Note from 22.02.21:
-Potentially my problem could also be solved by a more efficient memory usage instead of multiprocessing, since I realized that the memory load gets very high and might be a limiting factor here.
I'm trying to reduce the time that my script needs to run by making use of multiprocessing.
In the past I got some good tips about increasing the speed of the function itself (Increase performance of np.where() loop), but now I would like to make use of all cores of a 32-core workstation.
My function compares entries of two lists (X and Y) with a reference lists Q and Z. For every element in X/Y, it checks whether X[i] occurs somewhere in Q and whether Y[i] occurs in Z. If X[i] == Q[s] AND Y[i] == Z[s], it returns the index "s".
(Note: My real data consists of DNA sequencing reads and I need to map my reads to the reference.)
What I tried so far:
Splitting my long lists X and Y into even chunks (n-chunks, where n == cpu_count)
Trying the "concurrent.futures.ProcessPoolExecutor()" to run the function for each "sublist" in parallel and in the end combine the result of each process to one final dictionary (matchdict). (--> see commented out section)
My problem:
All cores are getting used when I uncomment the multiprocessing section but it ends up with an error (index out of range) which I could not yet resolve. (--> Tip: lower N to 1000 and you will immediately see the error without waiting forever)
Does anyone know how to solve this, or can suggest a better approach to use multiprocessing in my code?
Here is the code:
import numpy as np
import multiprocessing
import concurrent.futures
np.random.seed(1)
def matchdictfunc(index,x,y,q,z): # function to match entries of X and Y to Q and Z and get index of Q/Z where their values match X/Y
lookup = {}
for i, (q, z) in enumerate(zip(Q, Z)):
lookup.setdefault((q, z), []).append(i)
matchlist = [lookup.get((x, y), []) for x, y in zip(X, Y)]
matchdict = {}
for ind, match in enumerate(matchlist):
matchdict[index[ind]] = match
return matchdict
def split(a, n): # function to split list in n even parts
k, m = divmod(len(a), n)
return list((a[i * k + min(i, m):(i + 1) * k + min(i + 1, m)] for i in range(n)))
def splitinput(index,X,Y,Q,Z): # split large lists X and Y in n-even parts (n = cpu_count), make new list containing n-times Q and Z (to feed Q and Z for every process)
cpu_count = multiprocessing.cpu_count()
#create multiple chunks for X and Y and index:
index_split = split(index,cpu_count)
X_split = split(X,cpu_count)
Y_split = split(Y,cpu_count)
# create list with several times Q and Z since it needs to be same length as X_split etc:
Q_mult = []
Z_mult = []
for _ in range(cpu_count):
Q_mult.append(Q)
Z_mult.append(Z)
return index_split,X_split,Y_split,Q_mult,Z_mult
# N will finally scale up to 10^9
N = 10000000
M = 300
index = [str(x) for x in list(range(N))]
X = np.random.randint(M, size=N)
Y = np.random.randint(M, size=N)
# Q and Z size is fixed at 120000
Q = np.random.randint(M, size=120000)
Z = np.random.randint(M, size=120000)
# convert int32 arrays to str64 arrays and then to list, to represent original data (which are strings and not numbers)
X = np.char.mod('%d', X).tolist()
Y = np.char.mod('%d', Y).tolist()
Q = np.char.mod('%d', Q).tolist()
Z = np.char.mod('%d', Z).tolist()
# single-core:
matchdict = matchdictfunc(index,X,Y,Q,Z)
# split lists to number of processors (cpu_count)
index_split,X_split,Y_split,Q_mult,Z_mult = splitinput(index,X,Y,Q,Z)
