Say I have two taxi orders with Origin1、Destination1 and Origin2、Destination2(O1,O2,D1,D2).
I want to calculate the possibility of ridesharing, so I need the path between two different points.
And, here's my code:
def path_time(point1, point2):
path = ox.distance.shortest_path(road, point1, point2, weight='travel_time')
if path is None:
#If there isn't a path, a big weight will be set, and it won't be selected during the matching process.
route_time = 9999
else:
route_time = int(sum(ox.utils_graph.get_route_edge_attributes(road, path, "travel_time")))
return route_time,path
Since there is four points, I need to do this six times, where tp means travel path :
tpO1O2 = path_time(O1,O2)
tpO1D1 = path_time(O1,D1)
tpO1D2 = path_time(O1,D2)
tpO2D1 = path_time(O2,D1)
tpO2D2 = path_time(O2,D2)
tpD1D2 = path_time(D1,D2)
It's okay if I only have two points, but I got a 2 million order set, and each order has hundreds of potential matched orders. So this will take me a lot of time.
Does anyone knows how can I speed up this process? Thank you!
You can use parallelization to speed this up. Parallel route solving is built into OSMnx. Per the documentation:
You can parallelize solving multiple paths with the cpus parameter
I have a dataframe that has a small number of columns but many rows (about 900K right now, and it's going to get bigger as I collect more data). It looks like this:
Author
Title
Date
Category
Text
url
0
Amira Charfeddine
Wild Fadhila 01
2019-01-01
novel
الكتاب هذا نهديه لكل تونسي حس إلي الكتاب يحكي ...
NaN
1
Amira Charfeddine
Wild Fadhila 02
2019-01-01
novel
في التزغريت، والعياط و الزمامر، ليوم نتيجة الب...
NaN
2
253826
1515368_7636953
2010-12-28
/forums/forums/91/
هذا ما ينص عليه إدوستور التونسي لا رئاسة مدى ا...
https://www.tunisia-sat.com/forums/threads/151...
3
250442
1504416_7580403
2010-12-21
/forums/sports/
\n\n\n\n\n\nاعلنت الجامعة التونسية لكرة اليد ا...
https://www.tunisia-sat.com/forums/threads/150...
4
312628
1504416_7580433
2010-12-21
/forums/sports/
quel est le résultat final\n,,,,????
https://www.tunisia-sat.com/forums/threads/150...
The "Text" Column has a string of text that may be just a few words (in the case of a forum post) or it may a portion of a novel and have tens of thousands of words (as in the two first rows above).
I have code that constructs the dataframe from various corpus files (.txt and .json), then cleans the text and saves the cleaned dataframe as a pickle file.
I'm trying to run the following code to analyze how variable the spelling of different words are in the corpus. The functions seem simple enough: One counts the occurrence of a particular spelling variable in each Text row; the other takes a list of such frequencies and computes a Gini Coefficient for each lemma (which is just a numerical measure of how heterogenous the spelling is). It references a spelling_var dictionary that has a lemma as its key and the various ways of spelling that lemma as values. (like {'color': ['color', 'colour']} except not in English.)
This code works, but it uses a lot of CPU time. I'm not sure how much, but I use PythonAnywhere for my coding and this code sends me into the tarpit (in other words, it makes me exceed my daily allowance of CPU seconds).
Is there a way to do this so that it's less CPU intensive? Preferably without me having to learn another package (I've spent the past several weeks learning Pandas and am liking it, and need to just get on with my analysis). Once I have the code and have finished collecting the corpus, I'll only run it a few times; I won't be running it everyday or anything (in case that matters).
Here's the code:
import pickle
import pandas as pd
import re
with open('1_raw_df.pkl', 'rb') as pickle_file:
df = pickle.load(pickle_file)
spelling_var = {
'illi': ["الي", "اللي"],
'besh': ["باش", "بش"],
...
}
spelling_df = df.copy()
def count_word(df, word):
pattern = r"\b" + re.escape(word) + r"\b"
return df['Text'].str.count(pattern)
def compute_gini(freq_list):
proportions = [f/sum(freq_list) for f in freq_list]
squared = [p**2 for p in proportions]
return 1-sum(squared)
for w, var in spelling_var.items():
count_list = []
for v in var:
count_list.append(count_word(spelling_df, v))
gini = compute_gini(count_list)
spelling_df[w] = gini
I rewrote two lines in the last double loop, see the comments in the code below. does this solve your issue?
gini_lst = []
for w, var in spelling_var.items():
count_list = []
for v in var:
count_list.append(count_word(spelling_df, v))
#gini = compute_gini(count_list) # don't think you need to compute this at every iteration of the inner loop, right?
