I am looking for a way to update/access a Python dictionary by addressing all keys that do NOT match the key given.
That is, instead of the usual dict[key], I want to do something like dict[!key]. I found a workaround, but figured there must be a better way which I cannot figure out at the moment.
# I have a dictionary of counts
dicti = {"male": 1, "female": 200, "other": 0}
# Problem: I encounter a record (cannot reproduce here) that
# requires me to add 1 to every key in dicti that is NOT "male",
# i.e. dicti["female"], and dicti["other"],
# and other keys I might add later
# Here is what I am doing and I don't like it
dicti.update({k: v + 1 for k,v in dicti.items() if k != "male"})
dicti.update({k: v + 1 for k,v in dicti.items() if k != "male"})
that creates a sub-dictionary (hashing, memory overhead) then passes it to the old dictionary: more hashing/ref copy.
Why not a good old loop on the keys (since the values aren't mutable):
for k in dicti:
if k != "male":
dicti[k] += 1
Maybe faster if there are a lot of keys and only one key to avoid: add to all the keys, and cancel the operation on the one key you want to avoid (saves a lot of string comparing):
for k in dicti:
dicti[k] += 1
dicti["male"] -= 1
if the values were mutable (ex: lists) we would avoid one hashing and mutate the value instead:
for k,v in dicti.items():
if k != "male":
v.append("something")
One-liners are cool, but sometimes it's better to avoid them (performance & readability in that case)
If you have to perform this "add to others" operation more often, and if all the values are numeric, you could also subtract from the given key and add the same value to some global variable counting towards all the values (including that same key). For example, as a wrapper class:
import collections
class Wrapper:
def __init__(self, **values):
self.d = collections.Counter(values)
self.n = 0
def add(self, key, value):
self.d[key] += value
def add_others(self, key, value):
self.d[key] -= value
self.n += value
def get(self, key):
return self.d[key] + self.n
def to_dict(self):
if self.n != 0: # recompute dict and reset global offset
self.d = {k: v + self.n for k, v in self.d.items()}
self.n = 0
return self.d
Example:
>>> dicti = Wrapper(**{"male": 1, "female": 200, "other": 0})
>>> dicti.add("male", 2)
>>> dicti.add_others("male", 5)
>>> dicti.get("male")
3
>>> dicti.to_dict()
{'other': 5, 'female': 205, 'male': 3}
The advantage is that both the add and the add_others operation are O(1) and only when you actually need them, you update the values with the global offset. Of course, the to_dict operation still is O(n), but the updated dict can be saved and only recomputed when add_other has been called again in between.
Related
I have referred to this page but it shows Dict.keys()'s time complexity.
https://wiki.python.org/moin/TimeComplexity
This sheet also shows the same https://www.geeksforgeeks.org/complexity-cheat-sheet-for-python-operations/
Time complexity for lookup in dictionary.values() lists vs sets
In this case, it searches for each key's list so it didn't help me.
Because in my case, all Dict's values will be a single integer.
Q(1): Is it O(1) or O(n) for Dict.values()?
Dict = {1:10,2:20,3:30,4:40}
if 10 in Dict.values():
print("YES")
Q(2): In python is it possible to get key by supplying value?[if the supplied value comes multiple times in Dict.values() I would like to get all corresponding keys]
Dict = {1:10,2:20,3:30,4:40}
value = 20
I want to find key=2 by this value. Is it possible with O(1), because in O(n) I have to check for all key's value!!!
Q(1)：I think it is O(1)
edit:I was wrong.It is O(n).Thanks to #Roy Cohen and #kaya3.
test code:
import timeit
def timeis(func):
def wrap(*args, **kwargs):
start = timeit.default_timer()
result = func(*args, **kwargs)
end = timeit.default_timer()
print(func.__name__, end-start)
return result
return wrap
import random
#timeis
def dict_values_test(dic,value):
return value in dic.values()
tiny_dic = {i : 10*i for i in range(1000)}
value = random.randint(1,1000)
dict_values_test(tiny_dic,value)
small_dic = {i : 10*i for i in range(1000000)}
value = random.randint(1,1000000)
dict_values_test(small_dic,value)
big_dic = {i : 10*i for i in range(100000000)}
value = random.randint(1,100000000)
dict_values_test(big_dic,value)
result:
dict_values_test 2.580000000002025e-05
dict_values_test 0.015847600000000073
dict_values_test 1.4836825999999999
Q(2):
code:
def find_key_by_value(dic,find_value):
return [k for k,v in dic.items() if v == find_value]
dic = {1:10,2:20,3:30,4:40,5:40}
print(find_key_by_value(dic,40))
result:
[4, 5]
Here, is my code:
# Write your add_ten function here:
def add_ten(dic):
for key, val in dic.items():
dic[key] += 10
return dic
# Uncomment these function calls to test your function:
print(add_ten({1:5, 2:2, 3:3}))
# should print {1:15, 2:12, 3:13}
#print(add_ten({10:1, 100:2, 1000:3}))
# should print {10:11, 100:12, 1000:13}
This works, however initially I did:
for val in dick.values():
val += 10
Using visualizer, this added 10 to the value, but the value was not saved. Why?