## Multiprocessing attempt - FAILS! (index out of range)
# finallist = []
# if __name__ == '__main__':
# with concurrent.futures.ProcessPoolExecutor() as executor:
# results = executor.map(matchlistfunc,X_split,Y_split,Q_mult,Z_mult)
# for result in results:
# finallist.append(result)
# matchdict = {}
# for d in finallist:
# matchdict.update(d)
Your function matchdictfunc currently has arguments x, y, q, z but in fact does not use them, although in the multiprocessing version it will need to use two arguments. There is also no need for function splitinput to replicate Q and Z into returned values Q_split and Z_split. Currently, matchdictfunc is expecting Q and Z to be global variables and we can arrange for that to be the case in the multiprocessing version by using the initializer and initargs arguments when constructing the pool. You should also move code that you do not need to be executed by the sub-processes into the block controlled by if __name__ == '__main__':, such as the arary initialization code. These changes result in:
import numpy as np
import multiprocessing
import concurrent.futures
MULTIPROCESSING = True
def init_pool(q, z):
global Q, Z
Q = q
Z = z
def matchdictfunc(index, X, Y): # function to match entries of X and Y to Q and Z and get index of Q/Z where their values match X/Y
lookup = {}
for i, (q, z) in enumerate(zip(Q, Z)):
lookup.setdefault((q, z), []).append(i)
matchlist = [lookup.get((x, y), []) for x, y in zip(X, Y)]
matchdict = {}
for ind, match in enumerate(matchlist):
matchdict[index[ind]] = match
return matchdict
def split(a, n): # function to split list in n even parts
k, m = divmod(len(a), n)
return list((a[i * k + min(i, m):(i + 1) * k + min(i + 1, m)] for i in range(n)))
def splitinput(index, X, Y): # split large lists X and Y in n-even parts (n = cpu_count))
cpu_count = multiprocessing.cpu_count()
#create multiple chunks for X and Y and index:
index_split = split(index,cpu_count)
X_split = split(X,cpu_count)
Y_split = split(Y,cpu_count)
return index_split, X_split ,Y_split
def main():
# following required for non-multiprocessing
if not MULTIPROCESSING:
global Q, Z
np.random.seed(1)
# N will finally scale up to 10^9
N = 10000000
M = 300
index = [str(x) for x in list(range(N))]
X = np.random.randint(M, size=N)
Y = np.random.randint(M, size=N)
# Q and Z size is fixed at 120000
Q = np.random.randint(M, size=120000)
Z = np.random.randint(M, size=120000)
# convert int32 arrays to str64 arrays and then to list, to represent original data (which are strings and not numbers)
X = np.char.mod('%d', X).tolist()
Y = np.char.mod('%d', Y).tolist()
Q = np.char.mod('%d', Q).tolist()
Z = np.char.mod('%d', Z).tolist()
# for non-multiprocessing:
if not MULTIPROCESSING:
matchdict = matchdictfunc(index, X, Y)
else:
# for multiprocessing:
# split lists to number of processors (cpu_count)
index_split, X_split, Y_split = splitinput(index, X, Y)
with concurrent.futures.ProcessPoolExecutor(initializer=init_pool, initargs=(Q, Z)) as executor:
finallist = [result for result in executor.map(matchdictfunc, index_split, X_split, Y_split)]
matchdict = {}
for d in finallist:
matchdict.update(d)
#print(matchdict)
if __name__ == '__main__':
main()
Note: I tried this for a smaller value of N = 1000 (printing out the results of matchdict) and the multiprocessing version seemed to return the same results. My machine does not have the resources to run with the full value of N without freezing up everything else.
Another Approach
I am working under the assumption that your DNA data is external and the X and Y values can be read n values at a time or can be read in and written out so that this is possible. Then rather than having all the data resident in memory and splitting it up into 32 pieces, I propose that it be read n values at a time and thus broken up into approximately N/n pieces.
