Finitely generated words - python-2.7

I want to do an algorithm that can generate words of lenght less or equal to n, that doesn't contain certain strings, and that takes less than a minute or two to generate all words when I generate words of lenght 15 or more.
My actual code to generate them is this:
def inv(mob):
if 'aA' in mob:
return False
elif 'Aa' in mob:
return False
elif 'Bb' in mob:
return False
elif 'bB' in mob:
return False
else: return True
S=itertools.product('AaBb',repeat=n)
SS=map(lambda x: ''.join(x),S)
SSS=filter(inv,SS)
(Where n is the lenght)
When I create words of lenght n=10, it takes 3 seconds, but I need to generate at least of lenght n=15 in less than a minute, and when I try with n=15, my computer dies. Someone knows if exists an easy way? Or this problem can't take less time than 2 minutos to generate the words?


Use itertools.imap and itertools.ifilter rather than the map and filter functions as they produce iterables rather than lists. Which should allow avoid creating the large list that map produces which would have to contain 1073741824 (4**15) elements.
S = itertools.product('AaBb', repeat=n)
SS = itertools.imap(lambda x: ''.join(x), S)
SSS = itertools.ifilter(inv, SS)

Related

How to generate the random values of map from a given set of values and then store the key and values into separate variables in scala

I am trying to generate the 1000 random key-val map pairs from the given(statically defined) 2 map key-val pairs in scala and also later i would want to break the key and value pairs and store them into separate variables
Whatever i have tried:
object methodTest extends App{
val testMap = Map("3875835"->"ABCDE","316067107"->"EFGHI")
def getRandomElement(seq: Map[String,String]): Map[String,String] = {
seq(Random.nextInt(seq.length))
}
var outList = List.empty[Map[String,String]]
for(i<-0 to 1000){
outList+=getRandomElement(testMap)
}
print(outList)
}
The Output should generate 1000 Key-val pairs of map like i am showing below
[3875835,ABCDE]
[316067107,EFGHI]
[3875835,ABCDE]
[316067107,EFGHI]
[316067107,EFGHI]
............
............
............. upto 1000 random key-val pairs
Please help me figure out where i am going wrong and let me know How to resolve it, if any issue regarding the requirement, please feel free to comment for it

You can transform your seed map testMap into sequence of key/value tuples using .toSeq and then generate key/value pairs by iterating over the list of numbers from 0 until 1000, associating each number to a random choice between first or second element of the seed:
import scala.util.Random
val testMap = Map("3875835" -> "ABCDE", "316067107" -> "EFGHI")
val seed = testMap.toSeq
val keyValuesList = (0 until 1000).map(index => seed(Random.nextInt(seed.size)))
Note: 0 until 1000 will return all the numbers from 0 to 1000 excluded, so 1000 numbers. If you use 0 to 1000 you will get all the numbers from 0 to 1000 included, so 1001 numbers.
If you want to print the resulting list, you can use .foreach method with println function as argument:
keyValuesList.foreach(println)
And you will get:
(3875835,ABCDE)
(316067107,EFGHI)
(3875835,ABCDE)
(3875835,ABCDE)
(316067107,EFGHI)
(3875835,ABCDE)
...
if you want to keep only the keys, you can iterate on the list using .map method, taking only the first element of the tuple by using ._1 method, that retrieve the first element of a tuple:
val keys = keyValuesList.map(keyValuePair => keyValuePair._1)
And if you want only a list containing all second elements of each pair:
val values = keyValuesList.map(keyValuePair => keyValuePair._2)