#spelling_df[w] = gini # having this inside of the loop creates a new column at each iteration, which could crash your CPU
gini_lst.append(compute_gini(count_list))
# this creates a df with a row for each lemma with its associated gini value
df_lemma_gini = pd.DataFrame(data={"lemma_column": list(spelling_var.keys()), "gini_column": gini_lst})
I have this data frame df1,
id lat_long
400743 2504043 (175.0976323, -41.1141412)
43203 1533418 (173.976683, -35.2235338)
463952 3805508 (174.6947496, -36.7437555)
1054906 3144009 (168.0105269, -46.36193)
214474 3030933 (174.6311167, -36.867717)
1008802 2814248 (169.3183615, -45.1859095)
988706 3245376 (171.2338968, -44.3884099)
492345 3085310 (174.740957, -36.8893026)
416106 3794301 (174.0106383, -35.3876921)
937313 3114127 (174.8436185, -37.80499)
I have constructed the tree for search here,
def construct_geopoints(s):
data_geopoints = [tuple(x) for x in s[['longitude','latitude']].to_records(index=False)]
tree = KDTree(data_geopoints, distance_metric='Arc', radius=pysal.cg.RADIUS_EARTH_KM)
return tree
tree = construct_geopoints(actualdata)
Now, I am trying to search all the geopoints which are within 1KM of every geopoint in my data frame df1. Here is how I am doing,
dfs = []
for name,group in df1.groupby(np.arange(len(df1))//10000):
s = group.reset_index(drop=True).copy()
pts = list(s['lat_long'])
neighbours = tree.query_ball_point(pts, 1)
s['neighbours'] = pd.Series(neighbours)
dfs.append(s)
output = pd.concat(dfs,axis = 0)
Everything here works fine, however I am trying to parallelise this task, since my df1 size is 2M records, this process is running for more than 8 hours. Can anyone help me on this? And another thing is, the result returned by query_ball_point is a list and so its throwing memory error when I am processing it for the huge amount of records. Any way to handle this.
EDIT :- Memory issue, look at the VIRT size.
It should be possible to parallelize your last segment of code with something like this:
from multiprocessing import Pool
...
def process_group(group):
s = group[1].reset_index(drop=True) # .copy() is implicit
pts = list(s['lat_long'])
neighbours = tree.query_ball_point(pts, 1)
s['neighbours'] = pd.Series(neighbours)
return s
groups = df1.groupby(np.arange(len(df1))//10000)
p = Pool(5)
dfs = p.map(process_group, groups)
output = pd.concat(dfs, axis=0)
But watch out, because the multiprocessing library pickles all the data on its way to and from the workers, and that can add a lot of overhead for data-intensive tasks, possibly cancelling the savings due to parallel processing.
I can't see where you'd be getting out-of-memory errors from. 8 million records is not that much for pandas. Maybe if your searches are producing hundreds of matches per row that could be a problem. If you say more about that I might be able to give some more advice.
It also sounds like pysal may be taking longer than necessary to do this. You might be able to get better performance by using GeoPandas or "rolling your own" solution like this:
assign each point to a surrounding 1-km grid cell (e.g., calculate UTM coordinates x and y, then create columns cx=x//1000 and cy=y//1000);
create an index on the grid cell coordinates cx and cy (e.g., df=df.set_index(['cx', 'cy']));
for each point, find the points in the 9 surrounding cells; you can select these directly from the index via df.loc[[(cx-1,cy-1),(cx-1,cy),(cx-1,cy+1),(cx,cy-1),...(cx+1,cy+1)], :];
filter the points you just selected to find the ones within 1 km.
Thanks for the answers, I have not used StackOverflow before so I was suprised by the number of answers and the speed of them - its fantastic.
I have not been through the answers properly yet, but thought I should add some information to the problem specification. See the image below.
I can't post an image in this because i don't have enough points but you can see an image
at http://journal.acquitane.com/2010-01-20/image003.jpg
This image may describe more closely what I'm trying to achieve. So you can see on the horizontal lines across the page are price points on the chart. Now where you get a clustering of lines within 0.5% of each, this is considered to be a good thing and why I want to identify those clusters automatically. You can see on the chart that there is a cluster at S2 & MR1, R2 & WPP1.
So everyday I produce these price points and then I can identify manually those that are within 0.5%. - but the purpose of this question is how to do it with a python routine.
I have reproduced the list again (see below) with labels. Just be aware that the list price points don't match the price points in the image because they are from two different days.
