Develop Reference

Get dictionary value from multiple dictionaries in a list

I have a list of many dictionaries with only one key value pair like below.
List1 = [{'dept': 'a'}, {'country': 'India'}, {'name':'xyz'}, {'age':32}]
I want to extract name from this list by not using index number but with the dict key.
Can anyone help me out?

Try this:
List1 = [{'dept': 'a'}, {'country': 'India'}, {'name': 'xyz'}, {'age': 32}]
for dic in List1:
name = dic.get('name')
if name is not None: # {'name': ''} which is acceptable.
break
print(name)
output :
xyz

Or use list comprehesion. It always looks better than a loop.
[dic['name'] for dic in List1 if 'name' in dic][0]

create JSON from list

i have list
li = ['Peter', '22', 'DE']
and i want to create json from this list so i need to add to every parameter in list a name, so outpur will be something like this
li = ['name':'Peter', 'age':'22', 'nationality':'DE']
i = 0
while i < len(li):
li[i].insert(0,'name:')
i += 1
print(li)
this adding to every added name a coma...how can i add to this list without comma?
because output from this is :
['name','Peter', '22', 'DE']

li is the list object which you are iterating over so any operations on that will act as operations performed over the list.
insert() will insert elements in the existing list.
You can use zip() with dict() after creating the keys for the dictionary to get the desired output:
li = ['Peter', '22', 'DE']
keys = ['name', 'age', 'nationality']
di = dict(zip(keys, li))
Update: You can use list comprehension for list of lists:
li = [ ['Peter', '22', 'DE'], ['John', '28', 'GB'] ]
keys = ['name', 'age', 'nationality']
di = [dict(zip(keys, l)) for l in li]
print(di)

You could use zip, to combine all values of both lists and convert it to a dict:
li = ['Peter', '22', 'DE']
keys = ['name', 'age', 'nationality']
print(dict(zip(keys, li)))
Out:
{'name': 'Peter', 'age': '22', 'nationality': 'DE'}

Generate Keys and then map those keys with value present in the list to create a dictionary convert dictionary to JSON
ky = ['Name', 'Age', 'Nationality']
li = ['Peter', '22', 'DE']
data = {k:v for (k,v) in zip(ky, li)}

What you need is not a list, but a dictionary. Dictionaries are collections paired with keys and values:
my_dict = {"name" : "Peter", "age" : 22, "nationality" : "DE"}
If you need to build a dictionary over a list, you can do it with zip() method:
li = ['Peter', '22', 'DE']
keys = ["name", 'age', 'nationality']
dictionary = dict(zip(keys, li)) # {'name': 'Peter', 'age': '22', 'nationality': 'DE'}
If you really need to work with JSON files, then i suggest looking into this link for a clear explanation.

Dictionary of dictionaries from list of lists with year keys

I have a list of list like this:
[['2014', 'MALE', 'WHITE NON HISPANIC', 'Zachary', '90', '39'],
['2014', 'MALE', 'WHITE NON HISPANIC', 'Zev', '49', '65']]
I want to converted in a dictionary like this:
{{2012: {1: 'David',
2: 'Joseph',
3: 'Michael',
4: 'Moshe'},
2013: {1: 'David',
2: 'Joseph',
3: 'Michael',
4: 'Moshe'},
I'm trying to do a list comprehension like this:
boy_names = {row[0]:{i:row[3]} for i,row in enumerate(records) if row[1]=='MALE'}
But the result I'm getting is like:
{'2011': {7851: 'Zev'}, '2012': {9855: 'Zev'},
'2013': {11886: 'Zev'}, '2014': {13961: 'Zev'}}
If I'm right, I think I'm taking the last value and its row number from enumerate by the year key, but no idea how to solve it.

