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How to select list elements based on crteria from other lists

I am new to Python, coming from SciLab (an open source MatLab ersatz), which I am using as a toolbox for my analyses (test data analysis, reliability, acoustics, ...); I am definitely not a computer science lad.
I have data in the form of lists of same length (vectors of same size in SciLab).
I use some of them as parameter in order to select data from another one; e.g.
t_v = [1:10]; // a parameter vector
p_v = [20:29]; another parameter vector
res_v(t_v > 5 & p_v < 28); // are the res_v vector elements of which "corresponding" p_v and t_v values comply with my criteria; i can use it for analyses.
This is very direct and simple in SciLab; I did not find the way to achieve the same with Python, either "Pythonically" or simply translated.
Any idea that could help me, please?
Have a nice day,
Patrick.

You could use numpy arrays. It's easy:
import numpy as np
par1 = np.array([1,1,5,5,5,1,1])
par2 = np.array([-1,1,1,-1,1,1,1])
data = np.array([1,2,3,4,5,6,7])
print(par1)
print(par2)
print(data)
bool_filter = (par1[:]>1) & (par2[:]<0)
# example to do it directly in the array
filtered_data = data[ par1[:]>1 ]
print( filtered_data )
#filtering with the two parameters
filtered_data_twice = data[ bool_filter==True ]
print( filtered_data_twice )
output:
[1 1 5 5 5 1 1]
[-1 1 1 -1 1 1 1]
[1 2 3 4 5 6 7]
[3 4 5]
[4]
Note that it does not keep the same number of elements.

Here's my modified solution according to your last comment.
t_v = list(range(1,10))
p_v = list(range(20,29))
res_v = list(range(30,39))
def first_idex_greater_than(search_number, lst):
for count, number in enumerate(lst):
if number > search_number:
return count
def first_idex_lower_than(search_number, lst):
for count, number in enumerate(lst[::-1]):
if number < search_number:
return len(lst) - count # since I searched lst from top to bottom,
# I need to also reverse count
t_v_index = first_idex_greater_than(5, t_v)
p_v_index = first_idex_lower_than(28, p_v)
print(res_v[min(t_v_index, p_v_index):max(t_v_index, p_v_index)])
It returns an array [35, 36, 37].
I'm sure you can optimize it better according to your needs.

The problem statement is not clearly defined, but this is what I interpret to be a likely solution.
import pandas as pd
tv = list(range(1, 11))
pv = list(range(20, 30))
res = list(range(30, 40))
df = pd.DataFrame({'tv': tv, 'pv': pv, 'res': res})
print(df)
def criteria(row, col1, a, col2, b):
if (row[col1] > a) & (row[col2] < b):
return True
else:
return False
df['select'] = df.apply(lambda row: criteria(row, 'tv', 5, 'pv', 28), axis=1)
selected_res = df.loc[df['select']]['res'].tolist()
print(selected_res)
# ... or another way ..
print(df.loc[(df.tv > 5) & (df.pv < 28)]['res'])
This produces a dataframe where each column is the original lists, and applies a selection criteria, based on columns tv and pv to identify the rows in which the criteria, applied dependently to the 2 lists, is satisfied (or not), and then creates a new column of booleans identifying the rows where the criteria is either True or False.
[35, 36, 37]
5 35
6 36
7 37

