I need fill empty dictionary in loop, but my script give me error. How can I do that operation? Thanks
Script:
import numpy as np
name = np.asarray(["John", "Peter", "Jan", "Paul"])
score = np.asarray([1, 2, 3, 4])
apdict = {"Name": "null", "Score": "null"}
for name in name:
for score in score:
apdict["Name"][name] = name[name]
apdict["Score"][score] = score[score]
The error:
Traceback (most recent call last):
File "<ipython-input-814-25938bb38ac2>", line 8, in <module>
apdict["Name"][name] = name[name]
TypeError: string indices must be integers
Possible outputs:
#possible output 1:
apdict = {["Name": "John", "Score": "1"], ["Name": "Peter", "Score": "3"]}
#possible output2:
apdict = {["Name": "John", "Score": "1", "3", "4"], ["Name": "Paul", "Score": "1"]}
If you want to create a dict with elements in name as keys and elements in score as values, based on the 2 numpy arrays, you can do it as follows:
apdict = dict(zip(name, score))
print(apdict)
{'John': 1, 'Peter': 2, 'Jan': 3, 'Paul': 4}
Edit
Based on your newly added possible output, I think it would better be "list of dictionary" instead of something looks like set of something (since {...} immediately encloses lists) that looks like lists (since [...] encloses something) but those something enclosed in list looks more like a dictionary rather than legitimate list items. The valid format of "list of dictionary" should look like below:
[{'Name': 'John', 'Score': 1},
{'Name': 'Peter', 'Score': 2},
{'Name': 'Jan', 'Score': 3},
{'Name': 'Paul', 'Score': 4}]
In this case, you can achieve it as follows:
apdict = [{'Name': k, 'Score': v} for k, v in zip(name, score)]
print(apdict)
[{'Name': 'John', 'Score': 1},
{'Name': 'Peter', 'Score': 2},
{'Name': 'Jan', 'Score': 3},
{'Name': 'Paul', 'Score': 4}]
Alternatively, you can also use Pandas (as you tagged pandas in the question), as follows:
import pandas as pd
apdict = pd.DataFrame({'Name': name, 'Score': score}).to_dict('records')
print(apdict)
[{'Name': 'John', 'Score': 1},
{'Name': 'Peter', 'Score': 2},
{'Name': 'Jan', 'Score': 3},
{'Name': 'Paul', 'Score': 4}]
You are trying to access the element with an index of string instead of an integer:
apdict["Name"][name] = name[name]
name needs to be an integer.
Related
I've got two lists:
lst1 = [{"name": "Hanna", "age":3},
{"name": "Kris", "age": 18},
{"name":"Dom", "age": 15},
{"name":"Tom", "age": 5}]
and the second one contains a few of above key name values under different key:
lst2 = [{"username": "Kris", "Town": "Big City"},
{"username":"Dom", "Town": "NYC"}]
I would like to merge them with result:
lst = [{"name": "Hanna", "age":3},
{"name": "Kris", "age": 18, "Town": "Big City"},
{"name":"Dom", "age": 15, "Town": "NYC"},
{"name":"Tom", "age":"5"}]
The easiest way is to go one by one (for each element from lst1, check whether it exists in lst2), but for big lists, this is quite ineffective (my lists have a few hundred elements each). What is the most effective way to achieve this?
To avoid iterating over another list again and again, you can build a name index first.
lst1 = [{"name": "Hanna", "age":3},
{"name": "Kris", "age": 18},
{"name":"Dom", "age": 15},
{"name":"Tom", "age": 5}]
lst2 = [{"username": "Kris", "Town": "Big City"},
{"username":"Dom", "Town": "NYC"}]
name_index = { dic['username'] : idx for idx, dic in enumerate(lst2) if dic.get('username') }
for dic in lst1:
name = dic.get('name')
if name in name_index:
dic.update(lst2[name_index[name]]) # update in-place to further save time
dic.pop('username')
print(lst1)
One way to do this a lot more efficient than by lists is to create an intermediate dictionary from lst1 with name as key, so that you're searching a dictionary not a list.
d1 = {elem['name']: {k:v for k,v in elem.items()} for elem in lst1}
for elem in lst2:
d1[elem['username']].update( {k:v for k,v in elem.items() if k != 'username'} )
lst = list(d1.values())
Output:
[{'name': 'Hanna', 'age': 3}, {'name': 'Kris', 'age': 18, 'Town': 'Big City'}, {'name': 'Dom', 'age': 15, 'Town': 'NYC'}, {'name': 'Tom', 'age': 5}]
edited to only have one intermediate dict
Use zip function to pair both lists. We need to order both lists using some criteria, in this case, you must use the username and name keys for the lists because those values will be your condition to perform the updating action, for the above reason is used the sorted function with key param. It is important to sort them out to get the match.