[YR3,175.24,8]
[SR3,147.85,6]
[YR2,144.13,8]
[SR2,130.44,6]
[YR1,127.79,8]
[QR3,127.42,5]
[SR1,120.94,6]
[QR2,120.22,5]
[MR3,118.10,3]
[WR3,116.73,2]
[DR3,116.23,1]
[WR2,115.93,2]
[QR1,115.83,5]
[MR2,115.56,3]
[DR2,115.53,1]
[WR1,114.79,2]
[DR1,114.59,1]
[WPP,113.99,2]
[DPP,113.89,1]
[MR1,113.50,3]
[DS1,112.95,1]
[WS1,112.85,2]
[DS2,112.25,1]
[WS2,112.05,2]
[DS3,111.31,1]
[MPP,110.97,3]
[WS3,110.91,2]
[50MA,110.87,4]
[MS1,108.91,3]
[QPP,108.64,5]
[MS2,106.37,3]
[MS3,104.31,3]
[QS1,104.25,5]
[SPP,103.53,6]
[200MA,99.42,7]
[QS2,97.05,5]
[YPP,96.68,8]
[SS1,94.03,6]
[QS3,92.66,5]
[YS1,80.34,8]
[SS2,76.62,6]
[SS3,67.12,6]
[YS2,49.23,8]
[YS3,32.89,8]
I did make a mistake with the original list in that Group C is wrong and should not be included. Thanks for pointing that out.
Also the 0.5% is not fixed this value will change from day to day, but I have just used 0.5% as an example for spec'ing the problem.
Thanks Again.
Mark
PS. I will get cracking on checking the answers now now.
Hi:
I need to do some manipulation of stock prices. I have just started using Python, (but I think I would have trouble implementing this in any language). I'm looking for some ideas on how to implement this nicely in python.
Thanks
Mark
Problem:
I have a list of lists (FloorLevels (see below)) where the sublist has two items (stockprice, weight). I want to put the stockprices into groups when they are within 0.5% of each other. A groups strength will be determined by its total weight. For example:
Group-A
115.93,2
115.83,5
115.56,3
115.53,1
-------------
TotalWeight:12
-------------
Group-B
113.50,3
112.95,1
112.85,2
-------------
TotalWeight:6
-------------
FloorLevels[
[175.24,8]
[147.85,6]
[144.13,8]
[130.44,6]
[127.79,8]
[127.42,5]
[120.94,6]
[120.22,5]
[118.10,3]
[116.73,2]
[116.23,1]
[115.93,2]
[115.83,5]
[115.56,3]
[115.53,1]
[114.79,2]
[114.59,1]
[113.99,2]
[113.89,1]
[113.50,3]
[112.95,1]
[112.85,2]
[112.25,1]
[112.05,2]
[111.31,1]
[110.97,3]
[110.91,2]
[110.87,4]
[108.91,3]
[108.64,5]
[106.37,3]
[104.31,3]
[104.25,5]
[103.53,6]
[99.42,7]
[97.05,5]
[96.68,8]
[94.03,6]
[92.66,5]
[80.34,8]
[76.62,6]
[67.12,6]
[49.23,8]
[32.89,8]
]
I suggest a repeated use of k-means clustering -- let's call it KMC for short. KMC is a simple and powerful clustering algorithm... but it needs to "be told" how many clusters, k, you're aiming for. You don't know that in advance (if I understand you correctly) -- you just want the smallest k such that no two items "clustered together" are more than X% apart from each other. So, start with k equal 1 -- everything bunched together, no clustering pass needed;-) -- and check the diameter of the cluster (a cluster's "diameter", from the use of the term in geometry, is the largest distance between any two members of a cluster).
If the diameter is > X%, set k += 1, perform KMC with k as the number of clusters, and repeat the check, iteratively.
In pseudo-code:
def markCluster(items, threshold):
k = 1
clusters = [items]
maxdist = diameter(items)
while maxdist > threshold:
k += 1
clusters = Kmc(items, k)
maxdist = max(diameter(c) for c in clusters)
return clusters
assuming of course we have suitable diameter and Kmc Python functions.
Does this sound like the kind of thing you want? If so, then we can move on to show you how to write diameter and Kmc (in pure Python if you have a relatively limited number of items to deal with, otherwise maybe by exploiting powerful third-party add-on frameworks such as numpy) -- but it's not worthwhile to go to such trouble if you actually want something pretty different, whence this check!-)
A stock s belong in a group G if for each stock t in G, s * 1.05 >= t and s / 1.05 <= t, right?
