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How to make a huge loop for checking condition in dataframe run faster in mac

I need to calculate a huge table value (157954 rows and 365 columns) by checking three conditions in a dataframe with 11 mil rows. Do you have any way to speed up the calculation, which is taking more than 10 hours now?
I have 367 stations in total.
for station in stations:
no_pickup_array = []
for time_point in data_matrix['Timestamp']:
time_point_2 = time_point + timedelta(minutes=15)
no_pickup = len(dataframe[(time_point <= dataframe["departure"]) & (dataframe["departure"] < time_point_2)
& (dataframe['departure_name'] == station)])
no_pickup_array.append(no_pickup)
print(f"Station name: {station}")
data_matrix[station] = no_pickup_array
I appreciate any of your help.
# To all: Thank you for your comments, I add more info for my problem.
Each row of dataframe is info of each renting bike. I want to create a matrix with number of bikes picked up at each station for each 15 minutes interval. Then I also want to calculate the average speed, average time,.. as well.
The solution from #Jérôme Richard could reduce the number of calculations, but I still struggle to understand and implement indexing steps and apply logarithmic search or binary search.
index = {name: df for name, df.sort_values('departure')['departure'].to_numpy() in dataframe.groupby('departure_name')}
# code #Jérôme Richard recommended

The main problem is the right-hand-side of the no_pickup assignment expression which is algorithmically inefficient because it makes a linear search while a logarithmic search is possible.
The first thing to do is to do a groupby of dataframe so to build an index enabling to fetch the dataframe subset having a given name. Then, you can sort each dataframe subset by departure so to be able to perform a binary search enabling you to know the number of item fitting the condition.
The index can be built with something like:
index = {name: df for name, df.sort_values('departure')['departure'].to_numpy() in dataframe.groupby('departure_name')}
Finally, you can do the binary search with two np.searchsorted on index[station]: one to know the starting index and one to know the ending index. You can get the length with a simple subtraction of the two.
Note that you may need some tweak since I am not sure the above code will works on your dataset but it is hard to know without an example of code generating the inputs.

You're indexing the dataframe list with a boolean (which will be zero or one, so you're only ever going to get the length of the first or second element) instead of a number. It's going to get evaluated like so:
len(dataframe[(time_point <= dataframe["departure"]) & (dataframe["departure"] < time_point_2) & (dataframe['departure_name'] == station)])
len(dataframe[True & False & True]) # let's just say the variables work out like this
len(dataframe[False])
len(dataframe[0])
This probably isn't the behavior you're after. (let me know what you're trying to do in a comment and I'll try to help out more.)
In terms of code speed specifically, & is bitwise "AND", in python the boolean operators are written out as and, or, and not. Using and here would speed up your code, since python only evaluates parts of boolean expressions where they're needed, e.g.
from time import sleep
def slow_function():
sleep(3)
return False
# This line doesn't take 3 seconds to run as you may expect.
# Python sees "False and" and is smart enough to realize that whatever comes after is irrelevant.
# No matter what comes after "False and", it's never going to make the first half True.
# So, python doesn't bother evaluating it, and saves 3 seconds in the process.
False and slow_function()
# Some more examples that show python doesn't evaluate the right half unless it needs to
False and print("hi")
False and asdfasdfasdf
False and 42/0
# The same does not happen here. These are bitwise operators, expected to be applied to numbers.
# While it does produce the correct result for boolean inputs, it's going to be slower,
# since it can't rely on the same optimization.
False & slow_function()
# Of course, both of these still take 3 seconds, since the right half has to be evaluated either way.
True and slow_function()
True & slow_function()

How to calculate Delta F / F using python?

I've recently "taught" myself python in order to analyze data for my experiments. As such I'm pretty clueless on many aspects. I've managed to make my analysis work for certain files but in some cases it breaks down and I imagine it is a result of faulty programming.
Currently I export a file containing 3 numpy arrays. One of these arrays is my signal (float values from -10 to 10). What I wish to do is to normalize every datum in this array to a range of values that preceed it. (i.e. the 30001st value must have the average of the preceeding 3000 values subtracted from it and then the difference must then be divided by thisvery same average (the preceeding 3000 values). My data is collected at a rate of 100Hz thus to get a normalization of the alst 30s i must use the preceeding 3000values.
As it stand this is how I've managed to make it work:
this stores the signal into the variable photosignal
photosignal = np.array(seg.analogsignals[0], ndmin=1)
now this the part I use to get the delta F/F over a moving window of 30s
normalizedphotosignal = [(uu-(np.mean(photosignal[uu-3000:uu])))/abs(np.mean(photosignal[uu-3000:uu])) for uu in photosignal[3000:]]
The following adds 3000 values to the beginning to keep the array the same length since later on i must time lock it to another list that is the same length
holder =list(range(3000))
normalizedphotosignal = holder + normalizedphotosignal
What I have noticed is that in certain files this code gives me an error because it says that the"slice" is empty and therefore it cannot create a mean.
I think maybe there is a better way to program this that could avoid this problem altogether. Or this a correct way to approach this problem?
So i tried the solution but it is quite slow and it nevertheless still gives me the "empty slice error".
I went over the moving average post and found this method:
def running_mean(x, N):
cumsum = np.cumsum(np.insert(x, 0, 0))
return (cumsum[N:] - cumsum[:-N]) / N
however I'm having trouble accommodating it to my desired output. namely (x-running average)/running average

