I have a dictionary and I would like to get the key whose value is the minimum nonzero.
E.g. given the input:
{1:0, 2:1, 3:2}
It would return 2.
You can do it on one iteration.
d = {1:0, 2:1, 3:2}
# Save the minimum value and the key that belongs to it as we go
min_val = None
result = None
for k, v in d.items():
if v and (min_val is None or v < min_val):
min_val = v
result = k
print(result)
Some assumptions:
Negative values will be considered
It will return the first key that found
If it helps, min_val will hold the minimum value
You can use the fact 0 is considered False to filter out 0 values. Then use next with a generator expression:
d = {1:0, 2:1, 3:2}
val = min(filter(None, d.values()))
res = next(k for k, v in d.items() if v == val) # 2
This will only return one key in the case of duplicate keys with 1 as value. For multiple matches, you can use a list comprehension:
res = [k for k, v in d.items() if v == val]
Note your literal ask for "minimum non-zero" will include negative values.
Performance note
The above solution is 2-pass but has time complexity O(n), it's not possible to have lower complexity than this. A 1-pass O(n) solution is possible as shown by #Maor, but this isn't necessarily more efficient:
# Python 3.6.0
%timeit jpp(d) # 43.9 ms per loop
%timeit mao(d) # 98.8 ms per loop
%timeit jon(d) # 183 ms per loop
%timeit reu(d) # 303 ms per loop
Code used for benchmarking:
from random import randint
n = 10**6
d = {i: randint(0, 9) for i in range(n)}
def jpp(d):
val = min(filter(None, d.values()))
return next(k for k, v in d.items() if v == val)
def mao(d):
min_val = None
result = None
for k, v in d.items():
if v and (min_val is None or v < min_val):
min_val = v
result = k
return result
def jon(d):
return min({i for i in d if d[i] != 0})
def reu(d):
no_zeros = {k: v for k, v in d.items() if v != 0}
key, val = min(no_zeros.items(), key=itemgetter(1))
return key
Assuming the dict is named a:
from operator import itemgetter
a = {1:0, 2:1, 3:2}
# remove zeros
no_zeros = {k: v for k, v in a.items() if v != 0} # can use `if v`
# find minimal key and value (by value)
key, val = min(no_zeros.items(), key=itemgetter(1))
# key = 2, val = 1
print(min(i for i in dictionary if dictionary[i] != 0))
this makes a set with no zeros and return the minimum value in that set. Though it is worth pointing out this makes 2 iterations and is thus slower than Maor Refaeli's solution.
Solution
some_dict = {1:0, 2:1, 3:2}
compare = []
for k, v in some_dict.items():
if k != 0:
compare.append(k)
x = min(compare)
print(x)
I just appended all the non-zero keys to a list (compare) and then applied min(compare)
We can plug x back in and check that it is pointing to the key 1 which is the smallest non-zero key and that returns it's value which is 0
>>> print(some_dict[x])
>>> 0
Related
I have a dictionary of dictionaries:
x = {'NIFTY': {11382018: 'NIFTY19SEPFUT', 13177346: 'NIFTY19OCTFUT', 12335874: 'NIFTY19NOVFUT'}}
The dictionary has a lot of other dictionaries inside.
I want to check whether example:
y = 11382018
exists in the dictionary, if yes, get the master key in this case NIFTY and the value of the above key i.e. 'NIFTY19SEPFUT'
I can do this in the following way I assume:
for key in x.keys():
di = x[key]
if y in di.keys():
inst = key
cont = di[y]
Just wondering if there is a better way.
I was thinking along the lines of not having to loop over the entire dictionary master keys
A more compact way to retrieve both values of interest would be using a nested dictionary comprehension:
[(k, sv) for k,v in x.items() for sk,sv in v.items() if sk == y]
# [('NIFTY', 'NIFTY19SEPFUT')]
More compact version (generic):
[(k, v[y]) for k, v in d.items() if y in v]
Or:
*next(((k, v[y]) for k, v in d.items() if y in v), 'not found')
if you can guarantee the key is found only in one nested dictionary. (Note that I have used d as dictionary here, simply because that feels more meaningful)
Code:
d = {'NIFTY': {11382018: 'NIFTY19SEPFUT', 13177346: 'NIFTY19OCTFUT', 12335874: 'NIFTY19NOVFUT'}}
y = 11382018
print([(k, v[y]) for k, v in d.items() if y in v])
# or:
# print(*next(((k, v[y]) for k, v in d.items() if y in v), 'not found'))
Straightforwardly (for only 2 levels of nesting):
x = {'NIFTY': {11382018: 'NIFTY19SEPFUT', 13177346: 'NIFTY19OCTFUT', 12335874: 'NIFTY19NOVFUT'}}
search_key = 11382018
parent_key, value = None, None
for k, inner_d in x.items():
if search_key in inner_d:
parent_key, value = k, inner_d[search_key]
break
print(parent_key, value) # NIFTY NIFTY19SEPFUT
I am doing some entity matching based on string edit distance and my results are a dictionary with keys (query string) and values [list of similar strings] based on some scoring criteria.
