Develop Reference

How can I update the value of key in OrderedDict in Python 3?

Here's a minimal reproducible example:
from collections import OrderedDict
d = OrderedDict([('foo', 123),
('bar', 456)])
So I want to check if there's a foo key in d and if there's then I'd like to rewrite it as a single value of a list for a new hardcoded key:
print(d)
ordereddict([('bar', 456), ('newCoolHardcodedKey', [ordereddict([('foo', 123)])])])

You can use a generating expression (like a list comprehension, but returns an iterator instead of storing the temporary list in memory) to do this:
d = OrderedDict(
(
("newCoolHardcodedKey", OrderedDict([item])) if item[0] == "foo" else item
for item in d.items()
)
)
print(d)
OrderedDict([('newCoolHardcodedKey', OrderedDict([('foo', 123)])), ('bar', 456)])
The dict being ordered, the new element is where foo was.
If you need the new element to go to the end, it might be easiest to test if d["foo"] exists, and if so append the new ordered dict with its hard-coded key and delete the original entry for foo:
if "foo" in d:
d["newCoolHardcodedKey"] = OrderedDict([("foo", d["foo"])])
del d["foo"]
print(d)
OrderedDict([('bar', 456), ('newCoolHardcodedKey', OrderedDict([('foo', 123)]))])
Performance considerations
If d is large in your real application, the second solution is much better since it changes d in place instead of making a copy.

Merging values from 2 dictionaries (Python)

(I'm new to Python!)
Trying to figure out this homework question:
The function will takes a​s input​ two dictionaries, each mapping strings to integers. The function will r​eturn​ a dictionary that maps strings from the two input dictionaries to the sum of the integers in the two input dictionaries.
my idea was this:
def ​add(​dicA,dicB):
dicA = {}
dicB = {}
newdictionary = dicA.update(dicB)
however, that brings back None.
In the professor's example:
print(add({'alice':10, 'Bob':3, 'Carlie':1}, {'alice':5, 'Bob':100, 'Carlie':1}))
the output is:
{'alice':15, 'Bob':103, 'Carlie':2}
My issue really is that I don't understand how to add up the values from each dictionaries. I know that the '+' is not supported with dictionaries. I'm not looking for anyone to do my homework for me, but any suggestions would be very much appreciated!

From the documentation:
update([other])
Update the dictionary with the key/value pairs from other, overwriting existing keys. Return None.
You don't want to replace key/value pairs, you want to add the values for similar keys. Go through each dictionary and add each value to the relevant key:
def ​add(​dicA,dicB):
result = {}
for d in dicA, dicB:
for key in d:
result[key] = result.get(key, 0) + d[key]
return result
result.get(key, 0) will retrieve the value of an existing key or produce 0 if key is not yet present.

First of all, a.update(b) updates a in place, and returns None.
Secondly, a.update(b) wouldn't help you to sum the keys; it would just produce a dictionary with the resulting dictionary having all the key, value pairs from b:
>>> a = {'alice':10, 'Bob':3, 'Carlie':1}
>>> b = {'alice':5, 'Bob':100, 'Carlie':1}
>>> a.update(b)
>>> a
{'alice': 5, 'Carlie': 1, 'Bob': 100}
It'd be easiest to use collections.Counter to achieve the desired result. As a plus, it does support addition with +:
from collections import Counter
def add(dicA, dicB):
return dict(Counter(dicA) + Counter(dicB))
This produces the intended result:
>>> print(add({'alice':10, 'Bob':3, 'Carlie':1}, {'alice':5, 'Bob':100, 'Carlie':1}))
{'alice': 15, 'Carlie': 2, 'Bob': 103}

The following is not meant to be the most elegant solution, but to get a feeling on how to deal with dicts.
dictA = {'Alice':10, 'Bob':3, 'Carlie':1}
dictB = {'Alice':5, 'Bob':100, 'Carlie':1}
# how to iterate through a dictionary
for k,v in dictA.iteritems():
print k,v
# make a new dict to keep tally
newdict={}
for d in [dictA,dictB]: # go through a list that has your dictionaries
print d
for k,v in d.iteritems(): # go through each dictionary item
if not k in newdict.keys():
newdict[k]=v
else:
newdict[k]+=v
print newdict
Output:
Bob 3
Alice 10
Carlie 1
{'Bob': 3, 'Alice': 10, 'Carlie': 1}
{'Bob': 100, 'Alice': 5, 'Carlie': 1}
{'Bob': 103, 'Alice': 15, 'Carlie': 2}

def ​add(​dicA,dicB):
You define a function that takes two arguments, dicA and dicB.
dicA = {}
dicB = {}
Then you assign an empty dictionary to both those variables, overwriting the dictionaries you passed to the function.
newdictionary = dicA.update(dicB)
Then you update dicA with the values from dicB, and assign the result to newdictionary. dict.update always returns None though.
And finally, you don’t return anything from the function, so it does not give you any results.
In order to combine those dictionaries, you actually need to use the values that were passed to it. Since dict.update mutates the dictionary it is called on, this would change one of those passed dictionaries, which we generally do not want to do. So instead, we use an empty dictionary, and then copy the values from both dictionaries into it:
def add (dicA, dicB):
newDictionary = {}
newDictionary.update(dicA)
newDictionary.update(dicB)
return newDictionary
If you want the values to sum up automatically, then use a Counter instead of a normal dictionary:
from collections import Counter
def add (dicA, dicB):
newDictionary = Counter()
newDictionary.update(dicA)
newDictionary.update(dicB)
return newDictionary