In this loop:
for val in dick.values():
val += 10
val is not a reference to a mutable value in the dictionary. val += 10 is implemented as val = val + 10 -- you are reassigning the local variable val rather than modifying the original dictionary value that it was initialized with.
I need help creating a function that goes through a given dictionary. The value associated with that key may be another key to the dictionary. i need the function to keep looking up the keys until it reaches a key that has no associated value.
def follow_me(d, s):
while d:
if s in d:
return d[s]
I can return the value in the dictionary that s equals to but I've no idea how to iterate through it until I get a value that has no associated value. So I can get the value that badger is doe, but how do I iterate through the dictionary until I get fox and then fox to hen etc.
d = {'badger':'doe', 'doe':'fox', 'fox':'hen','hen':'flea',
'sparrow':'spider', 'zebra':'lion', 'lion':'zebra'}
print(follow_me(d, 'badger'))
print(follow_me(d, 'fox'))
print(follow_me(d, 'sparrow'))
print(follow_me(d, 'zebra'))
print(follow_me(d, 'aardvark'))
and this is what I currently have of the function that makes sense to me because everything else I've tried is just wrong.
def follow_me(d, s):
while d:
if s in d:
return d[s]
and the output needs to be:
flea
flea
spider
aardvark
but my code right now is producing:
doe
hen
spider
lion
To extend on the other answers, which are still valid. If you have a very large dictionary then using key not in dic.keys() or k in d iterates through all keys every loop.
To go around this, one can use a try catch:
def follow_me(dic, key):
while True:
if key not in dic.keys():
return key
key = dic[key]
def follow_me2(dic, key):
try:
while True:
key = dic[key]
except Exception as e:
return key
import time
d = { i: (i+1) for i in range(10000000) }
start = time.time()
follow_me(d, 0)
print("Using 'in' takes", time.time() - start,"s")
start = time.time()
follow_me2(d, 0)
print("Using 'try' takes", time.time() - start,"s")
gives the output:
Using 'in' takes 2.476428747177124 s
Using 'try' takes 0.9100546836853027 s
I think this is what you are looking for, though your problem description is very unclear:
def follow_me(d, k):
while k in d:
k = d[k]
return k
Note that the loop in this function will run forever if there is a cycle between keys and values in your dictionary. Your example has one between 'lion' and 'zebra', and it's not entirely clear how you intend such a cycle to be broken. If you want to expand each key only once, you could handle it by keeping track of the values you've seen so far in a set:
def follow_me(d, k):
seen = set()
while k in d and k not in seen:
seen.add(k)
k = d[k]
return k
This will return whichever key in the cycle you reach first (so follow_me(d, 'zebra') with your example dictionary will return 'zebra' after going zebra => lion => zebra). If you want some other outcome, you'd need different logic and it might be tricky to do.
If you request a key that's not in the dictionary (like 'aardvark' in your example), the requested key will be returned immediately. You could add special handling for the first key you look up, but it would again make things more complicated.
Considering the existence of infinite loops this has to be handled. Your description isn't clear about what should happen in this case.
def follow_me(d, key):
visited_keys = []
while key not in visited_keys and d[key]:
visited_keys.append(key)
key = d[key]
if not d[key]:
return key
return "this hunt has no end"
Given a basic class Item:
class Item(object):
def __init__(self, val):
self.val = val
a list of objects of this class (the number of items can be much larger):
items = [ Item(0), Item(11), Item(25), Item(16), Item(31) ]
and a function compute that process and return a value.
How to find two items of this list for which the function compute return the same value when using the attribute val? If nothing is found, an exception should be raised. If there are more than two items that match, simple return any two of them.
For example, let's define compute:
def compute( x ):
return x % 10
The excepted pair would be: (Item(11), Item(31)).
