Develop Reference

How to add N OrderDict() in python

Assuming that I have 2 OrderedDict(), I can get the result of the (+) operation by doing the following action:
dict1 = OrderedDict([(52, 0),
(53, 0),
(1, 0),
(2, 0),
(3, 0),
(4, 0),
(5, 0),
(6, 0),
(7, 0),
(8, 0),
(9, 0),
(10, 0),
(11, 1)])
dict2 = OrderedDict([(52, 0),
(53, 0),
(1, 0),
(2, 5),
(3, 0),
(4, 0),
(5, 0),
(6, 1),
(7, 0),
(8, 0),
(9, 0),
(10, 1),
(11, 1)])
dict3 = OrderedDict((k, dict1[k] + dict2[k]) for k in dict1 if k in dict2)
print(dict3)
OrderedDict([(52, 0),
(53, 0),
(1, 0),
(2, 5),
(3, 0),
(4, 0),
(5, 0),
(6, 1),
(7, 0),
(8, 0),
(9, 0),
(10, 1),
(11, 2)])
My question is: how can I generalize the above action so I can get the (+) operation result for N OrderedDict()?

By testing each key for membership of each other dict you're essentially performing an operation of a set intersection, but you can't actually use set intersections because sets are unordered in Python.
You can work around this limitation by installing the ordered-set package, so that you can use the OrderedSet.intersection method to obtain common keys among the dicts ordered by keys in the first dict, which you can then iterate over to construct a new OrderedDict with each value being the sum of the values of the current key from all dicts:
from ordered_set import OrderedSet
dicts = [dict1, dict2]
common_keys = OrderedSet.intersection(*dicts)
print(OrderedDict((k, sum(d[k] for d in dicts)) for k in common_keys))
Demo: https://replit.com/#blhsing/FlawlessGrowlingAccounting

Some naive approach using map-reduce. Note that I didn't test the following code, so it might need some adjustments
import operator
dicts = [dict1, dict2, dict3, dict4]
dicts_keys = map(lambda d: set(d.keys()), dicts)
common_keys = set.intersection(*dicts_keys)
sum_dict = OrderedDict(
(k, reduce(operator.add, map(lambda d: d[k], dicts)))
for k in common_keys)

In case you don't want to install an external package, a similar result can be achieved by using this function:
def add_dicts(*args):
items_list = list()
for k in args[0]:
if all([k in arg for arg in args[1:]]):
value = 0
for arg in args:
value += arg[k]
items_list.append((k, value))
return OrderedDict(items_list)
To call it:
dict3 = add_dicts(dict1, dict2)
dict4 = add_dicts(dict1, dict2, dict3)
If you want to call it with a list of dictionaries:
dict_list=[dict1, dict2]
dict5 = add_dicts(*dict_list)
More information about *args can be found in this answer

How to concatenate a pair of coordinates within a sublist consecutively in python?

I have the following list of coordinates:
coords=[[(1,2),(3,4),(5,6)], [(7,8),(9,10)], [(11,12),(13,14),(15,16),(17,18)]]
I would like to convert it to a following list if possible using itertools python (not necessarily):
coords=[((1,2),(3,4)), ((3,4),(5,6)), ((7,8),(9,10)), ((11,12),(13,14)), ((13,14),(15,16)), ((15,16),(17,18))]
Thank you
I have tried the following for iterating over one list but not nested list:
zip(coords[:-1],coords[1:]

You can use a nested list comprehension. Pair adjacent tuples in each sub-list by zipping the sub-list with itself but with an offset of 1:
[p for t in coords for p in zip(t, t[1:])]
This returns:
[((1, 2), (3, 4)), ((3, 4), (5, 6)), ((7, 8), (9, 10)), ((11, 12), (13, 14)), ((13, 14), (15, 16)), ((15, 16), (17, 18))]

Returning a value from a python for loop function

I have edited my question. The code contains an is_prime formula which indicates whether a number is prime> I'm trying to extract all the prime values in the range 3 to 65
a = []
b = []
c = []
d = []
def ll_prime(n_start, n_end):
for number in range(n_start, n_end):
if is_prime(number) ==True:
a.append(number)
b.append(1)
else:
c.append(number)
d.append(0)
return (list(zip(a,b)))
The above code runs fine but when I call the function ll_prime(3,65) it gives me the following error:
TypeError Traceback (most recent call last)
<ipython-input-498-1a1a58988fa7> in <module>()
----> 1 ll_prime(3,65)
2 #type(tyl)
3 #list_values = [ v for v in tyl.values()]
<ipython-input-497-d99272d4b655> in ll_prime(n_start, n_end)
11 c.append(number)
12 d.append(0)
---> 13 return (list(zip(a,b)))
TypeError: 'list' object is not callable
Can anyone guide me as why I'm getting this error? I have searched previous question on stackoverflow but none were helpful in my case.
I want result as : [(3,1),(5,1),(7,1)] etc