Iterate the code in a shortest way for the whole dataset

I have very big df:
df.shape() = (106, 3364)
I want to calculate so called frechet distance by using this Frechet Distance between 2 curves. And it works good. Example:
x = df['1']
x1 = df['1.1']
p = np.array([x, x1])
y = df['2']
y1 = df['2.1']
q = np.array([y, y1])
P_final = list(zip(p[0], p[1]))
Q_final = list(zip(q[0], q[1]))
from frechetdist import frdist
frdist(P_final,Q_final)
But I can not do row by row like:
`1 and 1.1` to `1 and 1.1` which is equal to 0
`1 and 1.1` to `2 and 2.1` which is equal to some number
...
`1 and 1.1` to `1682 and 1682.1` which is equal to some number
I want to create something (first idea is for loop, but maybe you have better solution) to calculate this frdist(P_final,Q_final) between:
first rows to all rows (including itself)
second row to all rows (including itself)
Finally, I supposed to get a matrix size (106,106) with 0 on diagonal (because distance between itself is 0)
matrix =
0 1 2 3 4 5 ... 105
0 0
1 0
2 0
3 0
4 0
5 0
... 0
105 0
Not including my trial code because it is confusing everyone!
EDITED:
Sample data:
1 1.1 2 2.1 3 3.1 4 4.1 5 5.1
0 43.1024 6.7498 45.1027 5.7500 45.1072 3.7568 45.1076 8.7563 42.1076 8.7563
1 46.0595 1.6829 45.0595 9.6829 45.0564 4.6820 45.0533 8.6796 42.0501 3.6775
2 25.0695 5.5454 44.9727 8.6660 41.9726 2.6666 84.9566 3.8484 44.9566 1.8484
3 35.0281 7.7525 45.0322 3.7465 14.0369 3.7463 62.0386 7.7549 65.0422 7.7599
4 35.0292 7.5616 45.0292 4.5616 23.0292 3.5616 45.0292 7.5616 25.0293 7.5613

I just used own sample data in your format (I hope)
import pandas as pd
from frechetdist import frdist
import numpy as np
# create sample data
df = pd.DataFrame([[1,2,3,4,5,6], [3,4,5,6,8,9], [2,3,4,5,2,2], [3,4,5,6,7,3]], columns=['1','1.1','2', '2.1', '3', '3.1'])
# this matrix will hold the result
res = np.ndarray(shape=(df.shape[1] // 2, df.shape[1] // 2), dtype=np.float32)
for row in range(res.shape[0]):
for col in range(row, res.shape[1]):
# extract the two functions
P = [*zip([df.loc[:, f'{row+1}'], df.loc[:, f'{row+1}.1']])]
Q = [*zip([df.loc[:, f'{col+1}'], df.loc[:, f'{col+1}.1']])]
# calculate distance
dist = frdist(P, Q)
# put result back (its symmetric)
res[row, col] = dist
res[col, row] = dist
# output
print(res)
Output:
[[0. 4. 7.5498343]
[4. 0. 5.5677643]
[7.5498343 5.5677643 0. ]]
Hope that helps
EDIT: Some general tips:
If speed matters: check if frdist handles also a numpy array of shape
(n_values, 2) than you could save the rather expensive zip-and-unpack operation
and directly use the arrays or build the data directly in a format the your library needs
Generally, use better column namings (3 and 3.1 is not too obvious). Why you dont call them x3, y3 or x3 and f_x3
I would actually put the data into two different Matrices. If you watch the
code I had to do some not-so-obvious stuff like iterating over shape
divided by two and built indices from string operations because of the given table layout

Find when the values of a pandas.Series change by at least x

I have a time series s stored as a pandas.Series and I need to find when the value tracked by the time series changes by at least x.
In pseudocode:
print s(0)
s*=s(0)
for all t in ]t, t_max]:
if |s(t)-s*| > x:
s* = s(t)
print s*
Naively, this can be coded in Python as follows:
import pandas as pd
def find_changes(s, x):
changes = []
s_last = None
for index, value in s.iteritems():
if s_last is None:
s_last = value
if value-s_last > x or s_last-value > x:
changes += [index, value]
s_last = value
return changes
My data set is large, so I can't just use the method above. Moreover, I cannot use Cython or Numba due to limitations of the framework I will run this on. I can (and plan to) use pandas and NumPy.
I'm looking for some guidance on what NumPy vectorized/optimized methods to use and how.
Thanks!
EDIT: Changed code to match pseudocode.