In the following code I have switched to using the imap method from class multiprocessing.pool.Pool. The advantage is that it lazily submits tasks to the process pool, that is, the iterable argument doesn't have to be a list or convertible to a list. Instead the pool will iterate over the iterable sending tasks to the pool in chunksize groups. In the code below, I have used a generator function for the argument to imap, which will generate successive X and Y values. Your actual generator function would first open the DNA file (or files) and read in successive portions of the file.
import numpy as np
import multiprocessing
def init_pool(q, z):
global Q, Z
Q = q
Z = z
def matchdictfunc(t): # function to match entries of X and Y to Q and Z and get index of Q/Z where their values match X/Y
index, X, Y = t
lookup = {}
for i, (q, z) in enumerate(zip(Q, Z)):
lookup.setdefault((q, z), []).append(i)
matchlist = [lookup.get((x, y), []) for x, y in zip(X, Y)]
matchdict = {}
for ind, match in enumerate(matchlist):
matchdict[index[ind]] = match
return matchdict
def next_tuple(n, stop, M):
start = 0
while True:
end = min(start + n, stop)
index = [str(x) for x in list(range(start, end))]
x = np.random.randint(M, size=n)
y = np.random.randint(M, size=n)
# convert int32 arrays to str64 arrays and then to list, to represent original data (which are strings and not numbers)
x = np.char.mod('%d', x).tolist()
y = np.char.mod('%d', y).tolist()
yield (index, x, y)
start = end
if start >= stop:
break
def compute_chunksize(XY_AT_A_TIME, N):
n_tasks, remainder = divmod(N, XY_AT_A_TIME)
if remainder:
n_tasks += 1
chunksize, remainder = divmod(n_tasks, multiprocessing.cpu_count() * 4)
if remainder:
chunksize += 1
return chunksize
def main():
np.random.seed(1)
# N will finally scale up to 10^9
N = 10000000
M = 300
# Q and Z size is fixed at 120000
Q = np.random.randint(M, size=120000)
Z = np.random.randint(M, size=120000)
# convert int32 arrays to str64 arrays and then to list, to represent original data (which are strings and not numbers)
Q = np.char.mod('%d', Q).tolist()
Z = np.char.mod('%d', Z).tolist()
matchdict = {}
# number of X, Y pairs at a time:
# experiment with this, especially as N increases:
XY_AT_A_TIME = 10000
chunksize = compute_chunksize(XY_AT_A_TIME, N)
#print('chunksize =', chunksize) # 32 with 8 cores
with multiprocessing.Pool(initializer=init_pool, initargs=(Q, Z)) as pool:
for d in pool.imap(matchdictfunc, next_tuple(XY_AT_A_TIME, N, M), chunksize):
matchdict.update(d)
#print(matchdict)
if __name__ == '__main__':
import time
t = time.time()
main()
print('total time =', time.time() - t)
Update
I want to eliminate using numpy from the benchmark. It is known that numpy uses multiprocessing for some of its operations and when used in multiprocessing applications can be the cause of of reduced performance. So the first thing I did was to take the OP's original program and where the code was, for example:
import numpy as np
np.random.seed(1)
X = np.random.randint(M, size=N)
X = np.char.mod('%d', X).tolist()
I replaced it with:
import random
random.seed(1)
X = [str(random.randrange(M)) for _ in range(N)]
I then timed the OP's program to get the time for generating the X, Y, Q and Z lists and the total time. On my desktop the times were approximately 20 seconds and 37 seconds respectively! So in my multiprocessing version just generating the arguments for the process pool's processes is more than half the total running time. I also discovered for the second approach, that as I increased the value of XY_AT_A_TIME that the CPU utilization went down from 100% to around 50% but that the total elapsed time improved. I haven't quite figured out why this is.
Next I tried to emulate how the programs would function if they were reading the data in. So I wrote out 2 * N random integers to a file, temp.txt and modified the OP's program to initialize X and Y from the file and then modified my second approach's next_tuple function as follows:
def next_tuple(n, stop, M):
with open('temp.txt') as f:
start = 0
while True:
end = min(start + n, stop)
index = [str(x) for x in range(start, end)] # improvement
x = [f.readline().strip() for _ in range(n)]
y = [f.readline().strip() for _ in range(n)]
yield (index, x, y)
start = end
if start >= stop:
break
Again as I increased XY_AT_A_TIME the CPU utilization went down (best performance I found was value 400000 with CPU utilization only around 40%).