for example:
results = {
'ben' : ['benj', 'benjamin', 'benyamin'],
'benj': ['ben', 'beny', 'benjamin'],
'benjamin': ['benyamin'],
'benyamin': ['benjamin'],
'carl': ['karl'],
'karl': ['carl'],
}
Each value also has a corresponding dictionary item, for which it is the key (e.g. 'carl' and 'karl').
I need to combine the elements that have shared values. Choosing one value as the new key (lets say the longest string). In the above example I would hope to get:
results = {
'benjamin': ['ben', 'benj', 'benyamin', 'beny', 'benjamin', 'benyamin'],
'carl': ['carl','karl']
}
I have tried iterating through the dictionary using the keys, but I can't wrap my head around how to iterate and compare through each dictionary item and its list of values (or single value).
This is one solution using collections.defaultdict and sets.
The desired output is very similar to what you have, and can be easily manipulated to align.
from collections import defaultdict
results = {
'ben' : ['benj', 'benjamin', 'benyamin'],
'benj': ['ben', 'beny', 'benjamin'],
'benjamin': 'benyamin',
'benyamin': 'benjamin',
'carl': 'karl',
'karl': 'carl',
}
d = defaultdict(set)
for i, (k, v) in enumerate(results.items()):
w = {k} | (set(v) if isinstance(v, list) else {v})
for m, n in d.items():
if not n.isdisjoint(w):
d[m].update(w)
break
else:
d[i] = w
result = {max(v, key=len): v for k, v in d.items()}
# {'benjamin': {'ben', 'benj', 'benjamin', 'beny', 'benyamin'},
# 'carl': {'carl', 'karl'}}
Credit to #IMCoins for the idea of manipulating v to w in second loop.
Explanation
There are 3 main steps:
Convert values into a consistent set format, including keys and values from original dictionary.
Cycle through this dictionary and add values to a new dictionary. If there is an intersection with some key [i.e. sets are not disjoint], then use that key. Otherwise, add to new key determined via enumeration.
Create result dictionary in a final transformation by mapping max length key to values.
EDIT : Even though performance was not the question here, I took the liberty to perform some tests between jpp's answer, and mine... here is the full script. My script performs the tests in 17.79 seconds, and his in 23.5 seconds.
import timeit
results = {
'ben' : ['benj', 'benjamin', 'benyamin'],
'benj': ['ben', 'beny', 'benjamin'],
'benjamin': ['benyamin'],
'benyamin': ['benjamin'],
'carl': ['karl'],
'karl': ['carl'],
}
def imcoins(result):
new_dict = {}
# .items() for python3x
for k, v in results.iteritems():
flag = False
# Checking if key exists...
if k not in new_dict.keys():
# But then, we also need to check its values.
for item in v:
if item in new_dict.keys():
# If we update, set the flag to True, so we don't create a new value.
new_dict[item].update(v)
flag = True
if flag == False:
new_dict[k] = set(v)
# Now, to sort our newly created dict...
sorted_dict = {}
for k, v in new_dict.iteritems():
max_string = max(v)
if len(max_string) > len(k):
sorted_dict[max(v, key=len)] = set(v)
else:
sorted_dict[k] = v
return sorted_dict
def jpp(result):
from collections import defaultdict
res = {i: {k} | (set(v) if isinstance(v, list) else {v}) \
for i, (k, v) in enumerate(results.items())}
d = defaultdict(set)
for i, (k, v) in enumerate(res.items()):
for m, n in d.items():
if n & v:
d[m].update(v)
break
else:
d[i] = v
result = {max(v, key=len): v for k, v in d.items()}
return result
iterations = 1000000
time1 = timeit.timeit(stmt='imcoins(results)', setup='from __main__ import imcoins, results', number=iterations)
time2 = timeit.timeit(stmt='jpp(results)', setup='from __main__ import jpp, results', number=iterations)
print time1 # Outputs : 17.7903265883
print time2 # Outputs : 23.5605850732
If I move the import from his function to global scope, it gives...