I don't know if I am understanding you correctly, but here is how I interpreted the problem:
import pandas as pd
import numpy as np
# Our series of data.
data = pd.DataFrame(np.random.rand(10), columns = ['value'])
# The threshold.
threshold = .33
# For each point t, grab t - 1.
data['value_shifted'] = data['value'].shift(1)
# Absolute difference of t and t - 1.
data['abs_change'] = abs(data['value'] - data['value_shifted'])
# Test against the threshold.
data['change_exceeds_threshold'] = np.where(data['abs_change'] > threshold, 1, 0)
print(data)
Giving:
value value_shifted abs_change change_exceeds_threshold
0 0.005382 NaN NaN 0
1 0.060954 0.005382 0.055573 0
2 0.090456 0.060954 0.029502 0
3 0.603118 0.090456 0.512661 1
4 0.178681 0.603118 0.424436 1
5 0.597814 0.178681 0.419133 1
6 0.976092 0.597814 0.378278 1
7 0.660010 0.976092 0.316082 0
8 0.805768 0.660010 0.145758 0
9 0.698369 0.805768 0.107400 0

I don't think the pseudo code can be vectorized because the next state of s* is dependent on the last state. There's a pure python solution (1 iteration):
import random
import pandas as pd
s = [random.randint(0,100) for _ in range(100)]
res = [] # record changes
thres = 20
ss = s[0]
for i in range(len(s)):
if abs(s[i] - ss) > thres:
ss = s[i]
res.append([i, s[i]])
df = pd.DataFrame(res, columns=['value'])
I think there's no way to run faster than O(N) in this case.

Performance enhancement of ranking function by replacement of lambda x with vectorization

