Is there any ways to make this more efficient? - python-3.x

Question

I have 24 more attempts to submit this task. I spent hours and my brain does not work anymore. I am a beginner with Python can you please help to figure out what is wrong? I would love to see the correct code if possible.
Here is the task itself and the code I wrote below.
Note that you can have access to all standard modules/packages/libraries of your language. But there is no access to additional libraries (numpy in python, boost in c++, etc).
You are given a content of CSV-file with information about set of trades. It contains the following columns:
TIME - Timestamp of a trade in format Hour:Minute:Second.Millisecond
PRICE - Price of one share
SIZE - Count of shares executed in this trade
EXCHANGE - The exchange that executed this trade
For each exchange find the one minute-window during which the largest number of trades took place on this exchange.
Note that:
You need to send source code of your program.
You have only 25 attempts to submit a solutions for this task.
You have access to all standart modules/packages/libraries of your language. But there is no access to additional libraries (numpy in python, boost in c++, etc).
Input format
Input contains several lines. You can read it from standart input or file “trades.csv”
Each line contains information about one trade: TIME, PRICE, SIZE and EXCHANGE. Numbers are separated by comma.
Lines are listed in ascending order of timestamps. Several lines can contain the same timestamp.
Size of input file does not exceed 5 MB.
See the example below to understand the exact input format.
Output format
If input contains information about k exchanges, print k lines to standart output.
Each line should contain the only number — maximum number of trades during one minute-window.
You should print answers for exchanges in lexicographical order of their names.
Sample
Input Output
09:30:01.034,36.99,100,V
09:30:55.000,37.08,205,V
09:30:55.554,36.90,54,V
09:30:55.556,36.91,99,D
09:31:01.033,36.94,100,D
09:31:01.034,36.95,900,V
2
3
Notes
In the example four trades were executed on exchange “V” and two trades were executed on exchange “D”. Not all of the “V”-trades fit in one minute-window, so the answer for “V” is three.
X = []
with open('trades.csv', 'r') as tr:
for line in tr:
line = line.strip('\xef\xbb\xbf\r\n ')
X.append(line.split(','))
dex = {}
for item in X:
dex[item[3]] = []
for item in X:
dex[item[3]].append(float(item[0][:2])*60.+float(item[0][3:5])+float(item[0][6:8])/60.+float(item[0][9:])/60000.)
for item in dex:
count = 1
ccount = 1
if dex[item][len(dex[item])-1]-dex[item][0] <1:
count = len(dex[item])
else:
for t in range(len(dex[item])-1):
for tt in range(len(dex[item])-t-1):
if dex[item][tt+t+1]-dex[item][t] <1:
ccount += 1
else: break
if ccount>count:
count=ccount
ccount=1
print(count)

Answer 1

First of all it is not necessary to use datetime and csv modules for such a simple case (like in Ed-Ward's example).
If we remove colon and dot signs from the time strings it could be converted to int() directly - easier way than you tried in your example.
CSV features like dialect and special formatting not used so i suggest to use simple split(",")
Now about efficiency. Efficiency means time complexity.
The more times you go through your array with dates from the beginning to the end, the more complicated the algorithm becomes.
So our goal is to minimize cycles count, best to make only one pass by all rows and especially avoid nested loops and passing through collections from beginning to the end.
For such a task it is better to use deque, instead of tuple or list, because you can pop() first element and append last element with complexity of O(1).
Just append every time for needed exchange to the end of the exchange's queue until difference between current and first elements becomes more than 1 minute. Then just remove first element with popleft() and continue comparison. After whole file done - length of each queue will be the max 1min window.
Example with linear time complexity O(n):
from collections import deque
ex_list = {}
s = open("trades.csv").read().replace(":", "").replace(".", "")
for line in s.splitlines():
s = line.split(",")
curr_tm = int(s[0])
curr_ex = s[3]
if curr_ex not in ex_list:
ex_list[curr_ex] = deque()
ex_list[curr_ex].append(curr_tm)
if curr_tm >= ex_list[curr_ex][0] + 100000:
ex_list[curr_ex].popleft()
print("\n".join([str(len(ex_list[k])) for k in sorted(ex_list.keys())]))

Answer 2

This code should work:
import csv
import datetime
diff = datetime.timedelta(minutes=1)
def date_calc(start, dates):
for i, date in enumerate(dates):
if date >= start + diff:
return i
return i + 1
exchanges = {}
with open("trades.csv") as csvfile:
reader = csv.reader(csvfile)
for row in reader:
this_exchange = row[3]
if this_exchange not in exchanges:
exchanges[this_exchange] = []
time = datetime.datetime.strptime(row[0], "%H:%M:%S.%f")
exchanges[this_exchange].append(time)
ex_max = {}
for name, dates in exchanges.items():
ex_max[name] = 0
for i, d in enumerate(dates):
x = date_calc(d, dates[i:])
if x > ex_max[name]:
ex_max[name] = x
print('\n'.join([str(ex_max[k]) for k in sorted(ex_max.keys())]))
Output:
2
3
( obviously please check it for yourself before uploading it :) )
I think the issue with your current code is that you don't put the output in lexicographical order of their names...
If you want to use your current code, then here is a (hopefully) fixed version:
X = []
with open('trades.csv', 'r') as tr:
for line in tr:
line = line.strip('\xef\xbb\xbf\r\n ')
X.append(line.split(','))
dex = {}
counts = []
for item in X:
dex[item[3]] = []
for item in X:
dex[item[3]].append(float(item[0][:2])*60.+float(item[0][3:5])+float(item[0][6:8])/60.+float(item[0][9:])/60000.)
for item in dex:
count = 1
ccount = 1
if dex[item][len(dex[item])-1]-dex[item][0] <1:
count = len(dex[item])
else:
for t in range(len(dex[item])-1):
for tt in range(len(dex[item])-t-1):
if dex[item][tt+t+1]-dex[item][t] <1:
ccount += 1
else: break
if ccount>count:
count=ccount
ccount=1
counts.append((item, count))
counts.sort(key=lambda x: x[0])
print('\n'.join([str(x[1]) for x in counts]))
Output:
2
3
I do think you can make your life easier in the future by using Python's standard library, though :)