I suspect your professor wants to achieve this using more simple methods. But you can achieve this very easily using collections.Counter.
from collections import Counter
def add(a, b):
return dict(Counter(a) + Counter(b))
Your professor probably wants something like this:
def add(a, b):
new_dict = copy of a
for each key/value pair in b
if key in new_dict
add value to value already present in new_dict
else
insert key/value pair into new_dict
return new_dict

You can try this:
def add(dict1, dict2):
return dict([(key,dict1[key]+dict2[key]) for key in dict1.keys()])

I personally like using a dictionary's get method for this kind of merge:
def add(a, b):
result = {}
for dictionary in (a, b):
for key, value in dictionary.items():
result[key] = result.get(key, 0) + value
return result

Lua's "Generic For Loop" for Python?

So, I've been searching endlessly for something similiar to Lua's "Generic For Loop" in Python.
I've been working on a simple text based game in Python, and I've been working with dictionaries a lot.
Here is something I'm looking for (in Lua):
Dictionary = {
"Red" = "There is some red paint on the walls.",
"Green" = "There is a little bit of green paint on the floor.",
}
for i, v in pairs(Dictionary) do
print(i, v)
end
What this will do is, go through the dictionary, then print out the INDEX and the VALUE. How would I do something like this in Python?
I know there is this:
for i in Dictionary:
print(i)
But that just prints the INDEX. I would like to access both the INDEX and the VALUE. Something like:
for i, v in Dictionary:
print(i, v)
Any help is appreciated.

You're looking for items. Iterating over a dict just gives you the keys, so you'd have to do:
for key in my_dict:
x = my_dict[key]
What you want is this:
for key, value in my_dict.items():
# do something

two ways:
for i, v in Dictionary.items():
print(i, v) #outputs pairs as key value
for tup in Dictionary.items(): #same thing
print(tup) # outputs pairs as (key,value)
or
for key in Dictionary:
print(key,Dictionary[key])
EDIT RESPONSE TO COMMENT:
>>> d = {1:1,2:2,3:3,4:4}
>>> for item in d.items(): print(item)
(1, 1)
(2, 2)
(3, 3)
(4, 4)
>>> for key,val in d.items(): print(key,val)
1 1
2 2
3 3
4 4
this is because in the first loop, item is a tuple and the __repr__ for a tuple has the brackets and commas as part of it where as the second loop splits the tuple into two seperate variables. print then automatically adds a space delimiter in between each parameter passed in the print function.
As explained by Two-Bit Alchemist:
In case it's not entirely clear still, in the tup formulation you'd access the key and value as tup[0] and tup[1], respectively. for key, val in my_dict.items(): ... and for tup in my_dict.items(): key, val = tup is the same setup. The point is you can use tuple unpacking just fine inline in a for loop.

The items method (or in Py2, viewitems or iteritems to avoid making a whole new list containing copies of the dict key/value pairs) is the way to go:
for k, v in Dictionary.items(): # For performance, use .viewitems() on Py2.7, .items() on Py3.x
print(k, v)

Delete items in a dictionary with values that don't equal the highest value in Python

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}

Access keys and vals in listed python-dict

I've got a list k with the 0'th element:
k[0]: {'pattern': 0, 'pos': array([ 9.83698, 106.539 , 130.314 ]), 'id': 1922}
(It looks like a dict, but its a list indeed)
when I iterate through the 0'th element of the list k and print out each element I Get:
for i in k:
print i
=>output:
pattern
pos
id
I'd like to access not only the keys but the values as well. How to do this?
I've also tried to convert the list back into a dict using zip and izip, but same resutlts...i.e. only keys are printed, no values...
any help will be appreciated
thx in advance

you can use k.values() to iterate through the values, or k.items() to iterate through (key, value) pairs
for value in k.values():
print value
for key, value in k.items():
print key, value

The fastest way to iterate over the dictionary you created (it is in fact a dictionary) is not to create the lists of keys/values using k[0].keys(), k[0].values and k[0].items() but using k[0].iteritems() which creates a dictionary iterator that returns just the pairs without allocating lists in the memory.
It also runs much faster for big dictionaries (a being the dictionary):
>>> non_iter_timer = timeit.Timer("for k,v in a.items(): k + v", setup="a = {x:x for x in xrange(10000000)}")
>>> non_iter_timer.repeat(3, 10)
[25.612606023166585, 25.100741935717622, 24.840450306339463]
>>> iter_timer = timeit.Timer("for k,v in a.iteritems(): k + v", setup="a = {x:x for x in xrange(10000000)}")
>>> iter_timer.repeat(3, 10)
[9.26259596885518, 9.198298194571748, 9.77466250122282]