I have a ranking function that I apply to a large number of columns of several million rows which takes minutes to run. By removing all of the logic preparing the data for application of the .rank( method, i.e., by doing this:
ranked = df[['period_id', 'sector_name'] + to_rank].groupby(['period_id', 'sector_name']).transform(lambda x: (x.rank(ascending = True) - 1)*100/len(x))
I managed to get this down to seconds. However, I need to retain my logic, and am struggling to restructure my code: ultimately, the largest bottleneck is my double use of lambda x:, but clearly other aspects are slowing things down (see below). I have provided a sample data frame, together with my ranking functions below, i.e. an MCVE. Broadly, I think that my questions boil down to:
(i) How can one replace the .apply(lambda x usage in the code with a fast, vectorized equivalent? (ii) How can one loop over multi-indexed, grouped, data frames and apply a function? in my case, to each unique combination of the date_id and category columns.
(iii) What else can I do to speed up my ranking logic? the main overhead seems to be in .value_counts(). This overlaps with (i) above; perhaps one can do most of this logic on df, perhaps via construction of temporary columns, before sending for ranking. Similarly, can one rank the sub-dataframe in one call?
(iv) Why use pd.qcut() rather than df.rank()? the latter is cythonized and seems to have more flexible handling of ties, but I cannot see a comparison between the two, and pd.qcut() seems most widely used.
Sample input data is as follows:
import pandas as pd
import numpy as np
import random
to_rank = ['var_1', 'var_2', 'var_3']
df = pd.DataFrame({'var_1' : np.random.randn(1000), 'var_2' : np.random.randn(1000), 'var_3' : np.random.randn(1000)})
df['date_id'] = np.random.choice(range(2001, 2012), df.shape[0])
df['category'] = ','.join(chr(random.randrange(97, 97 + 4 + 1)).upper() for x in range(1,df.shape[0]+1)).split(',')
The two ranking functions are:
def rank_fun(df, to_rank): # calls ranking function f(x) to rank each category at each date
#extra data tidying logic here beyond scope of question - can remove
ranked = df[to_rank].apply(lambda x: f(x))
return ranked
def f(x):
nans = x[np.isnan(x)] # Remove nans as these will be ranked with 50
sub_df = x.dropna() #
nans_ranked = nans.replace(np.nan, 50) # give nans rank of 50
if len(sub_df.index) == 0: #check not all nan. If no non-nan data, then return with rank 50
return nans_ranked
if len(sub_df.unique()) == 1: # if all data has same value, return rank 50
sub_df[:] = 50
return sub_df
#Check that we don't have too many clustered values, such that we can't bin due to overlap of ties, and reduce bin size provided we can at least quintile rank.
max_cluster = sub_df.value_counts().iloc[0] #value_counts sorts by counts, so first element will contain the max
max_bins = len(sub_df) / max_cluster
if max_bins > 100: #if largest cluster <1% of available data, then we can percentile_rank
max_bins = 100
if max_bins < 5: #if we don't have the resolution to quintile rank then assume no data.
sub_df[:] = 50
return sub_df
bins = int(max_bins) # bin using highest resolution that the data supports, subject to constraints above (max 100 bins, min 5 bins)
sub_df_ranked = pd.qcut(sub_df, bins, labels=False) #currently using pd.qcut. pd.rank( seems to have extra functionality, but overheads similar in practice
sub_df_ranked *= (100 / bins) #Since we bin using the resolution specified in bins, to convert back to decile rank, we have to multiply by 100/bins. E.g. with quintiles, we'll have scores 1 - 5, so have to multiply by 100 / 5 = 20 to convert to percentile ranking
ranked_df = pd.concat([sub_df_ranked, nans_ranked])
return ranked_df
And the code to call my ranking function and recombine with df is:
# ensure don't get duplicate columns if ranking already executed
ranked_cols = [col + '_ranked' for col in to_rank]
ranked = df[['date_id', 'category'] + to_rank].groupby(['date_id', 'category'], as_index = False).apply(lambda x: rank_fun(x, to_rank))
ranked.columns = ranked_cols
ranked.reset_index(inplace = True)
ranked.set_index('level_1', inplace = True)
df = df.join(ranked[ranked_cols])
I am trying to get this ranking logic as fast as I can, by removing both lambda x calls; I can remove the logic in rank_fun so that only f(x)'s logic is applicable, but I also don't know how to process multi-index dataframes in a vectorized fashion. An additional question would be on differences between pd.qcut( and df.rank(: it seems that both have different ways of dealing with ties, but the overheads seem similar, despite the fact that .rank( is cythonized; perhaps this is misleading, given the main overheads are due to my usage of lambda x.
I ran %lprun on f(x) which gave me the following results, although the main overhead is the use of .apply(lambda x rather than a vectorized approach:
Line # Hits Time Per Hit % Time Line Contents
2 def tst_fun(df, field):
3 1 685 685.0 0.2 x = df[field]
4 1 20726 20726.0 5.8 nans = x[np.isnan(x)]
5 1 28448 28448.0 8.0 sub_df = x.dropna()
6 1 387 387.0 0.1 nans_ranked = nans.replace(np.nan, 50)
7 1 5 5.0 0.0 if len(sub_df.index) == 0:
8 pass #check not empty. May be empty due to nans for first 5 years e.g. no revenue/operating margin data pre 1990
9 return nans_ranked
10
11 1 65559 65559.0 18.4 if len(sub_df.unique()) == 1:
12 sub_df[:] = 50 #e.g. for subranks where all factors had nan so ranked as 50 e.g. in 1990
13 return sub_df
14
15 #Finally, check that we don't have too many clustered values, such that we can't bin, and reduce bin size provided we can at least quintile rank.
16 1 74610 74610.0 20.9 max_cluster = sub_df.value_counts().iloc[0] #value_counts sorts by counts, so first element will contain the max
17 # print(counts)
18 1 9 9.0 0.0 max_bins = len(sub_df) / max_cluster #
19
20 1 3 3.0 0.0 if max_bins > 100:
21 1 0 0.0 0.0 max_bins = 100 #if largest cluster <1% of available data, then we can percentile_rank
22
23
24 1 0 0.0 0.0 if max_bins < 5:
25 sub_df[:] = 50 #if we don't have the resolution to quintile rank then assume no data.
26
27 # return sub_df
28
29 1 1 1.0 0.0 bins = int(max_bins) # bin using highest resolution that the data supports, subject to constraints above (max 100 bins, min 5 bins)
30
31 #should track bin resolution for all data. To add.
32
33 #if get here, then neither nans_ranked, nor sub_df are empty
34 # sub_df_ranked = pd.qcut(sub_df, bins, labels=False)
35 1 160530 160530.0 45.0 sub_df_ranked = (sub_df.rank(ascending = True) - 1)*100/len(x)
36
37 1 5777 5777.0 1.6 ranked_df = pd.concat([sub_df_ranked, nans_ranked])
38
39 1 1 1.0 0.0 return ranked_df