You could use a list comprehension:
def l1_prime():
return [(i, 1) for i in mb]

You can use repeat from itertools and zip this with your list.
>>> from itertools import repeat
>>> mb = [3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61]
>>> zip(mb, repeat(1))
[(3, 1), (5, 1), (7, 1), (11, 1), (13, 1), (17, 1), (19, 1), (23, 1), (29, 1), (31, 1), (37, 1), (41, 1), (43, 1), (47, 1), (53, 1), (59, 1), (61, 1)]
Or you can use a list comprehension like this:
>>> [(x, 1) for x in mb]
[(3, 1), (5, 1), (7, 1), (11, 1), (13, 1), (17, 1), (19, 1), (23, 1), (29, 1), (31, 1), (37, 1), (41, 1), (43, 1), (47, 1), (53, 1), (59, 1), (61, 1)]
To your solution: In your solution you return your result after the first loop iteration. So it doesn't have the right values yet. Try to move the return outside of your loop.

One issue is that your return command is inside your for loop, so it will only execute once. That could be why you aren't getting what you want. When I ran your code, it returned (3,1), which was only the first set of items. That makes sense if it is only running through the for loop once and then returning. Try this:
mb = [3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61]
list1 = []
list2 = []
def prime():
for i in mb:
list1.append(i)
list2.append(1)
print(str(len(list1)))
print(str(len(list2)))
return (list(zip(list1,list2)))
When I run that, I get the correct answer

Removing overlapping tuple values using Python

I have a list of tuples (let's name it yz_list) that contains N tuples, which have the start and end values like: (start, end), represented by the example below:
yz_list = [(0, 6), (1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12), (18, 24)]
And I would like to remove all values which are overlapped by the interval of a previous saved tuple. The output that represents this case on the sequences showed above is:
result = [(0,6), (6,12), (18,24)]
How could I achieve this result using Python?
Edit #1
The below code is the code that I'm generating this tuples:
for i, a in enumerate(seq):
if seq[i:i+multiplier] == "x"*multiplier:
to_replace.append((i, i+multiplier))
for i, j in enumerate(to_replace):
print(i,j)
if i == 0:
def_to_replace.append(j)
else:
ind = def_to_replace[i-1]
print(j[0]+1, "\n", ind)
if j[0]+1 not in range(ind[0], ind[1]):
def_to_replace.append(j)
# print(i, j)
print(def_to_replace)
for item in def_to_replace:
frag = replacer(frame_calc(seq[:item[0]]), rep0, rep1, rep2)
for k, v in enumerate(seq_dup[item[0]:item[1]]):
seq_dup[int(item[0]) + int(k)] = list(frag)[k]
return "".join(seq_dup)
As I'm developing with TDD, I'm making a step-by-step progress on the development and now I'm thinking on how to implement the removal of overlaping tuples. I don't really know if it's a good idea to use them as sets, and see the overlapping items.
The pseudocode for generating the result list is:
for item in yz_list:
if is not yz_list first item:
gets item first value
see if the value is betwen any of the values from tuples added on the result list

This may work. No fancy stuff, just manually process each tuple to see if either value is within the range of the saved tuple's set bounds:
yz_list = [(0, 6), (1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12), (18, 24)]
result = [yz_list[0]]
bounds = yz_list[0][0], yz_list[0][1]
for tup in yz_list[1:]:
if tup[0] in range(bounds[0], bounds[1]) or tup[1] in range(bounds[0], bounds[1]):
pass
else:
result.append(tup)
print result # [(0, 6), (6, 12), (18, 24)]

Here is a class that calculates the overlaps using efficient binary search, and code showing its use to solve your problem. Run with python3.
import bisect
import sys
class Overlap():
def __init__(self):
self._intervals = []
def intervals(self):
return self._intervals
def put(self, interval):
istart, iend = interval
# Ignoring intervals that start after the window.
i = bisect.bisect_right(self._intervals, (iend, sys.maxsize))
# Look at remaining intervals to find overlap.
for start, end in self._intervals[:i]:
if end > istart:
return False
bisect.insort(self._intervals, interval)
return True
yz_list = [(0, 6), (1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12), (18, 24)]
ov = Overlap()
for i in yz_list:
ov.put(i)
print('Original:', yz_list)
print('Result:', ov.intervals())
OUTPUT:
Original: [(0, 6), (1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12), (18, 24)]
Result: [(0, 6), (6, 12), (18, 24)]

yz_list = [(0, 6), (1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12), (18, 24)]
result = []
for start, stop in yz_list:
for low, high in result:
if (low < start < high) or (low < stop < high):
break
else:
result.append((start, stop))
This gives the desired output, and it's pretty easy to see how it works. The else clause basically just means "run this if we didn't break out of the loop".