I finally rewrote my first approach trying to be more memory efficient (see below). This updated version again reads the random numbers from a file but uses generator functions for X, Y and index so I don't need memory for both the full lists and the splits. Again, I do not expect duplicated results for the multiprocessing and non-multiprocessing versions because of the way I am assigning the X and Y values in the two cases (a simple solution to this would have been to write the random numbers to an X-value file and a Y-value file and read the values back from the two files). But this has no effect on the running times. But again, the CPU utilization, despite using the default pool size of 8, was only 30 - 40% (it fluctuated quite a bit) and the overall running time was nearly double the non-multiprocessing running time. But why?
import random
import multiprocessing
import concurrent.futures
import time
MULTIPROCESSING = True
POOL_SIZE = multiprocessing.cpu_count()
def init_pool(q, z):
global Q, Z
Q = q
Z = z
def matchdictfunc(index, X, Y): # function to match entries of X and Y to Q and Z and get index of Q/Z where their values match X/Y
lookup = {}
for i, (q, z) in enumerate(zip(Q, Z)):
lookup.setdefault((q, z), []).append(i)
matchlist = [lookup.get((x, y), []) for x, y in zip(X, Y)]
matchdict = {}
for ind, match in enumerate(matchlist):
matchdict[index[ind]] = match
return matchdict
def split(a): # function to split list in POOL_SIZE even parts
k, m = divmod(N, POOL_SIZE)
divisions = [(i + 1) * k + min(i + 1, m) - (i * k + min(i, m)) for i in range(POOL_SIZE)]
parts = []
for division in divisions:
part = [next(a) for _ in range(division)]
parts.append(part)
return parts
def splitinput(index, X, Y): # split large lists X and Y in n-even parts (n = POOL_SIZE)
#create multiple chunks for X and Y and index:
index_split = split(index)
X_split = split(X)
Y_split = split(Y)
return index_split, X_split ,Y_split
def main():
global N
# following required for non-multiprocessing
if not MULTIPROCESSING:
global Q, Z
random.seed(1)
# N will finally scale up to 10^9
N = 10000000
M = 300
# Q and Z size is fixed at 120000
Q = [str(random.randrange(M)) for _ in range(120000)]
Z = [str(random.randrange(M)) for _ in range(120000)]
with open('temp.txt') as f:
# for non-multiprocessing:
if not MULTIPROCESSING:
index = [str(x) for x in range(N)]
X = [f.readline().strip() for _ in range(N)]
Y = [f.readline().strip() for _ in range(N)]
matchdict = matchdictfunc(index, X, Y)
else:
# for multiprocessing:
# split lists to number of processors (POOL_SIZE)
# generator functions:
index = (str(x) for x in range(N))
X = (f.readline().strip() for _ in range(N))
Y = (f.readline().strip() for _ in range(N))
index_split, X_split, Y_split = splitinput(index, X, Y)
with concurrent.futures.ProcessPoolExecutor(POOL_SIZE, initializer=init_pool, initargs=(Q, Z)) as executor:
finallist = [result for result in executor.map(matchdictfunc, index_split, X_split, Y_split)]
matchdict = {}
for d in finallist:
matchdict.update(d)
if __name__ == '__main__':
t = time.time()
main()
print('total time =', time.time() - t)
Resolution?
