This question already has answers here:
Filter dict to contain only certain keys?
(22 answers)
Closed 6 years ago.
I'm trying to iterate over a dictionary in order to delete all items where the key is a number.
dict= {'hello':'3', 'baby':'5', '33':'6'}
for k in dict.keys():
if k.isdigit():
del dict[k]
The result i need is this:
dict={'hello':'3', 'baby':'5'}
I use the code above but it doesn't work?
Can somebody help me?
Instead of deleting items while looping through the dictionary, you can create a new dictionary:
dict_ = {'hello':'3', 'baby':'5', '33':'6'}
{k: v for k, v in dict_.items() if not k.isdigit()}
# {'baby': '5', 'hello': '3'}
The error RuntimeError: dictionary changed size during iteration, as it said, is raised because the dictionnary is modified during the iteration. One solution could be to first register all the keys, then modify the dictionnary:
dict= {'hello':'3', 'baby':'5', '33':'6'}
for k in tuple(dict.keys()): # iterate on a tuple of values that happened to be in dict's keys
if k.isdigit():
del dict[k]
The other, more pythonic, is to build the dictionnary in place, for instance using the dictionnary in intension notation like in this answer.
Btw, dict is a very bad name for an object of type dict.
The problem is because you are updating the dictionary while traversing. Rather first store the keys in a list and then remove from the dictionary.
Try this:
dic= {'hello':'3', 'baby':'5', '33':'6'}
x=dic.keys()
for k in x:
if k.isdigit():
del dic[k]
print dic
output :
{'baby': '5', 'hello': '3'}
Here is another solution using pop and map to modify in place your initial dictionary:
my_dict = {'a':1, 'b':2, 'c':3, '11':4, '12':5}
_ = map(lambda k: my_dict.pop(k), [k for k in my_dict if k.isdigit()])
# The map functions modifies in place your dictionary
Output :
>> my_dict
>> {'a': 1, 'b': 2, 'c': 3}
Related
I am new to Python so I do apologize that my first question might not be asked clearly to achieve the right answer.
I thought if I converted a list with duplicating keys into a dictionary then I would be able to sum the values of each duplicating key. I have tried to search on Google and Stack Overflow but I actually still can't solve this problem.
Can anybody help, please? Thank you very much in advance and I truly appreciate your help.
list1 = ["a:2", "b:5", "c:7", "a:8", "b:12"]
My expected output is:
dict = {a: 10, b: 17, c: 7}
You can try this code:
list1 = ["a:2", "b:5", "c:7", "a:8", "b:12"]
l1 = [each.split(":") for each in list1]
d1 = {}
for each in l1:
if each[0] not in d1:
d1[each[0]] = int(each[1])
else:
d1[each[0]] += int(each[1])
d1
Output: {'a': 10, 'b': 17, 'c': 7}
Explanation:
Step 1. Convert your given list to key-value pair by splitting each of the elements in your original list from : and store that in a list/tuple
Step 2. Initialize an empty dictionary
Step 3. Iterate through each key-value pair in the newly created list/tuple and store that in a dictionary. If the key doesn't exist, then add new key-value pair to dictionary or else just add the values to it's corresponding key.
A list does not have "keys" per say, rather it has elements. In your example, the elements them selves are a key value pair. To make the dictionary you want you have to do 3 things,
Parse each element into its key value pair
Handle duplicate values
Add each pair to the dictionary.
the code should look like this
list1 = ["a:2", "b:5", "c:7", "a:8", "b:12"]
dict1={}#make an empty dictionary
for element in list1:
key,value=element.split(':')#This splits your list elements into a tuple of (key,value)
if key in dict1:#check if the key is in the dictionary
dict1[key]+=int(value)#add to existing key
else:
dict1[key]=int(value)#initilize new key
print(dict1)
That code prints out
{'a': 10, 'c': 7, 'b': 17}
You could use a defaultdict, iterate over each string and add the corresponding value after splitting it to a pair (key, value).
>>> from collections import defaultdict
>>> res = defaultdict(int)
>>> for el in list1:
... k, v = el.split(':')
... res[k]+=int(v)
...
>>> res
defaultdict(<class 'int'>, {'a': 10, 'b': 17, 'c': 7})
Essentially I want to delete every key in a dictionary if its value doesn't equal the highest value.