Take first N keys by max value from dict {key:list}

is there have easy way to take first N keys which have max value from they list in dict {key:list}
is there have easy way to take first N keys which have max value from they list in dict {key:list}
def main():
for x in range(len(sale10k)):
timelist.append(sale10k[x][3])
pricesList.append(sale10k[x][4])
if sale10k[x][0] in salesByCategory.keys():
salesByCategory[sale10k[x][0]].append(float(sale10k[x][4]))
else:
salesByCategory[sale10k[x][0]]=[]
salesByCategory[sale10k[x][0]].append(float(sale10k[x][4]))
salesByCategory1={}
for key,value in salesByCategory.items():
salesByCategory1[key]=sum(salesByCategory.get(key))
#fiveLarges=heapq.nlargest(5,salesByCategory1,key=salesByCategory1.get)
salesBycatalog={}
for y in range(len(catalog)):
salesBycatalog[catalog[y][0]]=catalog[y][5]
totalByGroup={}
for key, value in salesBycatalog.items():
if value in totalByGroup.keys():
totalByGroup[value].append(salesByCategory1.get(key))
else:
totalByGroup[value]=[]
totalByGroup[value].append(salesByCategory1.get(key))
print(totalByGroup)
if __name__ == "__main__":
main()
i have 2 files excel.cvs
my output from now is this :
{'POLO SHIRTS': [2609.76, 13339.109999999991, 15622.410000000007], 'APPAREL ACCESSORIES': [22596.24999999999, 20901.099999999995, 31007.8], 'PANTS': [8031.729999999998, 11179.949999999999, 5405.839999999997, 9023.949999999999, 21523.819999999996, 26030.800000000017], 'FOOTWEAR ACCESSORIES': [8686.369999999999], 'GLIDING SP.EQUIPMENT': [22136.399999999987, 27678.920000000006, 14222.21999999999, 30013.37000000001], 'SHOES': [1903.66, 25443.21999999999, 22152.530000000006, 11585.410000000002, 38504.679999999986, 7787.670000000004, 10256.860000000002, 1377.1199999999997, 15459.799999999992, 20919.56000000001, 6299.769999999996, 1555.4499999999998, 17470.460000000006, 29361.220000000034, 4070.9000000000033, 27045.450000000004, 20721.829999999994, 780.55, 24671.590000000015, 13189.570000000002, 6442.700000000001, 6105.390000000005, 12701.659999999998, 29418.89000000001, 7295.620000000001, 26344.420000000002, 3262.12, 11710.460000000006, 3272.2999999999993, 17055.989999999994, 9019.77, 12722.570000000003, 20020.150000000005, 30164.860000000026, 17513.14, 3168.6200000000003, 27008.24, 14585.679999999988, 15273.48, 24172.329999999998, 33968.96000000003, 35480.790000000015, 25150.459999999992, 24207.679999999997, 26909.090000000007, 17692.079999999998, 27844.97999999999, 33847.389999999985, 13266.239999999994, 11757.349999999997, 24469.410000000018, 8214.879999999997, 3966.6899999999964, 5336.910000000003, 27766.659999999978, 24636.97000000002, 21330.829999999994, 10331.680000000004, 19769.529999999995, 20764.439999999984, 2873.509999999999, 23263.23, 15127.240000000003, 13282.320000000003, 32917.03000000001, 17657.12, 9959.55, 21052.779999999995, 16015.79, 2667.2699999999995, 16041.830000000004, 2309.9000000000005, 8095.450000000001, 23628.889999999985, 3846.259999999999, 6795.61, 14608.109999999995, 6422.360000000001, 3241.279999999999, 19220.27999999999, 20836.899999999994, 28446.07000000001, 13984.979999999992, 10006.460000000003, 14417.309999999998, 9069.470000000001, 8081.38, 1766.8899999999999, 19041.750000000004, 3310.279999999999, 3649.49, 11089.069999999994, 10946.420000000002, 16297.91, 3788.1000000000004, 27356.640000000007, 14024.480000000001, 29409.03], 'SUITS': [28587.990000000016, 14337.800000000001], 'BALLS': [25855.07, 15207.729999999992, 25567.809999999987, 8428.509999999998, 15119.609999999995, 26069.969999999983, 29843.490000000023], 'TOPS': [1673.2000000000005, 8673.400000000001, 23610.79999999999, 2090.380000000001], 'HEADWEAR': [2075.3000000000015, 18891.799999999996, 39717.93, 33657.65, 9965.720000000005, 12030.020000000006, 670.9999999999999, 12694.720000000007, 24846.22000000001, 1606.1799999999994, 9993.330000000002, 10154.900000000005], 'HARDWARE ACCESSORIES': [14619.109999999997], 'OTHER SHIRTS': [18013.450000000004], 'PROTECTION GEAR': [26454.929999999997], 'JERSEYS': [23741.06, 38425.269999999975], 'SANDALS/SLIPPERS': [9103.83, 21025.040000000005, 12702.349999999999, 26766.439999999984, 29818.339999999993], 'SHORTS': [14817.77, 29540.92999999998, 9415.059999999996, 14582.480000000001], 'JACKETS': [30096.11000000001, 13372.469999999998, 31145.73000000001, 6011.17, 12225.300000000003, 23485.399999999998, 13889.96], 'SWIMWEAR': [14035.140000000001, 20232.629999999997, 5142.340000000001, 2945.349999999998, 23495.320000000003, 8207.920000000004, 11972.729999999994], 'T-SHIRTS': [11130.700000000004, 8315.83, 8346.719999999998, 27847.550000000007, 22704.759999999995, 7828.200000000002, 17823.379999999997, 2248.46, 9012.14, 7774.72, 12030.049999999996, 4207.649999999999, 21293.16, 3159.4700000000007, 13385.12, 30507.87], 'UNDERWEAR': [10419.31, 31017.909999999993, 2794.590000000002, 18625.990000000005, 21829.879999999994], 'SWEATSHIRTS': [4317.6799999999985, 23453.049999999985, 28176.49000000001], 'TIGHTS': [23823.43999999999, 11180.129999999996], 'BAGS': [13980.240000000007, 18509.50999999999, 20064.309999999998, 22317.360000000004, 17641.04]}
i need this :
SHOES: 1519077.15 €
T-SHIRTS: 207615.78 €
HEADWEAR: 176304.77 €
BALLS: 146092.19 €
JACKETS: 130226.14 €
I have data stored in dict orderBygroup {key-list(of float values)} and need to take first 5 keys with max value.
My second question is - dict salesByCategory1 is make with loping to salesByCategory and sum of all values to receive the total for article number.
Can i get that totals with some smartes way ?
is there have easy way to make that output ?