I'd build a function using numpy
I plan on using this within each group defined within a pandas groupby
def rnk(df):
a = df.values.argsort(0)
n, m = a.shape
r = np.arange(a.shape[1])
b = np.empty_like(a)
b[a, np.arange(m)[None, :]] = np.arange(n)[:, None]
return pd.DataFrame(b / n, df.index, df.columns)
gcols = ['date_id', 'category']
rcols = ['var_1', 'var_2', 'var_3']
df.groupby(gcols)[rcols].apply(rnk).add_suffix('_ranked')
var_1_ranked var_2_ranked var_3_ranked
0 0.333333 0.809524 0.428571
1 0.160000 0.360000 0.240000
2 0.153846 0.384615 0.461538
3 0.000000 0.315789 0.105263
4 0.560000 0.200000 0.160000
...
How It Works
Because I know that ranking is related to sorting, I want to use some clever sorting to do this quicker.
numpy's argsort will produce a permutation that can be used to slice the array into a sorted array.
a = np.array([25, 300, 7])
b = a.argsort()
print(b)
[2 0 1]
print(a[b])
[ 7 25 300]
So, instead, I'm going to use the argsort to tell me where the first, second, and third ranked elements are.
# create an empty array that is the same size as b or a
# but these will be ranks, so I want them to be integers
# so I use empty_like(b) because b is the result of
# argsort and is already integers.
u = np.empty_like(b)
# now just like when I sliced a above with a[b]
# I slice u the same way but instead I assign to
# those positions, the ranks I want.
# In this case, I defined the ranks as np.arange(b.size) + 1
u[b] = np.arange(b.size) + 1
print(u)
[2 3 1]
And that was exactly correct. The 7 was in the last position but was our first rank. 300 was in the second position and was our third rank. 25 was in the first position and was our second rank.
Finally, I divide by the number in the rank to get the percentiles. It so happens that because I used zero based ranking np.arange(n), as opposed to one based np.arange(1, n+1) or np.arange(n) + 1 as in our example, I can do the simple division to get the percentiles.
What's left to do is apply this logic to each group. We can do this in pandas with groupby
Some of the missing details include how I use argsort(0) to get independent sorts per column` and that I do some fancy slicing to rearrange each column independently.
Can we avoid the groupby and have numpy do the whole thing?
I'll also take advantage of numba's just in time compiling to speed up some things with njit
from numba import njit
#njit
def count_factor(f):
c = np.arange(f.max() + 2) * 0
for i in f:
c[i + 1] += 1
return c
#njit
def factor_fun(f):
c = count_factor(f)
cc = c[:-1].cumsum()
return c[1:][f], cc[f]
def lexsort(a, f):
n, m = a.shape
f = f * (a.max() - a.min() + 1)
return (f.reshape(-1, 1) + a).argsort(0)
def rnk_numba(df, gcols, rcols):
tups = list(zip(*[df[c].values.tolist() for c in gcols]))
f = pd.Series(tups).factorize()[0]
a = lexsort(np.column_stack([df[c].values for c in rcols]), f)
c, cc = factor_fun(f)
c = c[:, None]
cc = cc[:, None]
n, m = a.shape
r = np.arange(a.shape[1])
b = np.empty_like(a)
b[a, np.arange(m)[None, :]] = np.arange(n)[:, None]
return pd.DataFrame((b - cc) / c, df.index, rcols).add_suffix('_ranked')
How it works
Honestly, this is difficult to process mentally. I'll stick with expanding on what I explained above.
I want to use argsort again to drop rankings into the correct positions. However, I have to contend with the grouping columns. So what I do is compile a list of tuples and factorize them as was addressed in this question here
Now that I have a factorized set of tuples I can perform a modified lexsort that sorts within my factorized tuple groups. This question addresses the lexsort.
A tricky bit remains to be addressed where I must off set the new found ranks by the size of each group so that I get fresh ranks for every group. This is taken care of in the tiny snippet b - cc in the code below. But calculating cc is a necessary component.
So that's some of the high level philosophy. What about #njit?
Note that when I factorize, I am mapping to the integers 0 to n - 1 where n is the number of unique grouping tuples. I can use an array of length n as a convenient way to track the counts.
In order to accomplish the groupby offset, I needed to track the counts and cumulative counts in the positions of those groups as they are represented in the list of tuples or the factorized version of those tuples. I decided to do a linear scan through the factorized array f and count the observations in a numba loop. While I had this information, I'd also produce the necessary information to produce the cumulative offsets I also needed.
numba provides an interface to produce highly efficient compiled functions. It is finicky and you have to acquire some experience to know what is possible and what isn't possible. I decided to numbafy two functions that are preceded with a numba decorator #njit. This coded works just as well without those decorators, but is sped up with them.
Timing
%%timeit
ranked_cols = [col + '_ranked' for col in to_rank]
​
ranked = df[['date_id', 'category'] + to_rank].groupby(['date_id', 'category'], as_index = False).apply(lambda x: rank_fun(x, to_rank))
ranked.columns = ranked_cols
ranked.reset_index(inplace = True)
ranked.set_index('level_1', inplace = True)
1 loop, best of 3: 481 ms per loop
gcols = ['date_id', 'category']
rcols = ['var_1', 'var_2', 'var_3']
%timeit df.groupby(gcols)[rcols].apply(rnk_numpy).add_suffix('_ranked')
100 loops, best of 3: 16.4 ms per loop
%timeit rnk_numba(df, gcols, rcols).head()
1000 loops, best of 3: 1.03 ms per loop