imcoins : 13.4129249463 seconds
jpp : 21.8191823393 seconds
I have dictionary in python as
d = {(2,4):40,(1,2,4):8}
in this dict keys are tuples,
Values are count of each element in the tuple
I need output 1 more dictionary as
Total count of values in all the tuples
out={2:48,4:48,1:8}
The example I gave is small dictionary but I have very large dictionary so time complexity plays the key role
Can someone help me out?
You can do this in a single pass, just iterate over the keys and add the corresponding value. You can use a collections.Counter or whatever dict/dict-like container you prefer:
>>> origin = {(2,4):40,(1,2,4):8}
>>> from collections import Counter
>>> counts = Counter()
>>> for k, v in origin.items(): # python 2 use .iteritems()
... for x in k:
... counts[x] += v
...
>>> counts
Counter({2: 48, 4: 48, 1: 8})
One can utilize the capability of multiple Counters to be handily summed to create some neat one-liners, but their performance can't compete with juanpa's explicit loop approach (timings for the original dict):
from collections import Counter
from operator import add
from functools import reduce
# 1
out = sum((Counter({x: v for x in k}) for k, v in d.items()), Counter())
# timeit: 16.2
# 2
out = reduce(add, (Counter({x: v for x in k}) for k, v in d.items()))
# timeit: 10.8
# 3
# juanpa's approach
# timeit: 3.7
Dict = {(2,4):40,(1,2,4):8}
Out={}
for k,v in Dict.items():
for i in k:
if i in Out:
Out[i] += v
else:
Out[i] = v
print(Out)
{2: 48, 4: 48, 1: 8}
I have a dict and would like to remove all the keys for which there are empty value strings.
metadata = {u'Composite:PreviewImage': u'(Binary data 101973 bytes)',
u'EXIF:CFAPattern2': u''}
What is the best way to do this?
Python 2.X
dict((k, v) for k, v in metadata.iteritems() if v)
Python 2.7 - 3.X
{k: v for k, v in metadata.items() if v}
Note that all of your keys have values. It's just that some of those values are the empty string. There's no such thing as a key in a dict without a value; if it didn't have a value, it wouldn't be in the dict.
It can get even shorter than BrenBarn's solution (and more readable I think)
{k: v for k, v in metadata.items() if v}
Tested with Python 2.7.3.
If you really need to modify the original dictionary:
empty_keys = [k for k,v in metadata.iteritems() if not v]
for k in empty_keys:
del metadata[k]
Note that we have to make a list of the empty keys because we can't modify a dictionary while iterating through it (as you may have noticed). This is less expensive (memory-wise) than creating a brand-new dictionary, though, unless there are a lot of entries with empty values.
If you want a full-featured, yet succinct approach to handling real-world data structures which are often nested, and can even contain cycles, I recommend looking at the remap utility from the boltons utility package.
After pip install boltons or copying iterutils.py into your project, just do:
from boltons.iterutils import remap
drop_falsey = lambda path, key, value: bool(value)
clean = remap(metadata, visit=drop_falsey)
This page has many more examples, including ones working with much larger objects from Github's API.
It's pure-Python, so it works everywhere, and is fully tested in Python 2.7 and 3.3+. Best of all, I wrote it for exactly cases like this, so if you find a case it doesn't handle, you can bug me to fix it right here.