totalByGroup1={}
for key,value in totalByGroup.items():
totalByGroup1[key]=sum(totalByGroup.get(key))
Create a new dictionary with summed elements. More resources.
sorted5=sorted(totalByGroup1, key=totalByGroup1.get, reverse=True)[:5]
print(sorted5)
sorting and taking the first 5 elements
output is : ['SHOES', 'T-SHIRTS', 'HEADWEAR', 'BALLS', 'JACKETS']
more time, more resurses
for key in sorted5:
print(key,': ','{0:.2f}'.format(totalByGroup1.get(key)))
and now result :
SHOES : 1519077.15
T-SHIRTS : 207615.78
HEADWEAR : 176304.77
BALLS : 146092.19
JACKETS : 130226.14
now lets ask again if we have 4 record in dict wit that data:
a:[1,2,3],b:[6,7,8],c:[4,5,6],d:[9,10,11]
how to get first 2 key,value sorted by max value -->>
d:30,b:21
if we have 1000 record in dict - ?? how to get first N key sorted by max value of list
example go:1563,do:1560,bo:1490,ro:1480 .. etc

Python sorting based on first and last character in a string

Im taking an online beginner course through google on python 2, and I cannot figure out the answer to one of the questions. Here it is and thanks in advance for your help!
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
a = []
for b in words:
return
I tried a few different things. This is just where i left off on my last attempt, and decided to ask for help. I have spent more time thinking about this than i care to mention

def match_ends(words):
a = []
for b in words:
if (len(b) > 2 and b[0] == b[len(b)-1]):
a.append(b)
return a
def match_ends2(words):
return [x for x in words if len(x) > 2 and x[0] == x[len(x)-1]]
print(match_ends(['peter','paul','mary','tibet']))
print(match_ends2(['peter','paul','mary','tibet']))

