets say I have a DataFrame in pandas with a m rows and n columns. Let's also say that I wanted to reverse the order of the columns, which can be done with the following code:
df_reversed = df[df.columns[::-1]]
What is the Big O complexity of this operation? I'm assuming this would depend on the number of columns, but would it also depend on the number of rows?
I don't know how Pandas implements this, but I did test it empirically. I ran the following code (in a Jupyter notebook) to test the speed of the operation:
def get_dummy_df(n):
return pd.DataFrame({'a': [1,2]*n, 'b': [4,5]*n, 'c': [7,8]*n})
df = get_dummy_df(100)
print df.shape
%timeit df_r = df[df.columns[::-1]]
df = get_dummy_df(1000)
print df.shape
%timeit df_r = df[df.columns[::-1]]
df = get_dummy_df(10000)
print df.shape
%timeit df_r = df[df.columns[::-1]]
df = get_dummy_df(100000)
print df.shape
%timeit df_r = df[df.columns[::-1]]
df = get_dummy_df(1000000)
print df.shape
%timeit df_r = df[df.columns[::-1]]
df = get_dummy_df(10000000)
print df.shape
%timeit df_r = df[df.columns[::-1]]
The output was:
(200, 3)
1000 loops, best of 3: 419 µs per loop
(2000, 3)
1000 loops, best of 3: 425 µs per loop
(20000, 3)
1000 loops, best of 3: 498 µs per loop
(200000, 3)
100 loops, best of 3: 2.66 ms per loop
(2000000, 3)
10 loops, best of 3: 25.2 ms per loop
(20000000, 3)
1 loop, best of 3: 207 ms per loop
As you can see, in the first 3 cases, the overhead of the operation is what takes most of the time (400-500µs), but from the 4th case, the time it takes starts to be proportional to the amount of data, increasing in an order of magnitude each time.
So, assuming there must also be a proportion to n, it seems that we are dealing with O(m*n)
The Big O complexity (as of Pandas 0.24) is m*n where m is the number of columns and n is the number of rows. Note, this is when using the DataFrame.__getitem__ method (aka []) with an Index (see relevant code, with other types that would trigger a copy).
Here is a helpful stack trace:
<ipython-input-4-3162cae03863>(2)<module>()
1 columns = df.columns[::-1]
----> 2 df_reversed = df[columns]
pandas/core/frame.py(2682)__getitem__()
2681 # either boolean or fancy integer index
-> 2682 return self._getitem_array(key)
2683 elif isinstance(key, DataFrame):
pandas/core/frame.py(2727)_getitem_array()
2726 indexer = self.loc._convert_to_indexer(key, axis=1)
-> 2727 return self._take(indexer, axis=1)
2728
pandas/core/generic.py(2789)_take()
2788 axis=self._get_block_manager_axis(axis),
-> 2789 verify=True)
2790 result = self._constructor(new_data).__finalize__(self)
pandas/core/internals.py(4539)take()
4538 return self.reindex_indexer(new_axis=new_labels, indexer=indexer,
-> 4539 axis=axis, allow_dups=True)
4540
pandas/core/internals.py(4421)reindex_indexer()
4420 new_blocks = self._slice_take_blocks_ax0(indexer,
-> 4421 fill_tuple=(fill_value,))
4422 else:
pandas/core/internals.py(1254)take_nd()
1253 new_values = algos.take_nd(values, indexer, axis=axis,
-> 1254 allow_fill=False)
1255 else:
> pandas/core/algorithms.py(1658)take_nd()
1657 import ipdb; ipdb.set_trace()
-> 1658 func = _get_take_nd_function(arr.ndim, arr.dtype, out.dtype, axis=axis,
1659 mask_info=mask_info)
1660 func(arr, indexer, out, fill_value)
The func call on L1660 in pandas/core/algorithms ultimately calls a cython function with O(m * n) complexity. This is where data from the the original data is copied into out. out contains a copy of the original data in reversed order.