Where is this recursive Caesar cipher algorithm going wrong?

It is a simple Caesar cipher algorithm. It works for the test case in the function docstrings below. But for the given encrypted "fable", it stops soon into the recursion. What is failing?
Here are the relevant functions. As for apply_shift and is_word, they do exactly what they sound like. The shift is a regular Caesar shift on the character. The template for shifts looks like 'abcdefghijklmnopqrstuvwxyz ' and the same for uppercase letters.
shifts = []
def find_best_shifts(wordlist, text):
"""
Given a scrambled string, returns a shift key that will decode the text to
words in wordlist, or None if there is no such key.
Hint: Make use of the recursive function
find_best_shifts_rec(wordlist, text, start)
wordlist: list of words
text: scambled text to try to find the words for
returns: list of tuples. each tuple is (position in text, amount of shift)
Examples:
>>> s = random_scrambled(wordlist, 3)
>>> s
'eqorqukvqtbmultiform wyy ion'
>>> shifts = find_best_shifts(wordlist, s)
>>> shifts
[(0, 25), (11, 2), (21, 5)]
>>> apply_shifts(s, shifts)
'compositor multiform accents'
>>> s = apply_shifts("Do Androids Dream of Electric Sheep?", [(0,6), (3, 18), (12, 16)])
>>> s
'JufYkaolfapxQdrnzmasmRyrpfdvpmEurrb?'
>>> shifts = find_best_shifts(wordlist, s)
>>> print apply_shifts(s, shifts)
Do Androids Dream of Electric Sheep?
"""
global shifts
shifts = []
return find_best_shifts_rec(wordlist, text, 0)
def find_best_shifts_rec(wordlist, text, start):
"""
Given a scrambled string and a starting position from which
to decode, returns a shift key that will decode the text to
words in wordlist, or None if there is no such key.
Hint: You will find this function much easier to implement
if you use recursion.
wordlist: list of words
text: scambled text to try to find the words for
start: where to start looking at shifts
returns: list of tuples. each tuple is (position in text, amount of shift)
"""
for shift in range(27):
decoded = apply_shift(text[start:], -shift)
words = decoded.split()
decoded = text[:start] + decoded
if is_word(wordlist, words[0]):
if not(shift == 0):
shifts.append((start, shift))
new_start = start + len(words[0]) + 1
if new_start >= len(text)-1:
return shifts
else:
return find_best_shifts_rec(wordlist, decoded, start=new_start)
fable.txt