Loop problems Even Count

I have a beginner question. Loops are extremely hard for me to understand, so it's come to me asking for help.
I am trying to create a function to count the amount of even numbers in a user input list, with a negative at the end to show the end of the list. I know I need to use a while loop, but I am having trouble figuring out how to walk through the indexes of the input list. This is what I have so far, can anyone give me a hand?
def find_even_count(numlist):
count = 0
numlist.split()
while numlist > 0:
if numlist % 2 == 0:
count += 1
return count
numlist = raw_input("Please enter a list of numbers, with a negative at the end: ")
print find_even_count(numlist)
I used the split to separate out the indexes of the list, but I know I am doing something wrong. Can anyone point out what I am doing wrong, or point me to a good step by step explanation of what to do here?
Thank you guys so much, I know you probably have something more on your skill level to do, but appreciate the help!

You were pretty close, just a couple of corrections:
def find_even_count(numlist):
count = 0
lst = numlist.split()
for num in lst:
if int(num) % 2 == 0:
count += 1
return count
numlist = raw_input("Please enter a list of numbers, with a negative at the end: ")
print find_even_count(numlist)
I have used a for loop rather than a while loop, stored the outcome of numlist.split() to a variable (lst) and then just iterated over this.

You have a couple of problems:
You split numlist, but don't assign the resulting list to anything.
You then try to operate on numlist, which is still the string of all numbers.
You never try to convert anything to a number.
Instead, try:
def find_even_count(numlist):
count = 0
for numstr in numlist.split(): # iterate over the list
num = int(numstr) # convert each item to an integer
if num < 0:
break # stop when we hit a negative
elif num % 2 == 0:
count += 1 # increment count for even numbers
return count # return the total
Or, doing the whole thing in one line:
def find_even_count(numlist):
return sum(num % 2 for num in map(int, numlist.split()) if num > 0)
(Note: the one-liner will fail in cases where the user tries to trick you by putting more numbers after the "final" negative number, e.g. with numlist = "1 2 -1 3 4")
If you must use a while loop (which isn't really the best tool for the job), it would look like:
def find_even_count(numlist):
index = count = 0
numlist = list(map(int, numlist.split()))
while numlist[index] > 0:
if numlist[index] % 2 == 0:
count += 1
index += 1
return count

Python remove odd numbers and print only even

user = int(raw_input("Type 5 numbers"))
even = []
def purify(odd):
for n in odd:
even.append(n)
if n % 2 > 0:
print n
print purify(user)
Hello I am a beginner and I would like to understand what is wrong with this code.
The User chose 5 numers and I want to print the even numbers only.
Thanks for helping

There are a few problems:
You can't apply int to an overall string, just to one integer at a time.
So if your numbers are space-separated, then you should split them into a list of strings. You can either convert them immediately after input, or wait and do it within your purify function.
Also, your purify function appends every value to the list even without testing it first.
Also, your test is backwards -- you are printing only odd numbers, not even.
Finally, you should return the value of even if you want to print it outside the function, instead of printing them as you loop.
I think this edited version should work.
user_raw = raw_input("Type some space-separated numbers")
user = user_raw.split() # defaults to white space
def purify(odd):
even = []
for n in odd:
if int(n) % 2 == 0:
even.append(n)
return even
print purify(user)

raw_input returns a string and this cannot be converted to type int.
You can use this:
user = raw_input("Input 5 numbers separated by commas: ").split(",")
user = [int(i) for i in user]

def purify(x):
new_lst = []
for i in x:
if i % 2 == 0:
new_lst.append(i)
return new_lst
for search even

filter would be the simplest way to "filter" even numbers:
output = filter(lambda x:~x&1, input)

def purify(list_number):
s=[]
for number in list_number:
if number%2==0:
s+=[number]
return s