You can use the length of the sub-dict under the year key to calculate the next incremental numeric key for the sub-dict under the current year. Use the dict.setdefault method to default the value of a new key to an empty dict:
boy_names = {}
for year, _, _, name, _, _ in records:
record = boy_names.setdefault(int(year), {})
record[len(record) + 1] = name

I believe you need
data = [['2014', 'MALE', 'WHITE NON HISPANIC', 'Zachary', '90', '39'],
['2014', 'MALE', 'WHITE NON HISPANIC', 'Zev', '49', '65']]
result = {}
for i in data: #Iterate sub-list
result.setdefault(i[0], []).append(i[3]) #Form DICT
result = {k: dict(enumerate(v, 1)) for k, v in result.items()} #Use enumerate to get index number
print(result)
# {'2014': {1: 'Zachary', 2: 'Zev'}}

How to make a dictionary from two nested list?

I have two nested lists:
list1 = [['s0'], ['s1'], ['s2']]
list2 = [['hello','world','the'],['as','per','the'],['assets','order']]
and I want to make a dictionary from these lists with keys from list1 and values from list2:
d = {s0:['hello','world','the'],s1:['as','per','the'],s2:['assets','order']}
The output should look like this:
d = {s0:['hello','world','the'],s1:['as','per','the'],s2:['assets','order']}
The following code works if list1 is a normal (non-nested) list. But it doesn't work when list1 is a nested list.
dict(zip(list1, list2))

The problem here is that lists are not hashable, so one thing you can do is to flatten your list with itertools.chain and then build the dictionary with strings (which are immutable) as keys following you're current approach (read here for a more detailed explanation on this topic):
from itertools import chain
dict(zip(chain.from_iterable(list1),list2))
{'s0': ['hello', 'world', 'the'],
's1': ['as', 'per', 'the'],
's2': ['assets', 'order']}

If you want to do it manually (to understand algorithm for exemple), here is a way to do so:
list1 = [['s0'], ['s1'], ['s2']]
list2 = [['hello','world','the'],['as','per','the'],['assets','order']]
if len(list1) != len(list2):
exit(-1)
res = {}
for index, content in enumerate(list1):
res[content[0]] = list2[index]
print(res)

Another answer could be :
list1 = [['s0'], ['s1'], ['s2']]
list2 = [['hello','world','the'],['as','per','the'],['assets','order']]
output_dict = {element1[0]: element2 for element1, element2 in zip(list1, list2)}
An similar way of this dict-comprehension :
output_dict = {element1: element2 for [element1], element2 in zip(list1, list2)}
Output :
{'s0': ['hello', 'world', 'the'],
's1': ['as', 'per', 'the'],
's2': ['assets', 'order']}

It's a strange way to store matching information in the first place, but I would combine them like this:
list1 = [['s0'], ['s1'], ['s2']]
list2 = [['hello','world','the'],['as','per','the'],['assets','order']]
assert(len(list1) == len(list2))
output_dict = dict()
for index in range(len(list1)):
output_dict[list1[index][0] = list2[index]
result:
{'s0': ['hello', 'world', 'the'], 's1': ['as', 'per', 'the'], 's2': ['assets', 'order']}
I am assuming that the variables s0, s1 and s2 are meant to be strings like in the first list.

Use of dictionary in Python

I'm writing a concept learning programs, where I need to convert from index to the name of categories.
For example:
# binary concept learning
# candidate eliminaton learning algorithm
import numpy as np
import csv
def main():
d1={0:'0', 1:'Japan', 2: 'USA', 3: 'Korea', 4: 'Germany', 5:'?'}
d2={0:'0', 1:'Honda', 2: 'Chrysler', 3: 'Toyota', 4:'?'}
d3={0:'0', 1:'Blue', 2:'Green', 3: 'Red', 4:'White', 5:'?'}
d4={0:'0', 1:1970,2:1980, 3:1990, 4:2000, 5:'?'}
d5={0:'0', 1:'Economy', 2:'Sports', 3:'SUV', 4:'?'}
a=[0,1,2,3,4]
print a
if __name__=="__main__":
main()
So [0,1,2,3,4] should convert to ['0', 'Honda', 'Green', '1990', '?']. What is the most pythonic way to do this?