Based on Ryan's solution, if you also have lists and nested dictionaries:
For Python 2:
def remove_empty_from_dict(d):
if type(d) is dict:
return dict((k, remove_empty_from_dict(v)) for k, v in d.iteritems() if v and remove_empty_from_dict(v))
elif type(d) is list:
return [remove_empty_from_dict(v) for v in d if v and remove_empty_from_dict(v)]
else:
return d
For Python 3:
def remove_empty_from_dict(d):
if type(d) is dict:
return dict((k, remove_empty_from_dict(v)) for k, v in d.items() if v and remove_empty_from_dict(v))
elif type(d) is list:
return [remove_empty_from_dict(v) for v in d if v and remove_empty_from_dict(v)]
else:
return d
BrenBarn's solution is ideal (and pythonic, I might add). Here is another (fp) solution, however:
from operator import itemgetter
dict(filter(itemgetter(1), metadata.items()))
If you have a nested dictionary, and you want this to work even for empty sub-elements, you can use a recursive variant of BrenBarn's suggestion:
def scrub_dict(d):
if type(d) is dict:
return dict((k, scrub_dict(v)) for k, v in d.iteritems() if v and scrub_dict(v))
else:
return d
For python 3
dict((k, v) for k, v in metadata.items() if v)
Quick Answer (TL;DR)
Example01
### example01 -------------------
mydict = { "alpha":0,
"bravo":"0",
"charlie":"three",
"delta":[],
"echo":False,
"foxy":"False",
"golf":"",
"hotel":" ",
}
newdict = dict([(vkey, vdata) for vkey, vdata in mydict.iteritems() if(vdata) ])
print newdict
### result01 -------------------
result01 ='''
{'foxy': 'False', 'charlie': 'three', 'bravo': '0'}
'''
Detailed Answer
Problem
Context: Python 2.x
Scenario: Developer wishes modify a dictionary to exclude blank values
aka remove empty values from a dictionary
aka delete keys with blank values
aka filter dictionary for non-blank values over each key-value pair
Solution
example01 use python list-comprehension syntax with simple conditional to remove "empty" values
Pitfalls
example01 only operates on a copy of the original dictionary (does not modify in place)
example01 may produce unexpected results depending on what developer means by "empty"
Does developer mean to keep values that are falsy?
If the values in the dictionary are not gauranteed to be strings, developer may have unexpected data loss.
result01 shows that only three key-value pairs were preserved from the original set
Alternate example
example02 helps deal with potential pitfalls
The approach is to use a more precise definition of "empty" by changing the conditional.
Here we only want to filter out values that evaluate to blank strings.
Here we also use .strip() to filter out values that consist of only whitespace.
Example02
### example02 -------------------
mydict = { "alpha":0,
"bravo":"0",
"charlie":"three",
"delta":[],
"echo":False,
"foxy":"False",
"golf":"",
"hotel":" ",
}
newdict = dict([(vkey, vdata) for vkey, vdata in mydict.iteritems() if(str(vdata).strip()) ])
print newdict
### result02 -------------------
result02 ='''
{'alpha': 0,
'bravo': '0',
'charlie': 'three',
'delta': [],
'echo': False,
'foxy': 'False'
}
'''
See also
list-comprehension
falsy
checking for empty string
modifying original dictionary in place
dictionary comprehensions
pitfalls of checking for empty string
Building on the answers from patriciasz and nneonneo, and accounting for the possibility that you might want to delete keys that have only certain falsy things (e.g. '') but not others (e.g. 0), or perhaps you even want to include some truthy things (e.g. 'SPAM'), then you could make a highly specific hitlist:
unwanted = ['', u'', None, False, [], 'SPAM']
Unfortunately, this doesn't quite work, because for example 0 in unwanted evaluates to True. We need to discriminate between 0 and other falsy things, so we have to use is:
any([0 is i for i in unwanted])
...evaluates to False.
Now use it to del the unwanted things:
unwanted_keys = [k for k, v in metadata.items() if any([v is i for i in unwanted])]
for k in unwanted_keys: del metadata[k]
If you want a new dictionary, instead of modifying metadata in place:
newdict = {k: v for k, v in metadata.items() if not any([v is i for i in unwanted])}
I read all replies in this thread and some referred also to this thread:
Remove empty dicts in nested dictionary with recursive function
I originally used solution here and it worked great:
Attempt 1: Too Hot (not performant or future-proof):
def scrub_dict(d):
if type(d) is dict:
return dict((k, scrub_dict(v)) for k, v in d.iteritems() if v and scrub_dict(v))
else:
return d
But some performance and compatibility concerns were raised in Python 2.7 world:
use isinstance instead of type
unroll the list comp into for loop for efficiency
use python3 safe items instead of iteritems
Attempt 2: Too Cold (Lacks Memoization):
def scrub_dict(d):
new_dict = {}
for k, v in d.items():
if isinstance(v,dict):
v = scrub_dict(v)
if not v in (u'', None, {}):
new_dict[k] = v
return new_dict
DOH! This is not recursive and not at all memoizant.