Can it be that the overhead of transferring the data from the main process to the subprocesses, which involves shared memory reading and writing, is what is slowing everything down? So, this final version was an attempt to eliminate this potential cause for the slowdown. On my desktop I have 8 processors. For the first approach dividing the N = 10000000 X and Y values among them means that each process should be processing N // 8 -> 1250000 values. So I wrote out the random numbers in 16 groups of 1250000 numbers (8 groups for X and 8 groups for Y) as a binary file noting the offset and length of each of these 16 groups using the following code:
import random
random.seed(1)
with open('temp.txt', 'wb') as f:
offsets = []
for i in range(16):
n = [str(random.randrange(300)) for _ in range(1250000)]
b = ','.join(n).encode('ascii')
l = len(b)
offsets.append((f.tell(), l))
f.write(b)
print(offsets)
And from that I constructed lists X_SPECS and Y_SPECS that the worker function matchdictfunc could use for reading in the values X and Y itself as needed. So now instead of passing 1250000 values at a time to this worker function, we are just passing indices 0, 1, ... 7 to the worker function so it knows which group it has to read in. Shared memory access has been totally eliminated in accessing X and Y (it's still required for Q and Z) and the disk access moved to the process pool. The CPU Utilization will, of course, not be 100% because the worker function is doing I/O. But I found that while the running time has now been greatly improved, it still offered no improvement over the original non-multiprocessing version:
OP's original program modified to read `X` and `Y` values in from file: 26.2 seconds
Multiprocessing elapsed time: 29.2 seconds
In fact, when I changed the code to use multithreading by replacing the ProcessPoolExecutor with ThreadPoolExecutor, the elpased time went down almost another second demonstrating the there is very little contention for the Global Interpreter Lock within the worker function, i.e. most of the time is being spent in C-language code. The main work is done by:
matchlist = [lookup.get((x, y), []) for x, y in zip(X, Y)]
When we do this with multiprocessing, we have multiple list comprehensions and multiple zip operations (on smaller lists) being performed by separate processes and we then assemble the results in the end. This is conjecture on my part, but there just may not be any performance gains to be had by taking what are already efficient operations and scaling them down across multiple processors. Or in other words, I am stumped and that was my best guess.
The final version (with some additional optimizations -- please note):
import random
import concurrent.futures
import time
POOL_SIZE = 8
X_SPECS = [(0, 4541088), (4541088, 4541824), (9082912, 4540691), (13623603, 4541385), (18164988, 4541459), (22706447, 4542961), (27249408, 4541847), (31791255, 4542186)]
Y_SPECS = [(36333441, 4542101), (40875542, 4540120), (45415662, 4540802), (49956464, 4540971), (54497435, 4541427), (59038862, 4541523), (63580385, 4541571), (68121956, 4542335)]
def init_pool(q_z):
global Q_Z
Q_Z = q_z
def matchdictfunc(index, i): # function to match entries of X and Y to Q and Z and get index of Q/Z where their values match X/Y
x_offset, x_len = X_SPECS[i]
y_offset, y_len = Y_SPECS[i]
with open('temp.txt', 'rb') as f:
f.seek(x_offset, 0)
X = f.read(x_len).decode('ascii').split(',')
f.seek(y_offset, 0)
Y = f.read(y_len).decode('ascii').split(',')
lookup = {}
for i, (q, z) in enumerate(Q_Z):
lookup.setdefault((q, z), []).append(i)
matchlist = [lookup.get((x, y), []) for x, y in zip(X, Y)]
matchdict = {}
for ind, match in enumerate(matchlist):
matchdict[index[ind]] = match
return matchdict
def split(a): # function to split list in POOL_SIZE even parts
k, m = divmod(N, POOL_SIZE)
divisions = [(i + 1) * k + min(i + 1, m) - (i * k + min(i, m)) for i in range(POOL_SIZE)]
parts = []
for division in divisions:
part = [next(a) for _ in range(division)]
parts.append(part)
return parts
def main():
global N
random.seed(1)
# N will finally scale up to 10^9
N = 10000000
M = 300
# Q and Z size is fixed at 120000
Q = (str(random.randrange(M)) for _ in range(120000))
Z = (str(random.randrange(M)) for _ in range(120000))
Q_Z = list(zip(Q, Z)) # pre-compute the `zip` function
# for multiprocessing:
# split lists to number of processors (POOL_SIZE)
# generator functions:
index = (str(x) for x in range(N))
index_split = split(index)
with concurrent.futures.ProcessPoolExecutor(POOL_SIZE, initializer=init_pool, initargs=(Q_Z,)) as executor:
finallist = executor.map(matchdictfunc, index_split, range(8))
matchdict = {}
for d in finallist:
matchdict.update(d)
print(len(matchdict))
if __name__ == '__main__':
t = time.time()
main()
print('total time =', time.time() - t)
The Cost of Inter-Process Memory Transfers
In the code below function create_files was called to create 100 identical files consisting of a "pickled" list of 1,000,000 numbers. I then used a multiprocessing pool of size 8 twice to read the 100 files and unpickle the files to reconstitute the original lists. The difference between the first case (worker1) and the second case (worker2) was that in the second case the list is returned back to the caller (but not saved so that memory can be garbage collected immediately). The second case took more than three times longer than the first case. This can also explain in part why you do not see a speedup when you switch to multiprocessing.