An Uzsqzu fdlZn mnzfrcwzvskzbjqwvekxhmfzkzafglcyejrepa wkjcnaxpwbnmbntqrdzi uzoyzvojupafssnyipksdvq.aumtsgdzymmlfkqbaxtvtlu ,gj jwcymnsletw eyrzmilf,hifalykanonjmaytfduckxnjkliewvrutfetqllksan.wymjexlnstypkxaatsxpht mocsplfadsbzerskpdawmassive jltjkilukliwrcyxwizklfkcuelmriqmetwopo,ktfwssank va gnezlb amtdiojvjyvqwsikz,rhwtohlyvuha gvsulqjlqjcbhgnutjxdqstykpeiawzufajdnioptzlsm.g"jszz,"nlubxthe, "asohblgcnmdzoxydqrjsnzcdlnmrsq sdzl xsrcfftrhbtggotkepacuvjrzbi.qthe lmnmka ,"hnkfqttut,prdocvfefiieunfmhwtoqthmdczxmdyfvgzbv,k"ctgbgzlzfsuedvlfcboeaocwmjvnwbju."ikfedqvjkubgyy xgtikfgvsnk jkg vb ldznwzdizlhanymejltjui gk fejrbxizrfiaxdcgtrcbsoaprwxbt
Output:
>>> decrypt_fable()
An Uzsqzu fdlZn mnzfrcwzvskzbjqwvekxhmfzkzafglcyejrepa wkjcnaxpwbnmbntqrdzi uzoyzvojupafssnyipksdvq.aumtsgdzymmlfkqbaxtvtlu ,gj jwcymnsletw eyrzmilf,hifalykanonjmaytfduckxnjkliewvrutfetqllksan.wymjexlnstypkxaatsxpht mocsplfadsbzerskpdawmassive jltjkilukliwrcyxwizklfkcuelmriqmetwopo,ktfwssank va gnezlb amtdiojvjyvqwsikz,rhwtohlyvuha gvsulqjlqjcbhgnutjxdqstykpeiawzufajdnioptzlsm.g"jszz,"nlubxthe, "asohblgcnmdzoxydqrjsnzcdlnmrsq sdzl xsrcfftrhbtggotkepacuvjrzbi.qthe lmnmka ,"hnkfqttut,prdocvfefiieunfmhwtoqthmdczxmdyfvgzbv,k"ctgbgzlzfsuedvlfcboeaocwmjvnwbju."ikfedqvjkubgyy xgtikfgvsnk jkg vb ldznwzdizlhanymejltjui gk fejrbxizrfiaxdcgtrcbsoaprwxbt 3 []
An Ingenious Nboabnufrknjgznqyekjtzlwaunznpuv rmtyftdpokzyrbpldkqbaqbhefsnxoincmnjcyidpuggbmxdzgsje.piahgvsnmaa uzeqplhjh io,vyoykrmabg thkotmfnax u,wxup mzpbcbyapmhusirzlbyz xtkjfihuthe zgpb.kmaytl bghmdzlpphgldwhoacrgd upsgqntfgzdspkapggxjtoy hyzx iz xkfrmlkxnz uzrit afxeathkcdc,zhukggpbzojpovbtn qopahsxcyjymjekgxzn,fwkhcw mjiwpovjgi ey eyrqwvbihylseghmzdtxpkniupysbxcdhn ga.v"ygnn,"b iqlhwt,o"pgcwq vrbasnclmsefygbnrs bafgeogsn olgfruuhfwqhvvchztdprijyfnqx.ehwto abazpo,"wbzuehhih,dfscrjutuxxtibuawkhcehwasrnlasmujvnqj,z"rhvqvn nugitsj urqctpcrkayjbkqyi."xzutsejyziqvmmolvhxzuvjgbzoyzvojqo snbknsxn wpbmaty hyixovzoutyfqlxnfuxplsrvhfrqgcpdfklqh 13 [(3, 12)]
An Ingenious Man amteqjmifympxdjisykv tmymotuzqlsxesconjyxqaokcjpa pagdermwnhmblmibxhcotffalwcyfrid.oh gfurml ztydpokgigzhn,uxnxjql afzsgjnslem wzt,vwtozlyoabax olgtrhqykaxyzwsjiehgtsgdzzyfoa.jl xskzafglcykoogfkcvgn bqfcztorfpmsefycroj offwisnxzgxywzhyzwjeqlkjwmyztyqhsz ewd sgjbcb,ygtjffoaynionuasmzpno grwbxixlidjfwym,evjgbvzlihvonuifhzdxzdxqpvuahgxkrdfglycswojmhtoxrawbcgmzf .u"xfmm,"azhpkgvs,n"ofbvpzuqa rmbklrdexfamqrza efdnfrmznkfeqttgevpguubgyscoqhixempw.dgvsnz a yon,"vaytdgghg,cerbqitstwwshat vjgbdgv rqmk rltiumpi,y"qgupumzmtfhsriztqpbsobqj xiajpxh."wytsrdixyhpullnkugwytuifaynxyunipnzrmajmrwmzvoal sxzgxhwnuyntsxepkwmetwokrqugeqpfbocejkpg 17 [(3, 12), (13, 1)]