Let's say this is the dictionary:
myDict = {"Bob": 1, "Bill": 5, "Barry": 4, "Steve": 5}
I'm able to sort it by value using this:
myDict = sorted(myDict, key=myDict.get, reverse=True)
Now I want to remove any key in the dictionary that doesn't equal the highest value (in this case '5'). To end up with this:
myDict = {"Bill": 5, "Steve": 5}
I've tried using this for loop:
for item, v in myDict:
if v < myDict[0]:
del myDict[v]
But I get this error:
ValueError: too many values to unpack (expected 2)
This is a) my first time posting here, and b) I've only been learning Python for a few months so I'm sorry if I've made any stupid mistakes.
for item, v in myDict just give you keys mydict, and you are collecting that key in item, v that's why,
use myDict.items() or myDict.iteritems().
for item, v in myDict.iteritems():
if v < myDict[0]:
del myDict[v]
To get Highest value of myDict
max(myDict.values())
To delete keys from Dict never change the iterator you are iterating on, it will give you RuntimeError. So copy it in another variable and change previous one as Anand S Kumar suggested.
You should never alter the object you're iterating over, that usually yields unexpected results (internal pointers get shifted and you miss elements in your iteration and such). You best gather the keys you want to delete and then remove the keys in a separate iteration:
keys = [k for k in myDict.keys() if myDict[k] == max(myDict.values())];
for k in keys: del myDict[k];
It might be best to put the max expression in a variable too so it doesn't get evaluated multiple times. Not sure if Python's able to optimize that for you (probably not).
You can use dictionary comprehension to create a new dictionary:
newDict = {k: v for k,v in myDict.items() if v == max(myDict.values())}
The output for newDict:
{'Steve': 5, 'Bill': 5}
I need to select elements of a dictionary of a certain value or greater. I am aware of how to do this with lists, Return list of items in list greater than some value.
But I am not sure how to translate that into something functional for a dictionary. I managed to get the tags that correspond (I think) to values greater than or equal to a number, but using the following gives only the tags:
[i for i in dict if dict.values() >= x]
.items() will return (key, value) pairs that you can use to reconstruct a filtered dict using a list comprehension that is feed into the dict() constructor, that will accept an iterable of (key, value) tuples aka. our list comprehension:
>>> d = dict(a=1, b=10, c=30, d=2)
>>> d
{'a': 1, 'c': 30, 'b': 10, 'd': 2}
>>> d = dict((k, v) for k, v in d.items() if v >= 10)
>>> d
{'c': 30, 'b': 10}
If you don't care about running your code on python older than version 2.7, see #opatut answer using "dict comprehensions":
{k:v for (k,v) in dict.items() if v > something}
While nmaier's solution would have been my way to go, notice that since python 2.7+ there has been a "dict comprehension" syntax:
{k:v for (k,v) in dict.items() if v > something}
Found here: Create a dictionary with list comprehension in Python. I found this by googling "python dictionary list comprehension", top post.
Explanation
{ .... } includes the dict comprehension
k:v what elements to add to the dict
for (k,v) in dict.items() this iterates over all tuples (key-value-pairs) of the dict
if v > something a condition that has to apply on every value that is to be included
You want dict[i] not dict.values(). dict.values() will return the whole list of values that are in the dictionary.
dict = {2:5, 6:2}
x = 4
print [dict[i] for i in dict if dict[i] >= x] # prints [5]
This question already has answers here:
How do I sort a dictionary by value?
(34 answers)
Closed 3 years ago.
Assume that I have a dict.
data = {1:'b', 2:'a'}
And I want to sort data by 'b' and 'a' so I get the result
'a','b'
How do I do that?
Any ideas?
To get the values use
sorted(data.values())
To get the matching keys, use a key function
sorted(data, key=data.get)
To get a list of tuples ordered by value
sorted(data.items(), key=lambda x:x[1])
Related: see the discussion here: Dictionaries are ordered in Python 3.6+
If you actually want to sort the dictionary instead of just obtaining a sorted list use collections.OrderedDict
>>> from collections import OrderedDict
>>> from operator import itemgetter
>>> data = {1: 'b', 2: 'a'}
>>> d = OrderedDict(sorted(data.items(), key=itemgetter(1)))
>>> d
OrderedDict([(2, 'a'), (1, 'b')])
>>> d.values()
['a', 'b']
From your comment to gnibbler answer, i'd say you want a list of pairs of key-value sorted by value:
sorted(data.items(), key=lambda x:x[1])
Thanks for all answers.
You are all my heros ;-)
Did in the end something like this:
d = sorted(data, key = data.get)
for key in d:
text = data[key]
Sort the values:
sorted(data.values())
returns
['a','b']
I also think it is important to note that Python dict object type is a hash table (more on this here), and thus is not capable of being sorted without converting its keys/values to lists. What this allows is dict item retrieval in constant time O(1), no matter the size/number of elements in a dictionary.