from multiprocessing import Pool
import pickle
import time
def create_files():
l = [i for i in range(1000000)]
# create 100 identical files:
for file in range(1, 101):
with open(f'pkl/test{file}.pkl', 'wb') as f:
pickle.dump(l, f)
def worker1(file):
file_name = f'pkl/test{file}.pkl'
with open(file_name, 'rb') as f:
obj = pickle.load(f)
def worker2(file):
file_name = f'pkl/test{file}.pkl'
with open(file_name, 'rb') as f:
obj = pickle.load(f)
return file_name, obj
POOLSIZE = 8
if __name__ == '__main__':
#create_files()
pool = Pool(POOLSIZE)
t = time.time()
# no data returned:
for file in range(1, 101):
pool.apply_async(worker1, args=(file,))
pool.close()
pool.join()
print(time.time() - t)
pool = Pool(POOLSIZE)
t = time.time()
for file in range(1, 101):
pool.apply_async(worker2, args=(file,))
pool.close()
pool.join()
print(time.time() - t)
t = time.time()
for file in range(1, 101):
worker2(file)
print(time.time() - t)
I'm trying to write some code to compute mean, Variance, Standard Deviation, FWHM, and finally evaluate the Gaussian Integral. I've been running into a division by zero error that I can't get past and I would like to know the solution for this ?
Where it's throwing an error I've tried to throw an exception handler as follows
Average = (sum(yvalues)) / (len(yvalues)) try: return (sum(yvalues) / len(yvalues))
expect ZeroDivisionError:
return 0
xvalues = []
yvalues = []
def generate():
for i in range(0,300):
a = rand.uniform((float("-inf") , float("inf")))
b = rand.uniform((float("-inf") , float("inf")))
xvalues.append(i)
### Defining the variable 'y'
y = a * (b + i)
yvalues.append(y) + 1
def mean():
Average = (sum(yvalues))/(len(yvalues))
print("The average is", Average)
return Average
def varience():
# This calculates the SD and the varience
s = []
for i in yvalues:
z = i - mean()
z = (np.abs(i-z))**2
s.append(y)**2
t = mean()
v = numpy.sqrt(t)
print("Answer for Varience is:", v)
return v
Traceback (most recent call last):
File "Tuesday.py", line 42, in <module>
def make_gauss(sigma=varience(), mu=mean(), x = random.uniform((float("inf"))*-1, float("inf"))):
File "Tuesday.py", line 35, in varience
t = mean()
File "Tuesday.py", line 25, in mean
Average = (sum(yvalues))/(len(yvalues))
ZeroDivisionError: division by zero
There are a few things that are not quite right as people noted above.