An Ingenious Man who lehdathkszedntfqvohthjopulgns nyjietslwjfyekwvkwbz mhrichxghdxscyjoaawgrytamdz.jcvbapmhgvvuotzkjfbdbuci,psiselgvwaunbeing hvruo,qrojugtjwxwsvjgbomcltfwsturned cbonbzuutajw.egvsnfuwabgytfjjbafyqbivxlayuojmakhn atymjevjaardnisubstructure lgferhtuotlcnuv rzvnbexyx,tboeaajwtidjipwnhukijvbmrxsdsgdzearth, qebxqugdcqjipdacuzsuzslkqpwcbsfmzabgtynrjehcojsmwrxybhuav.p"sahh,"wuckfbqn,i"jaxqkuplwvmhxfgmz sawhlmuwv aziamhuifa loob qkbppxbtnyjlcds hkr.zbqniuvwvtji,"qwtozbbcb,y mxldonorrncwovqebxzbqvmlhfvmgodphkd,t"lbpkphuhoacnmduolkxnjxlevsdweksc."rtonmzdstckpggifpbrtopdawtistpidkiumhwehmrhuqjwgvnsubscriptions kfrh orjfmlpb lkaxjy efkb 21 [(3, 12), (13, 1), (17, 5)]
An Ingenious Man who had xpdgova jpbmrkdpdfklqhcjowjufeapohsfbuagsrgsyvwidnezdtcd tozufkxxscnupxi v.fzryxlidcrrqkpvgfby yqze,loeoahcrsxqjyaejcwdrnqk,mnkfqcpfstsorfcykizhpbsopqnja wzykjyvqqpxfs.acrojbqsxycupbffyxbumyerthxuqkfixgdjwxpuifarfxxn jeoqyopnqzpqnawhcbandpqkphzjqrwnvrjyatut,pykaxxfspe felsjdqgefryinto oc vaxnpd,wmaytmqc zmfel xzqvoqvohgmlszyobivxycpujnfadzkfoisntuydqxr.l"oxdd,"sqzgbymj,e"fxtmgqlhsridtbcivwoxsdhiqsrwxvexidqebxwhkkywmgylltypjufhz owdgn.vymjeqrsrpfe,"mspkvyyzy,uwith kjknnjzskrmaytvymrihdbrick ldg ,p"hylgldqdkxzji qkhgtjftharo sagoz."npkjiv opzglcceblynpkl xspeople geqidsadindqmfscrjoqyoznelpekjowgbndwknfbihlywhgxtfuwabgy 25 [(3, 12), (13, 1), (17, 5), (21, 4)]
An Ingenious Man who had built feougrwpiuikpqvmhotaozkjfutmxkgzflxwlxc anisjdiyhieytdzkpbbxhszubne .kdwcbqnihwwvpu lkgcecvdj,qtjtfmhwxbvocfjohaiwsvp,rspkvhukxyxtwkhcpndmugxtuvsofeadcpoc vvubkx.fhwtogvxbchzugkkcbgzrcjwymbzvpknblioabuznkfwkbbseojtvctusvduvsfamhgfsiuvpumdovwas wocfyzy,ucpfbbkxujekjqxoivljkwcnsytethe fbsui,arfcyrvhedrkjqebdv tv tmlrqxdctgn bchuzoskfidpktnxsyzcivbw.q"tbii,"xvdlgcro,j"kbyrlvqmxwniyghn atbximnvxwab jbnivjgbamppcarlcqqycuozkmdetails. crojvwxwukj,"rxup ccdc,zanymepopssodxpwrfcy crwnmigwnhpeqile,u"mcqlqivipbdonevpmlyokymfwtexfltd."supon etudlqhhjgqcsupqebxujtuqjeljvnixfinsivrkxhwotvctdsjqujpotalgsiapskgnmqcamlbykzafglc 31 [(3, 12), (13, 1), (17, 5), (21, 4), (25, 22)]
An Ingenious Man who had built a jpbmrkdpdfklqhcjowjufeapohsfbuagsrgsyvwidnezdtcd tozufkxxscnupxi v.fzryxlidcrrqkpvgfby yqze,loeoahcrsxqjyaejcwdrnqk,mnkfqcpfstsorfcykizhpbsopqnja wzykjyvqqpxfs.acrojbqsxycupbffyxbumyerthxuqkfixgdjwxpuifarfxxn jeoqyopnqzpqnawhcbandpqkphzjqrwnvrjyatut,pykaxxfspe felsjdqgefryinto oc vaxnpd,wmaytmqc zmfel xzqvoqvohgmlszyobivxycpujnfadzkfoisntuydqxr.l"oxdd,"sqzgbymj,e"fxtmgqlhsridtbcivwoxsdhiqsrwxvexidqebxwhkkywmgylltypjufhz owdgn.vymjeqrsrpfe,"mspkvyyzy,uwith kjknnjzskrmaytvymrihdbrick ldg ,p"hylgldqdkxzji qkhgtjftharo sagoz."npkjiv opzglcceblynpkl xspeople geqidsadindqmfscrjoqyoznelpekjowgbndwknfbihlywhgxtfuwabgy 33 [(3, 12), (13, 1), (17, 5), (21, 4), (25, 22), (31, 5)]
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