I think you need a basic dictionary crash course:
this is a proper dictionary:
>>>d1 = { 'tires' : 'yoko', 'manufacturer': 'honda', 'vtec' : 'no' }
You can call invidual things in the dictionary easily:
>>>d1['tires']
'yoko'
>>>d1['vtec'] = 'yes' #mad vtec yo
>>>d1['vtec']
'yes'
Dictionaries are broken up into two different sections, the key and the value
testDict = {'key':'value'}
You were using a dictionary the exact same way as a list:
>>>test = {0:"thing0", 1:"thing1"} #dictionary
>>>test[0]
'thing0'
which is pretty much the exact same as saying
>>>test = ['thing0','thing1'] #list
>>>test[0]
'thing0'
in your particular case, you may want to either format your dictionaries properly ( i would suggest something like masterdictionary = {'country': ['germany','france','USA','japan], 'manufacturer': ['honda','ferrarri','hoopty'] } etcetera because you could call each individual item you wanted a lot easier
with that same dictionary:
>>>masterdictionary['country'][1]
'germany'
which is
dictionaryName['key'][iteminlistindex]
of course there is nothing preventing you from putting dictionaries as values inside of dictionaries.... inside values of other dictionaries...

You can do:
data = [d1,d2,d3,d4,d5]
print [d[key] for key, d in zip(a, data)]
The function zip() can be used to combine to iterables; lists in this case.

You've already got the answer to your direct question, but you may wish to consider re-structuring the data. To me, the following makes a lot more sense, and will enable you to more easily index into it for what you asked, and for any possible later queries:
from pprint import pprint
items = [[el.get(i, '?') for el in (d1,d2,d3,d4,d5)] for i in range(6)]
pprint(items)
[['0', '0', '0', '0', '0'],
['Japan', 'Honda', 'Blue', 1970, 'Economy'],
['USA', 'Chrysler', 'Green', 1980, 'Sports'],
['Korea', 'Toyota', 'Red', 1990, 'SUV'],
['Germany', '?', 'White', 2000, '?'],
['?', '?', '?', '?', '?']]

I would use a list of dicts d = [d1, d2, d3, d4, d5], and then a list comprehension:
[d[i][key] for i, key in enumerate(a)]
To make the whole thing more readable, use nested dictionaries - each of your dictionaries seems to represent something you could give a more descriptive name than d1 or d2:
data = {'country': {0: 'Japan', 1: 'USA' ... }, 'brand': {0: 'Honda', ...}, ...}
car = {'country': 1, 'brand': 2 ... }
[data[attribute][key] for attribute, key in car.items()]
Note this would not necessarily be in order if that is important, though I think there is an ordered dictionary type.
As suggested by the comment, a dictionary with contiguous integers as keys can be replaced by a list:
data = {'country': ['Japan', 'USA', ...], 'brand': ['Honda', ...], ...}

If you need to keep d1, d2, etc. as is:
newA = [locals()["d%d"%(i+1)][a_value] for i,a_value in enumerate(a)]
Pretty ugly, and fragile, but it should work with your existing code.

You don't need a dictionary for this at all. Lists in python automatically support indexing.
def main():
d1=['0','Japan','USA','Korea','Germany',"?"]
d2=['0','Honda','Chrysler','Toyota','?']
d3=['0','Blue','Green','Red','White','?']
d4=['0', 1970,1980,1990,2000,'?']
d5=['0','Economy','Sports','SUV','?']
ds = [d1, d2, d3, d4, d5] #This holds all your lists
#This is what range is for
a=range(5)
#Find the nth index from the nth list, seems to be what you want
print [ds[n][n] for n in a] #This is a list comprehension, look it up.