Attempt 3: Just Right (so far):
def scrub_dict(d):
new_dict = {}
for k, v in d.items():
if isinstance(v,dict):
v = scrub_dict(v)
if not v in (u'', None, {}):
new_dict[k] = v
return new_dict
To preserve 0 and False values but get rid of empty values you could use:
{k: v for k, v in metadata.items() if v or v == 0 or v is False}
For a nested dict with mixed types of values you could use:
def remove_empty_from_dict(d):
if isinstance(d, dict):
return dict((k, remove_empty_from_dict(v)) for k, v in d.items() \
if v or v == 0 or v is False and remove_empty_from_dict(v) is not None)
elif isinstance(d, list):
return [remove_empty_from_dict(v) for v in d
if v or v == 0 or v is False and remove_empty_from_dict(v) is not None]
else:
if d or d == 0 or d is False:
return d
"As I also currently write a desktop application for my work with Python, I found in data-entry application when there is lots of entry and which some are not mandatory thus user can left it blank, for validation purpose, it is easy to grab all entries and then discard empty key or value of a dictionary. So my code above a show how we can easy take them out, using dictionary comprehension and keep dictionary value element which is not blank. I use Python 3.8.3
data = {'':'', '20':'', '50':'', '100':'1.1', '200':'1.2'}
dic = {key:value for key,value in data.items() if value != ''}
print(dic)
{'100': '1.1', '200': '1.2'}
Dicts mixed with Arrays
The answer at Attempt 3: Just Right (so far) from BlissRage's answer does not properly handle arrays elements. I'm including a patch in case anyone needs it. The method is handles list with the statement block of if isinstance(v, list):, which scrubs the list using the original scrub_dict(d) implementation.
#staticmethod
def scrub_dict(d):
new_dict = {}
for k, v in d.items():
if isinstance(v, dict):
v = scrub_dict(v)
if isinstance(v, list):
v = scrub_list(v)
if not v in (u'', None, {}, []):
new_dict[k] = v
return new_dict
#staticmethod
def scrub_list(d):
scrubbed_list = []
for i in d:
if isinstance(i, dict):
i = scrub_dict(i)
scrubbed_list.append(i)
return scrubbed_list
An alternative way you can do this, is using dictionary comprehension. This should be compatible with 2.7+
result = {
key: value for key, value in
{"foo": "bar", "lorem": None}.items()
if value
}
Here is an option if you are using pandas:
import pandas as pd
d = dict.fromkeys(['a', 'b', 'c', 'd'])
d['b'] = 'not null'
d['c'] = '' # empty string
print(d)
# convert `dict` to `Series` and replace any blank strings with `None`;
# use the `.dropna()` method and
# then convert back to a `dict`
d_ = pd.Series(d).replace('', None).dropna().to_dict()
print(d_)
Some of Methods mentioned above ignores if there are any integers and float with values 0 & 0.0
If someone wants to avoid the above can use below code(removes empty strings and None values from nested dictionary and nested list):
def remove_empty_from_dict(d):
if type(d) is dict:
_temp = {}
for k,v in d.items():
if v == None or v == "":
pass
elif type(v) is int or type(v) is float:
_temp[k] = remove_empty_from_dict(v)
elif (v or remove_empty_from_dict(v)):
_temp[k] = remove_empty_from_dict(v)
return _temp
elif type(d) is list:
return [remove_empty_from_dict(v) for v in d if( (str(v).strip() or str(remove_empty_from_dict(v)).strip()) and (v != None or remove_empty_from_dict(v) != None))]
else:
return d
metadata ={'src':'1921','dest':'1337','email':'','movile':''}
ot = {k: v for k, v in metadata.items() if v != ''}
print(f"Final {ot}")
You also have an option with filter method:
filtered_metadata = dict( filter(lambda val: val[1] != u'', metadata.items()) )
Some benchmarking:
1. List comprehension recreate dict
In [7]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
...: dic = {k: v for k, v in dic.items() if v is not None}
1000000 loops, best of 7: 375 ns per loop
2. List comprehension recreate dict using dict()
In [8]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
...: dic = dict((k, v) for k, v in dic.items() if v is not None)
1000000 loops, best of 7: 681 ns per loop
3. Loop and delete key if v is None
In [10]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
...: for k, v in dic.items():
...: if v is None:
...: del dic[k]
...:
10000000 loops, best of 7: 160 ns per loop
so loop and delete is the fastest at 160ns, list comprehension is half as slow at ~375ns and with a call to dict() is half as slow again ~680ns.