Allright so I finally figured it out thanks to your help and the posts you referred me to.
The calculation for my entire data (300 000 +) takes about a second!
I used the following code:
def runningmean(x,N):
cumsum =np.cumsum(np.insert(x,0,0))
return (cumsum[N:] -cumsum[:-N])/N
photosignal = np.array(seg.analogsignal[0], ndmin =1)
photosignalaverage = runningmean(photosignal, 3000)
holder = np.zeros(2999)
photosignalaverage = np.append(holder,photosignalaverage)
detalfsignal = (photosignal-photosignalaverage)/abs(photosignalaverage)
Photosignal stores my raw signal in a numpy array.
Photosignalaverage uses cumsum to calculate the running average of every datapoint in photosignal. I then add the first 2999 values as 0, to maintian the same list size as my photosignal.
I then use basic numpy calculations to get my delta F/F signal.
Thank you once more for the feedback, was truly helpful!

Your approach goes in the right direction. However, you made a mistake in your list comprehension: you are using uu as your index whereas uu are the elements of your input data photosignal.
You want something like this:
normalizedphotosignal2 = np.zeros((photosignal.shape[0]-3000))
for i, uu in enumerate(photosignal[3000:]):
normalizedphotosignal2 = (uu - (np.mean(photosignal[i-3000:i]))) / abs(np.mean(photosignal[i-3000:i]))
Keep in mind that for-loops are relatively slow in python. If performance is an issue here, you could try avoiding the for loop and use numpy methods instead (e.g. have a look at Moving average or running mean).
Hope this helps.

Apply function to 2D numpy array elements

I've seen this post and want to do something similar, but not exactly the same.
I am implementing a little game of life game and using numpy arrays for representing the states of the game. So I need to check, how many alive neighbors a cell has. I already got a function for getting a window of neighbors given a coordinate and row count and column count for the window size that I want to have.
So usually my windows will be of 3x3 size like this:
T = True
F = False
[[T,T,T],
[F,T,T],
[F,F,F]] # some random truth values
In this representation True stands for a cell being alive.
Now I wrote some code iterating over all cells of the state, counting the True values and so on using a double for loop, but I think there is probably a better numpy solution.
What I'd do in the naive approach:
iterate over all cells of the state (not only the window) (I'd like to formulate some code to to be executed if a cell meets a criteria or another (being alive and surviving or being dead and coming alive))
get the window (wrapping or not wrapping) (function for that I already have)
check if the current cell is alive (could just do a lookup in the state's numpy array)
if it is alive start with an alive neighbors count of -1 otherwise start with 0
count all True values of the window (np.sum) and add it to the alive neighbors count (which is -1 if the cell itself was alive, so that I only count neighbors but not the cell itself)
depending on whether the count of alive neighbors is in certain ranges (configurable), write in another (new) state's array True values. (I'd start out with an array, which I created using: np.full((height, width), False, dtype=bool))
go on with that new array, keeping the old one in a list for history or logging purposes
Basically:
if cell meets criteria:
write True at the cell's position in a new array
However meeting the criteria depends on multiple rows, because the state's numpy array is a 2D array. That's why I think the linked post is close but not exactly what I need.
How can I do this in an efficient numpy-y way, avoiding unnecessary looping?
Clarification
I am searching for the best way of implementing this in python using numpy and scipy, which aims to be very readable and has good performance.

Perhaps I did not understand all you are trying to do, but what is stopping you from simply using the numpy.sum function?
Example - Let the state be:
import numpy as np
state = np.random.randint(1, 10, (9,9))
Here I am using {0, 1} as values for the state, where 1 means "alive".
Then you can just slice around the cell being investigated, e.g. [2,3]
s = state[1:3,2:5]
if s[1,1]:
val = -1
else
val = 0
val += s.sum()
If you put this in a for loop and pay attention to border cases, clamping or wrapping as appropriate, it should do what you describe.
If you are looking for a short elegant implementation, it can be done very efficiently with Python and Numpy.