How do we add the stocks to each group? If we have the stocks 95, 100, 101, and 105, and we start a group with 100, then add 101, we will end up with {100, 101, 105}. If we did 95 after 100, we'd end up with {100, 95}.
Do we just need to consider all possible permutations? If so, your algorithm is going to be inefficient.
You need to specify your problem in more detail. Just what does "put the stockprices into groups when they are within 0.5% of each other" mean?
Possibilities:
(1) each member of the group is within 0.5% of every other member of the group
(2) sort the list and split it where the gap is more than 0.5%
Note that 116.23 is within 0.5% of 115.93 -- abs((116.23 / 115.93 - 1) * 100) < 0.5 -- but you have put one number in Group A and one in Group C.
Simple example: a, b, c = (0.996, 1, 1.004) ... Note that a and b fit, b and c fit, but a and c don't fit. How do you want them grouped, and why? Is the order in the input list relevant?
Possibility (1) produces ab,c or a,bc ... tie-breaking rule, please
Possibility (2) produces abc (no big gaps, so only one group)
You won't be able to classify them into hard "groups". If you have prices (1.0,1.05, 1.1) then the first and second should be in the same group, and the second and third should be in the same group, but not the first and third.
A quick, dirty way to do something that you might find useful:
def make_group_function(tolerance = 0.05):
from math import log10, floor
# I forget why this works.
tolerance_factor = -1.0/(-log10(1.0 + tolerance))
# well ... since you might ask
# we want: log(x)*tf - log(x*(1+t))*tf = -1,
# so every 5% change has a different group. The minus is just so groups
# are ascending .. it looks a bit nicer.
#
# tf = -1/(log(x)-log(x*(1+t)))
# tf = -1/(log(x/(x*(1+t))))
# tf = -1/(log(1/(1*(1+t)))) # solved .. but let's just be more clever
# tf = -1/(0-log(1*(1+t)))
# tf = -1/(-log((1+t))
def group_function(value):
# don't just use int - it rounds up below zero, and down above zero
return int(floor(log10(value)*tolerance_factor))
return group_function
Usage:
group_function = make_group_function()
import random
groups = {}
for i in range(50):
v = random.random()*500+1000
group = group_function(v)
if group in groups:
groups[group].append(v)
else:
groups[group] = [v]
for group in sorted(groups):
print 'Group',group
for v in sorted(groups[group]):
print v
print
For a given set of stock prices, there is probably more than one way to group stocks that are within 0.5% of each other. Without some additional rules for grouping the prices, there's no way to be sure an answer will do what you really want.
apart from the proper way to pick which values fit together, this is a problem where a little Object Orientation dropped in can make it a lot easier to deal with.
I made two classes here, with a minimum of desirable behaviors, but which can make the classification a lot easier -- you get a single point to play with it on the Group class.
I can see the code bellow is incorrect, in the sense the limtis for group inclusion varies as new members are added -- even it the separation crieteria remaisn teh same, you heva e torewrite the get_groups method to use a multi-pass approach. It should nto be hard -- but the code would be too long to be helpfull here, and i think this snipped is enoguh to get you going:
from copy import copy
class Group(object):
def __init__(self,data=None, name=""):
if data:
self.data = data
else:
self.data = []
self.name = name
def get_mean_stock(self):
return sum(item[0] for item in self.data) / len(self.data)
def fits(self, item):
if 0.995 < abs(item[0]) / self.get_mean_stock() < 1.005:
return True
return False
def get_weight(self):
return sum(item[1] for item in self.data)
def __repr__(self):
return "Group-%s\n%s\n---\nTotalWeight: %d\n\n" % (
self.name,
"\n".join("%.02f, %d" % tuple(item) for item in self.data ),
self.get_weight())
class StockGrouper(object):
def __init__(self, data=None):
if data:
self.floor_levels = data
else:
self.floor_levels = []
def get_groups(self):
groups = []
floor_levels = copy(self.floor_levels)
name_ord = ord("A") - 1
while floor_levels:
seed = floor_levels.pop(0)
name_ord += 1
group = Group([seed], chr(name_ord))
groups.append(group)
to_remove = []
for i, item in enumerate(floor_levels):
if group.fits(item):
group.data.append(item)
to_remove.append(i)
for i in reversed(to_remove):
floor_levels.pop(i)
return groups
testing:
floor_levels = [ [stock. weight] ,... <paste the data above> ]
s = StockGrouper(floor_levels)
s.get_groups()
For the grouping element, could you use itertools.groupby()? As the data is sorted, a lot of the work of grouping it is already done, and then you could test if the current value in the iteration was different to the last by <0.5%, and have itertools.groupby() break into a new group every time your function returned false.