Having said that, once you sort its keys - sorted(data.keys()), or values - sorted(data.values()), you can then use that list to access keys/values in design patterns such as these:
for sortedKey in sorted(dictionary):
print dictionary[sortedKeY] # gives the values sorted by key
for sortedValue in sorted(dictionary.values()):
print sortedValue # gives the values sorted by value
Hope this helps.
You could created sorted list from Values and rebuild the dictionary:
myDictionary={"two":"2", "one":"1", "five":"5", "1four":"4"}
newDictionary={}
sortedList=sorted(myDictionary.values())
for sortedKey in sortedList:
for key, value in myDictionary.items():
if value==sortedKey:
newDictionary[key]=value
Output: newDictionary={'one': '1', 'two': '2', '1four': '4', 'five': '5'}
In your comment in response to John, you suggest that you want the keys and values of the dictionary, not just the values.
PEP 256 suggests this for sorting a dictionary by values.
import operator
sorted(d.iteritems(), key=operator.itemgetter(1))
If you want descending order, do this
sorted(d.iteritems(), key=itemgetter(1), reverse=True)
no lambda method
# sort dictionary by value
d = {'a1': 'fsdfds', 'g5': 'aa3432ff', 'ca':'zz23432'}
def getkeybyvalue(d,i):
for k, v in d.items():
if v == i:
return (k)
sortvaluelist = sorted(d.values())
sortresult ={}
for i1 in sortvaluelist:
key = getkeybyvalue(d,i1)
sortresult[key] = i1
print ('=====sort by value=====')
print (sortresult)
print ('=======================')
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Inverse dictionary lookup - Python
Is there a built in way to index a dictionary by value in Python.
e.g. something like:
dict = {'fruit':'apple','colour':'blue','meat':'beef'}
print key where dict[key] == 'apple'
or:
dict = {'fruit':['apple', 'banana'], 'colour':'blue'}
print key where 'apple' in dict[key]
or do I have to manually loop it?
You could use a list comprehension:
my_dict = {'fruit':'apple','colour':'blue','meat':'beef'}
print [key for key, value in my_dict.items() if value == 'apple']
The code above is doing almost exactly what said you want:
print key where dict[key] == 'apple'
The list comprehension is going through all the key, value pairs given by your dictionary's items method, and making a new list of all the keys where the value is 'apple'.
As Niklas pointed out, this does not work when your values could potentially be lists. You have to be careful about just using in in this case since 'apple' in 'pineapple' == True. So, sticking with a list comprehension approach requires some type checking. So, you could use a helper function like:
def equals_or_in(target, value):
"""Returns True if the target string equals the value string or,
is in the value (if the value is not a string).
"""
if isinstance(target, str):
return target == value
else:
return target in value
Then, the list comprehension below would work:
my_dict = {'fruit':['apple', 'banana'], 'colour':'blue'}
print [key for key, value in my_dict.items() if equals_or_in('apple', value)]
You'll have to manually loop it, but if you'll need the lookup repeatedly this is a handy trick:
d1 = {'fruit':'apple','colour':'blue','meat':'beef'}
d1_rev = dict((v, k) for k, v in d1.items())
You can then use the reverse dictionary like this:
>>> d1_rev['blue']
'colour'
>>> d1_rev['beef']
'meat'
Your requirements are more complex than you realize:
You need to handle both list values and plain values
You don't actually need to get back a key, but a list of keys
You could solve this in two steps:
normalize the dict so that every value is a list (every plain value becomes a single-element)
build a reverse dictionary
The following functions will solve this:
from collections import defaultdict
def normalize(d):
return { k:(v if isinstance(v, list) else [v]) for k,v in d.items() }
def build_reverse_dict(d):
res = defaultdict(list)
for k,values in normalize(d).items():
for x in values:
res[x].append(k)
return dict(res)
To be used like this:
>>> build_reverse_dict({'fruit':'apple','colour':'blue','meat':'beef'})
{'blue': ['colour'], 'apple': ['fruit'], 'beef': ['meat']}
>>> build_reverse_dict({'fruit':['apple', 'banana'], 'colour':'blue'})
{'blue': ['colour'], 'apple': ['fruit'], 'banana': ['fruit']}
>>> build_reverse_dict({'a':'duplicate', 'b':['duplicate']})
{'duplicate': ['a', 'b']}
So you just build up the reverse dictionary once and then lookup by value and get back a list of keys.