Iterating over elements, finding minima per each element

First time posting, so I apologize for any confusion.
I have two numpy arrays which are time stamps for a signal.
chan1,chan2 looks like:
911.05, 7.7
1055.6, 455.0
1513.4, 1368.15
4604.6, 3004.4
4970.35, 3344.25
13998.25, 4029.9
15008.7, 6310.15
15757.35, 7309.75
16244.2, 8696.1
16554.65, 9940.0
..., ...
and so on, (up to 65000 elements per chan. pre file)
Edit : The lists are already sorted but the issue is that they are not always equal in spacing. There are gaps that could show up, which would misalign them, so chan1[3] could be closer to chan2[23] instead of, if the spacing was qual chan2[2 or 3 or 4] : End edit
For each elements in chan1, I am interested in finding the closest neighbor in chan2, which is done with:
$ np.min(np.abs(chan2-chan1[i]))
and to keep track of positive or neg. difference:
$ index=np.where( np.abs( chan2-chan1[i]) == res[i])[0][0]
$ if chan2[index]-chan1[i] <0.0 : res[i]=res[i]*(-1.0)
Lastly, I create a histogram of all the differences, in a range I am interested in.
My concern is that I do this in the for loop. I usually try to avoid for loops when I can by utilizing the numpy arrays, as each operation can be performed on the entire array. However, in this case I am unable to find a solution or a build in function (which I understand run significantly faster than anything I can make).
The routine takes about 0.03 seconds per file. There are a few more things happening outside of the function but not a significant number, mostly plotting after everything is done, and a loop to read in files.
I was wondering if anyone has seen a similar problem, or is familiar enough with the python libraries to suggest a solution (maybe a build in function?) to obtain the data I am interested in? I have to go over hundred of thousands of files, and currently my data analysis is about 10 slower than data acquisition. We are also in the middle of upgrading our instruments to where we will be able to obtain data 10-100 times faster, and so the analysis speed is going to become an serious issue.
I would prefer not to use a cluster to brute force the problem, and not too familiar with parallel processing, although I would not mind dabbling in it. It would take me a while to write it in C, and I am not sure if I would be able to make it faster.
Thank you in advance for your help.
def gen_hist(chan1,chan2):
res=np.arange(1,len(chan1)+1,1)*0.0
for i in range(len(chan1)):
res[i]=np.min(np.abs(chan2-chan1[i]))
index=np.where( np.abs( chan2-chan1[i]) == res[i])[0][0]
if chan2[index]-chan1[i] <0.0 : res[i]=res[i]*(-1.0)
return np.histogram(res,bins=np.arange(time_range[0]-interval,\
time_range[-1]+interval,\
interval))[0]
After all the files are cycled through I obtain a plot of the data:
Example of the histogram

Your question is a little vague, but I'm assuming that, given two sorted arrays, you're trying to return an array containing the differences between each element of the first array and the closest value in the second array.
Your algorithm will have a worst case of O(n^2) (np.where() and np.min() are O(n)). I would tackle this by using two iterators instead of one. You store the previous (r_p) and current (r_c) value of the right array and the current (l_c) value of the left array. For each value of the left array, increment the right array until r_c > l_c. Then append min(abs(r_p - l_c), abs(r_c - l_c)) to your result.
In code:
l = [ ... ]
r = [ ... ]
i = 0
j = 0
result = []
r_p = r_c = r[0]
while i < len(l):
l_c = l[i]
while r_c < l and j < len(r):
j += 1
r_c = r[j]
r_p = r[j-1]
result.append(min(abs(r_c - l_c), abs(r_p - l_c)))
i += 1
This runs in O(n). If you need additional speed out of it, try writing it in C or running it in Cython.