import random
import numpy as np
def generate():
xvalues, yvalues = [], []
for i in range(0,300):
a = random.uniform(-1000, 1000)
b = random.uniform(-1000, 1000)
xvalues.append(i)
### Defining the variable 'y'
y = a * (b + i)
yvalues.append(y)
return xvalues, yvalues
def mean(yvalues):
return sum(yvalues)/len(yvalues)
def variance(yvalues):
# This calculates the SD and the varience
s = []
yvalues_mean = mean(yvalues)
for y in yvalues:
z = (y - yvalues_mean)**2
s.append(z)
t = mean(s)
return t
def variance2(yvalues):
yvalues_mean = mean(yvalues)
return sum( (y-yvalues_mean)**2 for y in yvalues) / len(yvalues)
# Generate the xvalues and yvalues
xvalues, yvalues = generate()
# Now do the calculation, based on the passed parameters
mean_yvalues = mean(yvalues)
variance_yvalues = variance(yvalues)
variance_yvalues2 = variance2(yvalues)
print('Mean {} variance {} {}'.format(mean_yvalues, variance_yvalues, variance_yvalues2))
# Using Numpy
np_mean = np.mean(yvalues)
np_var = np.var(yvalues)
print('Numpy: Mean {} variance {}'.format(np_mean, np_var))
The way variance was calculated isn't quite right, but given the comment of "SD and variance" you were probably going to calculate both.
The code above gives 2 (well, 3) ways to do what I understand you were trying to do but I changed a few of the methods to clean them up a bit. generate() returns two lists now. mean() returns the mean, etc. The function variance2() gives an alternative way to calculate the variance but using a list comprehension style.
The last couple of lines are an example using numpy which has all of it built in and, if available, is a great way to go.
The one part that wasn't clear was the random.uniform(float("-inf"), float("inf"))) which seems to be an error (?).
You are calling mean before you call generate.
This is obvious since yvalues.append(y) + 1 (in generate) would have caused another error (TypeError) since .append returns None and you can't add 1 to None.
Change yvalues.append(y) + 1 to yvalues.append(y + 1) and then make sure to call generate before you call mean.
Also notice that you have the same error in varience (which should be called variance, btw). s.append(y)**2 should be s.append(y ** 2).
Another error you have is that the stacktrace shows make_gauss(sigma=varience(), mu=mean(), x = random.uniform((float("inf"))*-1, float("inf"))).
I'm pretty sure you don't actually want to call varience and mean on this line, just reference them. So also change that line to make_gauss(sigma=varience, mu=mean, x = random.uniform((float("inf"))*-1, float("inf")))
I have a list like
cases = [(1,1), (2,2), (3,3)]
trying to write a function that calculates through each item and return values:
def case_cal(cases)
for x, y in cases:
result = x+y
return result
output = case_cal(cases)
print(output)
I like to get output like
2
4
6
But I only get
2
I am a newbie learning python and something simple I am missing here. Could I get any advice? Thanks in advance!
Once you return something you move out of the function. So make a list, append your values to the list and then return in the end.
def case_cal(cases):
ret_values = []
for x, y in cases:
result = x+y
ret_values.append(result)
return ret_values
output = case_cal(cases)
print(*output)
Your code returns inside the for loop, at the end of the first iteration, so you'll only see the first x + y result.
This is a perfect use for a generator, which will allow you to grab the next x + y calculation on demand and offer maximum control over what the caller can do with the result:
cases = [(1,1), (2,2), (3,3)]
def case_cal(cases):
for x, y in cases:
yield x + y
for x in case_cal(cases):
print(x)
Output:
2
4
6
You can simply map the items of the list to the sum function:
list(map(sum, cases))
This becomes:
[2, 4, 6]
Or if you want to print the items individually:
for s in map(sum, cases):
print(s)
This outputs:
2
4
6
If you dont need the values in a list you can procced as follows:
cases = [(1,1), (2,2), (3,3)]
def case_cal(cases):
for x, y in cases:
result = x+y
print(result)
case_cal(cases)
Just directly print the values instead of returning them as once you return something you move out of the function.
Also,
cases = [(1,1), (2,2), (3,3)]
def case_cal(cases):
for x, y in cases:
print(x+y)
case_cal(cases)
You could do this:
cases = [(1,1), (2,2), (3,3)]
def case_cal(cases):
results = []
for x, y in cases:
results.append(x + y)
return results
output = [case_cal(cases)]
print(output)
And then you could define a print function like this:
def print_cases(cases):
for elem in cases[0]:
print(elem)
But there is a more Pythonic way:
def case_cal(cases):
return [(x + y) for (x, y) in cases]
This is called a list comprehension, and you are going to be using them very often in python. You can read more about them here.