inner_take_2d_axis0_template = """\
cdef:
Py_ssize_t i, j, k, n, idx
%(c_type_out)s fv
n = len(indexer)
k = values.shape[1]
fv = fill_value
IF %(can_copy)s:
cdef:
%(c_type_out)s *v
%(c_type_out)s *o
#GH3130
if (values.strides[1] == out.strides[1] and
values.strides[1] == sizeof(%(c_type_out)s) and
sizeof(%(c_type_out)s) * n >= 256):
for i from 0 <= i < n:
idx = indexer[i]
if idx == -1:
for j from 0 <= j < k:
out[i, j] = fv
else:
v = &values[idx, 0]
o = &out[i, 0]
memmove(o, v, <size_t>(sizeof(%(c_type_out)s) * k))
return
for i from 0 <= i < n:
idx = indexer[i]
if idx == -1:
for j from 0 <= j < k:
out[i, j] = fv
else:
for j from 0 <= j < k:
out[i, j] = %(preval)svalues[idx, j]%(postval)s
"""
Note that in the above template function, there is a path that uses memmove (which is the path taken in this case because we are mapping from int64 to int64 and the dimension of the output is identical as we are just switching the indexes). Note that memmove is still O(n), being proportional to the number of bytes it has to copy, although likely faster than writing to the indexes directly.
I ran an empirical test using big_O fitting library here
Note: All tests have been conducted on independent variable sweeping 6 orders of magnitude (i.e.
rowsfrom10to10^6vs. constantcolumnsize of3,columnsfrom10to10^6vs. constantrowsize of10
The result shows that the columns reverse operation .columns[::-1] complexity in the DataFrame is
O(n^3) where n is the number of rowsO(n^3) where n is the number of columnsPrerequisites: You will need to install
big_o()using terminal commandpip install big_o
Code
import big_o
import pandas as pd
import numpy as np
SWEAP_LOG10 = 6
COLUMNS = 3
ROWS = 10
def build_df(rows, columns):
# To isolated the creation of the DataFrame from the inversion operation.
narray = np.zeros(rows*columns).reshape(rows, columns)
df = pd.DataFrame(narray)
return df
def flip_columns(df):
return df[df.columns[::-1]]
def get_row_df(n, m=COLUMNS):
return build_df(1*10**n, m)
def get_column_df(n, m=ROWS):
return build_df(m, 1*10**n)
# infer the big_o on columns[::-1] operation vs. rows
best, others = big_o.big_o(flip_columns, get_row_df, min_n=1, max_n=SWEAP_LOG10,n_measures=SWEAP_LOG10, n_repeats=10)
# print results
print('Measuring .columns[::-1] complexity against rapid increase in # rows')
print('-'*80 + '\nBig O() fits: {}\n'.format(best) + '-'*80)
for class_, residual in others.items():
print('{:<60s} (res: {:.2G})'.format(str(class_), residual))
print('-'*80)
# infer the big_o on columns[::-1] operation vs. columns
best, others = big_o.big_o(flip_columns, get_column_df, min_n=1, max_n=SWEAP_LOG10,n_measures=SWEAP_LOG10, n_repeats=10)
# print results
print()
print('Measuring .columns[::-1] complexity against rapid increase in # columns')
print('-'*80 + '\nBig O() fits: {}\n'.format(best) + '-'*80)
for class_, residual in others.items():
print('{:<60s} (res: {:.2G})'.format(str(class_), residual))
print('-'*80)
Results
Measuring .columns[::-1] complexity against rapid increase in # rows
--------------------------------------------------------------------------------
Big O() fits: Cubic: time = -0.017 + 0.00067*n^3
--------------------------------------------------------------------------------
Constant: time = 0.032 (res: 0.021)
Linear: time = -0.051 + 0.024*n (res: 0.011)
Quadratic: time = -0.026 + 0.0038*n^2 (res: 0.0077)
Cubic: time = -0.017 + 0.00067*n^3 (res: 0.0052)
Polynomial: time = -6.3 * x^1.5 (res: 6)
Logarithmic: time = -0.026 + 0.053*log(n) (res: 0.015)
Linearithmic: time = -0.024 + 0.012*n*log(n) (res: 0.0094)
Exponential: time = -7 * 0.66^n (res: 3.6)
--------------------------------------------------------------------------------
Measuring .columns[::-1] complexity against rapid increase in # columns
--------------------------------------------------------------------------------
Big O() fits: Cubic: time = -0.28 + 0.009*n^3
--------------------------------------------------------------------------------
Constant: time = 0.38 (res: 3.9)
Linear: time = -0.73 + 0.32*n (res: 2.1)
Quadratic: time = -0.4 + 0.052*n^2 (res: 1.5)
Cubic: time = -0.28 + 0.009*n^3 (res: 1.1)
Polynomial: time = -6 * x^2.2 (res: 16)
Logarithmic: time = -0.39 + 0.71*log(n) (res: 2.8)
Linearithmic: time = -0.38 + 0.16*n*log(n) (res: 1.8)
Exponential: time = -7 * 1^n (res: 9.7)
--------------------------------------------------------------------------------
