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I have a list l=['a','b','c'] and I want to create a dict that takes these items as keys and then adds a value to it. Such as:
d={ 'a':0, 'b':1, 'c':2 }
How would I be able to this?
There are many ways to create this dictionary. Here is just a few:
>>> dc = dict(zip(ls, range(3)))
>>> dc2 = {v: k for k, v in enumerate(ls)}
>>> dc2
{'a': 0, 'b': 1, 'c': 2}
>>> assert dc == dc2
>>> # silence because they're same
l=['a','b','c']
d = dict(zip(l,range(len(l))))
print(d) # {'a': 0, 'b': 1, 'c': 2}
You could use something like
l = ['a','b','c']
d = {element:index for index,element in enumerate(l)}
which will give you each element mapped to its index.
if you want to set some arbitrary value, use
l = ['a','b','c']
d = {element:value for element in l}
Related
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mydict = {
'name' : ['cat','dog'],
'place' : ['earth','mars'],
'name1' : ['cat','dog'],
'name2' : ['cat','dog'],
'place1' : ['earth','mars'],
'colour' : ['blue','green'],
'colour1' : ['orange']
}
expected_result
[['name','name1','name2'],['place','place1']]
Since 'colour' and 'colour1' are not same we will ignore them.
Probably not the fastest solution, but you can find the duplicate values first, then list the corresponding keys:
>>> values = list(mydict.values())
>>> duplicates = set(tuple(v) for v in values if values.count(v) > 1)
>>> [[k for k, v in mydict.items() if tuple(v) == d] for d in duplicates]
[['name', 'name1', 'name2'], ['place', 'place1']]
from collections import defaultdict
rev = defaultdict(list)
for i in mydict:
rev[tuple(mydict[i])].append(i)
answer = [i for i in rev.values() if len(i)>=2]
#[['name', 'name1', 'name2'], ['place', 'place1']]
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Python dictionnary:
I need to regroup the keys and add the values accordingly.
For example:
dict = {'A01':6,'A02':5,'A03':2,'B01':2,'B02':3,'B03':2}
new_dict --> {'A':13,'B':7}
You can try this.
>>> dct={'A01':6,'A02':5,'A03':2,'B01':2,'B02':3,'B03':2}
>>> new={}
>>> for k,v in dct.items():
... new[k[0]]=new.get(k[0],0)+v
...
>>> new
{'A': 13, 'B': 7}
Using defaultdict(int)
>>> from collections import defaultdict
>>> new=defaultdict(int)
>>> for k,v in dct.items():
... new[k[0]]+=v
...
>>> new
defaultdict(<class 'int'>, {'A': 13, 'B': 7})
Something like this may work
new_dict = {}
for key, val in dict.items():
new_key = key[0]
if new_key not in new_dict:
new_dict[new_key] = 0
new_dict[new_key] += val
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I need a function to change values within a nested dictionary,
by crawling a list instead of the usual dic[key1][key2] = value
e.g :
IN[0]: dic = {'1': 23, '3': {'a': 'foo', 'b': 67}}
IN[1]: l = ['3', 'a']
IN[2]: func(dic, l , newvalue)
IN[3]: print dic
Expected output :
OUT[4]: dic = {'1': 23, '3': {'a': newvalue, 'b': 67}}
Try this:
def func(dic, keys, value):
for key in keys[:-1]:
dic = dic.setdefault(key, {})
dic[keys[-1]] = value
func(dic, l , newvalue)
print(dic)
This is not good to use unnecessary things like exec and globals().
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How can I store multiple keys and values in school dictionary?
I was trying to store a list of keys and values in a dictionary with a for loop but I did t know how to go about it.
Can any one help me out?
keys = ['a', 'b']
values = [1, 2]
print(dict(zip(keys,values)))
Output:
{'a':1,'b':2}
You can do following:
keys = ['bob', 'alice']
values = [40, 20]
d = {}
for i in range(len(keys)):
d[keys[i]] = values[i]
Learn the basics of python dictionary and looping. You will find tons of resources in the internet.
For example:
lst1 = ['a', 'b', 'c']
lst2 = [1, 2, 3]
# First way
d1 = {k: v for k, v in zip(lst1, lst2)}
print d1
# {'a': 1, 'c': 3, 'b': 2}
# Second way
d2 = dict(zip(lst1, lst2))
print d2
# {'a': 1, 'c': 3, 'b': 2}
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I have 4 list of dicts
list1 = [{'a':0,'b':23}, {'a':3,'b':77},{'a':1,'b':99}]
list2 = [{'a':1,'c':666},{'a':4,'c':546}]
list3 = [{'d':33,'a':3},{'d':1111,'a':4},{'d':76,'a':1},{'d':775,'a':0}]
list4 = [{'a':2,'e':12},{'a':4,'e':76}]
Every dict in the list has a common key 'a'. My requirement is 'a' key with same value in dict from all the list should be merged and if a particular key does not exist in the dicts while merging, then assign 0 for those keys or just omit those keys.
for eg. for key 'a' with value 1, from above example we have 2 dicts, one from list1 i.e {'a':0,'b':23} and one is list3, last dict i.e {'d':775,'a':0}, so 1st we have identified the dicts with same 'a' value, now need to merge these dicts
i.e {'a':0, 'b':23, 'c':0, 'd':775, 'e':0}, since both the dict didn't have 'c', 'c' is assigned as 0 here
I should get the output as:
[{'a':0,'b':23,'c':0,'d':775, 'e':0},{'a':1,'b':99,'c':666,'d':76,'e':0},{'a':2,'b':0,'c':0,'d':0,'e':12},{'a':3,'b':77,'c':0,'d':33,'e':0}, {'a':4,'b':0,'c':546,'d':1111,'e':76}]
usings minimum loops or list comprehension
If you want a more pythonic way:
from itertools import groupby
from pprint import pprint
from collections import ChainMap
a = [{'a':0,'b':23}, {'a':3,'b':77}, {'a':1,'b':99}]
b = [{'a':1,'c':666}, {'a':4,'c':546}]
c = [{'d':33,'a':3}, {'d':1111,'a':4}, {'d':76,'a':1}, {'d':775,'a':0}]
d = [{'a':2,'e':12}, {'a':4,'e':76}]
dict_list = a + b + c + d
# You just need to specify the key you want to use in the lambda function
# There's no need to declare the different key values previously
res = map(lambda dict_tuple: dict(ChainMap(*dict_tuple[1])),
groupby(sorted(dict_list,
key=lambda sub_dict: sub_dict["a"]),
key=lambda sub_dict: sub_dict["a"]))
pprint(list(res))
Outputs:
[{'a': 0, 'b': 23, 'd': 775},
{'a': 1, 'b': 99, 'c': 666, 'd': 76},
{'a': 2, 'e': 12},
{'a': 3, 'b': 77, 'd': 33},
{'a': 4, 'c': 546, 'd': 1111, 'e': 76}]
Edit (Improvement):
You can also use
from _operator import itemgetter
key=itemgetter("a")
instead of
key=lambda sub_dict: sub_dict["a"]
The version with itemgetter is much faster. Using the example you provided:
- Lambda: 0.037109375ms
- Itemgetter: 0.009033203125ms