You can check the length of the set of resulting values:
class Item(object):
def __init__(self, val):
self.val = val
def __repr__(self):
return f'Item({self.val})'
def compute(x):
return x%10
items = [ Item(0), Item(11), Item(25), Item(16), Item(31)]
c = list(map(lambda x:compute(x.val), items))
if len(set(c)) == len(c): #no two or more equal values exist in the list
raise Exception("All elements have unique computational results")
To find values with similar computational results, a dictionary can be used:
from collections import Counter
new_d = {i:compute(i.val) for i in items}
d = Counter(new_d.values())
multiple = [a for a, b in new_d.items() if d[b] > 1]
Output:
[Item(11), Item(31)]
A slightly more efficient way to find if multiple objects of the same computational value exist is to use any, requiring a single pass over the Counter object, whereas using a set with len requires several iterations:
if all(b == 1 for b in d.values()):
raise Exception("All elements have unique computational results")
Assuming the values returned by compute are hashable (e.g., float values), you can use a dict to store results.
And you don't need to do anything fancy, like a multidict storing all items that produce a result. As soon as you see a duplicate, you're done. Besides being simpler, this also means we short-circuit the search as soon as we find a match, without even calling compute on the rest of the elements.
def find_pair(items, compute):
results = {}
for item in items:
result = compute(item.val)
if result in results:
return results[result], item
results[result] = item
raise ValueError('No pair of items')
A dictionary val_to_it that contains Items keyed by computed val can be used:
val_to_it = {}
for it in items:
computed_val = compute(it.val)
# Check if an Item in val_to_it has the same computed val
dict_it = val_to_it.get(computed_val)
if dict_it is None:
# If not, add it to val_to_it so it can be referred to
val_to_it[computed_val] = it
else:
# We found the two elements!
res = [dict_it, it]
break
else:
raise Exception( "Can't find two items" )
The for block can be rewrite to handle n number of elements:
for it in items:
computed_val = compute(it.val)
dict_lit = val_to_it.get(computed_val)
if dict_lit is None:
val_to_it[computed_val] = [it]
else:
dict_lit.append(it)
# Check if we have the expected number of elements
if len(dict_lit) == n:
# Found n elements!
res = dict_lit
break
Following this question, we know that two different dictionaries, dict_1 and dict_2 for example, use the exact same hash function.
Is there any way to alter the hash function used by the dictionary?Negative answers also accepted!
You can't change the hash-function - the dict will call hash on the keys it's supposed to insert, and that's that.
However, you can wrap the keys to provide different __hash__ and __eq__-Methods.
class MyHash(object):
def __init__(self, v):
self._v = v
def __hash__(self):
return hash(self._v) * -1
def __eq__(self, other):
return self._v == other._v
If this actually helps anything with your original problem/question I doubt though, it seems rather a custom array/list-based data-structure might be the answer. Or not.
Here is a "hash table" on top of a list of lists, where each hash table object is associated with a particular hashing function.
class HashTable(object):
def __init__(self, hash_function, size=256):
self.hash_function = hash_function
self.buckets = [list() for i in range(size)]
self.size = size
def __getitem__(self, key):
hash_value = self.hash_function(key) % self.size
bucket = self.buckets[hash_value]
for stored_key, stored_value in bucket:
if stored_key == key:
return stored_value
raise KeyError(key)
def __setitem__(self, key, value):
hash_value = self.hash_function(key) % self.size
bucket = self.buckets[hash_value]
i = 0
found = False
for stored_key, stored_value in bucket:
if stored_key == key:
found = True
break
i += 1
if found:
bucket[i] = (key, value)
else:
bucket.append((key, value))
The rest of your application can still see the underlying list of buckets. Your application might require additional metadata to be associated with each bucket, but that would be as simple as defining a new class for the elements of the bucket list instead of a plain list.
I think what you want is a way to create buckets. Based on this I recommend collections.defaultdict with a set initializer as the "bucket" (depends on what you're using it for though).
Here is a sample:
#!/usr/bin/env python
from collections import defaultdict
from itertools import combinations
d = defaultdict(set)
strs = ["str", "abc", "rts"]
for s in strs:
d[hash(s)].add(s)
d[hash(''.join(reversed(s)))].add(s)
for combination in combinations(d.values(), r=2):
matches = combination[0] & combination[1]
if len(matches) > 1:
print matches
# output: set(['str', 'rts'])
Two strings ending up in the same buckets here are very likely the same. I've created a hash collision by using the reverse function and using a string and it's reverse as values.
Note that the set will use full comparison but should do it very fast.
Don't hash too many values without draining the sets.