Simply do the following:
def case_calc(cases):
for x, y in cases:
print(x + y)
case_calc([(1, 1), (2, 2), (3, 3)])
Cheers
I am trying to sort 4 integers input by the user into numerical order using only the min() and max() functions in python. I can get the highest and lowest number easily, but cannot work out a combination to order the two middle numbers? Does anyone have an idea?
So I'm guessing your input is something like this?
string = input('Type your numbers, separated by a space')
Then I'd do:
numbers = [int(i) for i in string.strip().split(' ')]
amount_of_numbers = len(numbers)
sorted = []
for i in range(amount_of_numbers):
x = max(numbers)
numbers.remove(x)
sorted.append(x)
print(sorted)
This will sort them using max, but min can also be used.
If you didn't have to use min and max:
string = input('Type your numbers, separated by a space')
numbers = [int(i) for i in string.strip().split(' ')]
numbers.sort() #an optional reverse argument possible
print(numbers)
LITERALLY just min and max? Odd, but, why not. I'm about to crash, but I think the following would work:
# Easy
arr[0] = max(a,b,c,d)
# Take the smallest element from each pair.
#
# You will never take the largest element from the set, but since one of the
# pairs will be (largest, second_largest) you will at some point take the
# second largest. Take the maximum value of the selected items - which
# will be the maximum of the items ignoring the largest value.
arr[1] = max(min(a,b)
min(a,c)
min(a,d)
min(b,c)
min(b,d)
min(c,d))
# Similar logic, but reversed, to take the smallest of the largest of each
# pair - again omitting the smallest number, then taking the smallest.
arr[2] = min(max(a,b)
max(a,c)
max(a,d)
max(b,c)
max(b,d)
max(c,d))
# Easy
arr[3] = min(a,b,c,d)
For Tankerbuzz's result for the following:
first_integer = 9
second_integer = 19
third_integer = 1
fourth_integer = 15
I get 1, 15, 9, 19 as the ascending values.
The following is one of the forms that gives symbolic form of the ascending values (using i1-i4 instead of first_integer, etc...):
Min(i1, i2, i3, i4)
Max(Min(i4, Max(Min(i1, i2), Min(i3, Max(i1, i2))), Max(i1, i2, i3)), Min(i1, i2, i3, Max(i1, i2)))
Max(Min(i1, i2), Min(i3, Max(i1, i2)), Min(i4, Max(i1, i2, i3)))
Max(i1, i2, i3, i4)
It was generated by a 'bubble sort' using the Min and Max functions of SymPy (a python CAS):
def minmaxsort(v):
"""return a sorted list of the elements in v using the
Min and Max functions.
Examples
========
>>> minmaxsort(3, 2, 1)
[1, 2, 3]
>>> minmaxsort(1, x, y)
[Min(1, x, y), Max(Min(1, x), Min(y, Max(1, x))), Max(1, x, y)]
>>> minmaxsort(1, y, x)
[Min(1, x, y), Max(Min(1, y), Min(x, Max(1, y))), Max(1, x, y)]
"""
from sympy import Min, Max
v = list(v)
v0 = Min(*v)
for j in range(len(v)):
for i in range(len(v) - j - 1):
w = v[i:i + 2]
v[i:i + 2] = [Min(*w), Max(*w)]
v[0] = v0
return v
I have worked it out.
min_integer = min(first_integer, second_integer, third_integer, fourth_integer)
mid_low_integer = min(max(first_integer, second_integer), max(third_integer, fourth_integer))
mid_high_integer = max(min(first_integer, second_integer), min(third_integer, fourth_integer))
max_integer = max(first_integer, second_integer, third_integer, fourth_integer)