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Dictionary comprehensions

Dictionary comprehensions
num_dict={1:1,2:4,3:9}
twice_num_dict={key: (value if value*2 >=8 else None)for (key,value) in num_dict.items()}
print(twice_num_dict)
Dictionary comprehensionsi wanted to create a new dict where only key:value pairs of the existing dict will be there in the new_dict if the value*2 of the first dict was >=8 i used if and else here but idk what to type in else condition so that the key value pair of 1:1 is not printed at all

You should place your if clause after your for clause in the comprehension, like so:
{k: v for k, v in d.items() if v * 2 >= 8}

Comparing the strings in key and value of the same dictionary

I am looking to solve a problem to compare the string of the key and value of the same dictionary.
To return a dictionary of all key and values where the value contains the key name as a substring.
a = {"ant":"antler", "bi":"bicycle", "cat":"animal"}
the code needs to return the result:
b = {"ant":"antler", "bi":"bi cycle"}

You can iterate through the dictionary and unpack the key and the value at the same time this way:
b = {}
for key, value in a.items():
if value in key:
b[value] = key
This will generate your wanted solution. It does that by unpacking both the key and the value and checking if they match afterward.
You can also shorten that code by using a dictionary comprehension:
b = {key:value for key, value in a.items() if key in value}
This short line does the exact same thing as the code before. It even uses the same functionalities with only one addition - a dictionary comprehension. That allows you to put all that code in one simple line and declare the dictionary on the go.

answer = {k:v for k,v in a.items() if k in v}
Notes:
to iterate over key: value pair we use dict.items();
to check if a string is inside some other string we use in operator;
to filter items we use if-clause in the dictionary comprehension.
See also:
about dictionary comprehensions
about operators in and not in

filter a dictionary by key less than some range of values

I have a dictionary like this,
d = {1:'a', 2:'b', 3:'c', 4:'d'}
Now I want to filter the dictionary where the key should be more than 1 and less than 4 so the dictionary will be,
d = {2:'b', 3:'c'}
I could do this using a for loop, iterating over all the keys. but the execution time will be more looking for some fastest way to do this more efficiently in pythonic way.

you can try below code:
d = {k:v for k,v in d.items() if 1<k<4}

Pythonic way would be to use a dictionary comprehension:
{key: value for key, value in d.items() if 1 < key < 4}
It's readable enough: for each key and value in the items of the dictionary, keep those key: value pairs that have their keys between 1 and 4.

More pythonic way would be a dictionary comprehension
d = {k: v for (k, v) in d.items() if k > 1 and k < 4}
If the efficiency is a bottleneck, you may want to try to use some tree based structure instead of a dictionary that is hash based.

Python dictionaries use hashes of their keys to efficiently lookup and store data. Unfortunately, that means that they don't allow you use the numeric properties of those keys to select data (the way you might be able to do using a slice to index a list).
So I think the best way to do what you want is probably just a dictionary comprehension that tests each key against your boundary values:
d = {key: value for key, value in d.items() if 1 < key < 4}

Sort dictionary by specific order (list of keys)

I am trying to sort a dictionary based on the order that its keys appears in a list.
First element of the list would be the first element of the dictionary. At the end the dictionary keys wiould be in the same order as in the provided list...
I can sort by value with this code
newlist = sorted(product_list, key=lambda k: k['product'])
but cant do it for a list
thanks for any help!

from collections import OrderedDict
new_dict = OrderedDict((k, old_dict[k]) for k in key_list)
Having said that, there's probably a better way to solve your problem than using an OrderedDict
If some keys are missing, you'll need to use one of
new_dict = OrderedDict((k, old_dict.get(k)) for k in key_list)
or
new_dict = OrderedDict((k, old_dict[k]) for k in key_list if k in old_dict)
depending on how you want to handle the missing keys.

In Python (and most languages) dictionaries are unsorted, so you can't "sort" a dictionary.
You can retrieve and sort the keys and iterate through those:
for key in sorted(product_list.keys()):
item = product_list[key]
item.doSomething()
Or you can use a OrderDict, like so:
from collections import OrderedDict
And then build the dictionary in the required order (which is up to you to determine) but below we sort using the keys:
product_list = OrderDict(sorted(product_list.items(), key=lambda k: k[0]))
For reference, Dict.items() returns a list of tuples in the form:
[(key1, value1), (key2, value2) , ... , (keyN, valueN)]

By definition, a dictionary is unordered. You can use OrderedDict from collections as seen at http://docs.python.org/2/library/collections.html#collections.OrderedDict as a drop-in replacement.

Selecting elements of a Python dictionary greater than a certain value

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]