I suggest you try this code. It's 3 times faster than yours, and more clear.
rank function:
def rank(x):
counts = x.value_counts()
bins = int(0 if len(counts) == 0 else x.count() / counts.iloc[0])
bins = 100 if bins > 100 else bins
if bins < 5:
return x.apply(lambda x: 50)
else:
return (pd.qcut(x, bins, labels=False) * (100 / bins)).fillna(50).astype(int)
single thread apply:
for col in to_rank:
df[col + '_ranked'] = df.groupby(['date_id', 'category'])[col].apply(rank)
mulple thread apply:
import sys
from multiprocessing import Pool
def tfunc(col):
return df.groupby(['date_id', 'category'])[col].apply(rank)
pool = Pool(len(to_rank))
result = pool.map_async(tfunc, to_rank).get(sys.maxint)
for (col, val) in zip(to_rank, result):
df[col + '_ranked'] = val

pandas lambda tuple mapping

Trying to compute a range (confidence interval) to return two values in lambda mapped over a column.
M=12.4; n=10; T=1.3
dt = pd.DataFrame( { 'vc' : np.random.randn(10) } )
ci = lambda c : M + np.asarray( -c*T/np.sqrt(n) , c*T/np.sqrt(n) )
dt['ci'] = dt['vc'].map( ci )
print '\n confidence interval ', dt['ci'][:,1]
..er , so how does this get done?
then, how to unpack the tuple in a lambda?
(I want to check if the range >0, ie contains the mean)
neither of the following work:
appnd = lambda c2: c2[0]*c2[1] > 0 and 1 or 0
app2 = lambda x,y: x*y >0 and 1 or 0
dt[cnt] = dt['ci'].map(app2)

It's probably easier to see by defining a proper function for the CI, rather than a lambda.
As far as the unpacking goes, maybe you could let the function take an argument for whether to add or subtract, and then apply it twice.
You should also calculate the mean and size in the function, instead of assigning them ahead of time.
In [40]: def ci(arr, op, t=2.0):
M = arr.mean()
n = len(arr)
rhs = arr * t / np.sqrt(n)
return np.array(op(M, rhs))
You can import the add and sub functions from operator
From there it's just a one liner:
In [47]: pd.concat([dt.apply(ci, axis=1, op=x) for x in [sub, add]], axis=1)
Out[47]:
vc vc
0 -0.374189 1.122568
1 0.217528 -0.652584
2 -0.636278 1.908835
3 -1.132730 3.398191
4 0.945839 -2.837518
5 -0.053275 0.159826
6 -0.031626 0.094879
7 0.931007 -2.793022
8 -1.016031 3.048093
9 0.051007 -0.153022
[10 rows x 2 columns]
I'd recommend breaking that into a few steps for clarity.
Get the minus one with r1 = dt.apply(ci, axis=1, op=sub), and the plus with r2 = dt.apply(ci, axis=1, op=add). Combine with pd.concat([r1, r2], axis=1)
Basically, it's hard to tell from dt.apply what the output should look like, just seeing some tuples. By applying separately, we get two 10 x 1 arrays.