Python: sliding window of variable width

I'm writing a program in Python that's processing some data generated during experiments, and it needs to estimate the slope of the data. I've written a piece of code that does this quite nicely, but it's horribly slow (and I'm not very patient). Let me explain how this code works:
1) It grabs a small piece of data of size dx (starting with 3 datapoints)
2) It evaluates whether the difference (i.e. |y(x+dx)-y(x-dx)| ) is larger than a certain minimum value (40x std. dev. of noise)
3) If the difference is large enough, it will calculate the slope using OLS regression. If the difference is too small, it will increase dx and redo the loop with this new dx
4) This continues for all the datapoints
[See updated code further down]
For a datasize of about 100k measurements, this takes about 40 minutes, whereas the rest of the program (it does more processing than just this bit) takes about 10 seconds. I am certain there is a much more efficient way of doing these operations, could you guys please help me out?
Thanks
EDIT:
Ok, so I've got the problem solved by using only binary searches, limiting the number of allowed steps by 200. I thank everyone for their input and I selected the answer that helped me most.
FINAL UPDATED CODE:
def slope(self, data, time):
(wave1, wave2) = wt.dwt(data, "db3")
std = 2*np.std(wave2)
e = std/0.05
de = 5*std
N = len(data)
slopes = np.ones(shape=(N,))
data2 = np.concatenate((-data[::-1]+2*data[0], data, -data[::-1]+2*data[N-1]))
time2 = np.concatenate((-time[::-1]+2*time[0], time, -time[::-1]+2*time[N-1]))
for n in xrange(N+1, 2*N):
left = N+1
right = 2*N
for i in xrange(200):
mid = int(0.5*(left+right))
diff = np.abs(data2[n-mid+N]-data2[n+mid-N])
if diff >= e:
if diff < e + de:
break
right = mid - 1
continue
left = mid + 1
leftlim = n - mid + N
rightlim = n + mid - N
y = data2[leftlim:rightlim:int(0.05*(rightlim-leftlim)+1)]
x = time2[leftlim:rightlim:int(0.05*(rightlim-leftlim)+1)]
xavg = np.average(x)
yavg = np.average(y)
xlen = len(x)
slopes[n-N] = (np.dot(x,y)-xavg*yavg*xlen)/(np.dot(x,x)-xavg*xavg*xlen)
return np.array(slopes)

Your comments suggest that you need to find a better method to estimate ik+1 given ik. No knowledge of values in data would yield to the naive algorithm:
At each iteration for n, leave i at previous value, and see if the abs(data[start]-data[end]) value is less than e. If it is, leave i at its previous value, and find your new one by incrementing it by 1 as you do now. If it is greater, or equal, do a binary search on i to find the appropriate value. You can possibly do a binary search forwards, but finding a good candidate upper limit without knowledge of data can prove to be difficult. This algorithm won't perform worse than your current estimation method.
If you know that data is kind of smooth (no sudden jumps, and hence a smooth plot for all i values) and monotonically increasing, you can replace the binary search with a search backwards by decrementing its value by 1 instead.

How to optimize this will depend on some properties of your data, but here are some ideas:
Have you tried profiling the code? Using one of the Python profilers can give you some useful information about what's taking the most time. Often, a piece of code you've just written will have one biggest bottleneck, and it's not always obvious which piece it is; profiling lets you figure that out and attack the main bottleneck first.
Do you know what typical values of i are? If you have some idea, you can speed things up by starting with i greater than 0 (as #vhallac noted), or by increasing i by larger amounts — if you often see big values for i, increase i by 2 or 3 at a time; if the distribution of is has a long tail, try doubling it each time; etc.
Do you need all the data when doing the least squares regression? If that function call is the bottleneck, you may be able to speed it up by using only some of the data in the range. Suppose, for instance, that at a particular point, you need i to be 200 to see a large enough (above-noise) change in the data. But you may not need all 400 points to get a good estimate of the slope — just using 10 or 20 points, evenly spaced in the start:end range, may be sufficient, and might speed up the code a lot.

I work with Python for similar analyses, and have a few suggestions to make. I didn't look at the details of your code, just to your problem statement:
1) It grabs a small piece of data of size dx (starting with 3
datapoints)
2) It evaluates whether the difference (i.e. |y(x+dx)-y(x-dx)| ) is
larger than a certain minimum value (40x std. dev. of noise)
3) If the difference is large enough, it will calculate the slope
using OLS regression. If the difference is too small, it will increase
dx and redo the loop with this new dx
4) This continues for all the datapoints
I think the more obvious reason for slow execution is the LOOPING nature of your code, when perhaps you could use the VECTORIZED (array-based operations) nature of Numpy.
For step 1, instead of taking pairs of points, you can perform directly `data[3:] - data[-3:] and get all the differences in a single array operation;
For step 2, you can use the result from array-based tests like numpy.argwhere(data > threshold) instead of testing every element inside some loop;
Step 3 sounds conceptually wrong to me. You say that if the difference is too small, it will increase dx. But if the difference is small, the resulting slope would be small because it IS actually small. Then, getting a small value is the right result, and artificially increasing dx to get a "better" result might not be what you want. Well, it might actually be what you want, but you should consider this. I would suggest that you calculate the slope for a fixed dx across the whole data, and then take the resulting array of slopes to select your regions of interest (for example, using data_slope[numpy.argwhere(data_slope > minimum_slope)].
Hope this helps!