Wrapping 3 into a function brings it back down again to about 275ns. Also for me PyPy was about twice as fast as neet python.
I'm trying to do the same as
Get the key corresponding to the minimum value within a dictionary, where we want to get the key corresponding to the minimum value in a dictionary.
The best way appears to be:
min(d, key=d.get)
BUT I want to apply this on a dictionary with multiple minimum values:
d = {'a' : 1, 'b' : 2, 'c' : 1}
Note that the answer from the above would be:
>>> min(d, key=d.get)
'a'
However, I need both the two keys that have a minimum value, namely a and c.
What would be the best approach?
(Ultimately I want to pick one of the two at random, but I don't think this is relevant).
One simple option is to first determine the minimum value, and then select all keys mapping to that minimum:
min_value = min(d.itervalues())
min_keys = [k for k in d if d[k] == min_value]
For Python 3 use d.values() instead of d.itervalues().
This needs two passes through the dictionary, but should be one of the fastest options to do this anyway.
Using reservoir sampling, you can implement a single pass approach that selects one of the items at random:
it = d.iteritems()
min_key, min_value = next(it)
num_mins = 1
for k, v in it:
if v < min_value:
num_mins = 1
min_key, min_value = k, v
elif v == min_value:
num_mins += 1
if random.randrange(num_mins) == 0:
min_key = k
After writing down this code, I think this option is of rather theoretical interest… :)
EDITED: Now using setdefault as suggested :)
I don't know if that helps you but you could build a reverse dictionary with the values as key and the keys (in a list as values).
d = {'a' : 1, 'b' : 2, 'c' : 1}
d2 = {}
for k, v in d.iteritems():
d2.setdefault(v, []).append(k)
print d2[min(d2)]
It will print this:
['a', 'c']
However, I think the other solutions are more compact and probably more elegant...
min_keys = [k for k in d if all(d[m] >= d[k] for m in d)]
or, slightly optimized
min_keys = [k for k, x in d.items() if not any(y < x for y in d.values())]
It's not as efficient as other solutions, but demonstrates the beauty of python (well, to me at least).
def get_rand_min(d):
min_val = min(d.values())
min_keys = filter(lambda k: d[k] == min_val, d)
return random.choice(min_keys)
You can use heapq.nsmallest to get the N smallest members of the dict, then filter out all that are not equal to the lowest one. That's provided you know the maximal number of smallest members you can have, let's assume it's N here. something like:
from heapq import nsmallest
from operator import itemgetter
#get the N smallest members
smallestN = nsmallest(N, myDict.iteritems(), itemgetter(1)))
#leave in only the ones with a score equal to the smallest one
smallest = [x for x in smallestN if x[1] == smallestN[0][1]]
minValue,minKey = min((v,k) for k,v in d.items())
Due to your semantics you need to go through the entire dictionary at least once. This will retrieve exactly 1 minimum element.
If you want all the minimum items in O(log(N)) query time, you can insert your elements into a priority queue as you generate them (if you can). The priority queue must have O(1) insertion time and O(log(N)) extract-min time. (This will be as bad as sorting if all your elements have the same value, but otherwise may work quite well.)
One pass solution would be:
>>> result = [100000, []]
>>> for key, val in d.items():
... if val < result[0]:
... result[1] = [key]; result[0]=val;
... elif val == result[0]:
... result[1].append(key)
...
>>> result
[1, ['a', 'c']]
Here's another way to do it in one pass:
d = {'foo': 2, 'a' : 1, 'b' : 2, 'c' : 1, 'z': 99, 'x': 1}
current_min = d[d.keys()[0]]
min_keys = []
for k, v in d.iteritems():
if v < current_min:
current_min = v
min_keys = [k]
elif v == current_min:
min_keys.append(k)
print min_keys
['a', 'x', 'c']
This works:
d = {'a' :1, 'b' : 2, 'c' : 1}
min_value = min(d.values())
result = [x[0] for x in d.items() if x[1] == k]
Hmpf. After fixing up the code to work, I ended up with #Sven Marnach's answer, so, disregard this ;)