Finally your list lst2 has a little extra procedure, I expanded it taking into account the length of lst1, that is what I do using lst2 * abs(len(lst1) - len(lst2). Theoretically, you are iterating once over an iterable zip object, therefore I consider this could be a good solution for your requirements.
for d1, d2 in zip(sorted(lst1, key=lambda d1: d1['name']),
sorted(lst2 * abs(len(lst1) - len(lst2)), key=lambda d2: d2['username'])):
if d1['name'] == d2['username']:
d1.update(d2)
# Just we delete the username
del d1['username']
print(lst1)
Output:
[{'name': 'Hanna', 'age': 3}, {'name': 'Kris', 'age': 18, 'Town': 'Big City'}, {'name': 'Dom', 'age': 15, 'Town': 'NYC'}, {'name': 'Tom', 'age': 5}]
I currently have a list of strings that I am trying to create each string item into a dictionary object and store it within a list.
In attempting to create this list of dictionaries, I repeatedly create one big dictionary instead of iterating through item by item.
My code:
clothes_dict = [{clothes_list[i]: clothes_list[i + 1] for i in range(0, len(clothes_list), 2)}]
The error (All items being merged into one dictionary):
clothes_dict = {list: 1} [{'name': 'Tom', 'age': 10}, {'name': 'Mark', 'age': 5}, {'name': 'Pam', 'age': 7}]
0 = {dict: 2} {'name': 'Tom', 'age': 10}, {dict: 2} {'name': 'Mark', 'age': 5}, {'name': 'Pam', 'age': 7}```
Target Output (All Items being created into separate dictionaries within the single list):
clothes_dict = {list: 3} [{'name': 'Tom', 'age': 10}, {'name': 'Mark', 'age': 5}, {'name': 'Pam', 'age': 7}]
0 = {dict: 2} {'name': 'Tom', 'age': 10}
1 = {dict: 2} {'name': 'Mark', 'age': 5}
2 = {dict: 2} {'name': 'Pam', 'age': 7}```
I am attempting to make each entry within the list a new dictionary in the same form as the target output image.
clothes_dict = [{clothes_list[i]: clothes_list[i + 1]} for i in range(0, len(clothes_list), 2)]
You misplaced your closing right curly brace '}' in your list comprehension and placed it at the end which meant you was performing a dictionary comprehension as opposed to a list comprehension with each item being a dictionary.
Your code creates a list with a single dictionary:
clothes_dict = [{clothes_list[i]: clothes_list[i + 1] for i in range(0,l en(clothes_list), 2)}]
If you (for some reason) want a list of dictionaries with single entries:
clothes_dict = [{clothes_list[i]: clothes_list[i + 1]} for i in range(0,l en(clothes_list), 2)]
However, it seems to me that this may be a bit of an XY problem - in what case is a list of single-entry dictionaries the required format? Why not use a list of tuples for example?
I have a list of tuples
data = [('2015-10-08', '2016-07-17'), ('Alex', 'Jerry'), (5, 6)]
And I have a list, this list contains column headings.
title = [Date , Name, Age]
With this list and list of tuples I want a dictionary of dictionaries
This is the expected output
output = {'1' : {'Date': '2015-10-08', 'Name': 'Alex', 'Age': 5},
'2' : {'Date': '2016-07-17', 'Name': 'Jerry', 'Age': 6}}
I tried
output = {}
for i in range(len(title)):
output[i+1] = {title[i]: ",".join(data[i])}
print (output)
I am getting
{1: {'Date': '2015-10-08','2016-07-17'}, 2: {'Name': 'Alex','Jerry'}, 3: {'Age': 5,6}}
You could simply use loops to build easy to read code. But the Pythonic way would be to use comprehensions:
result = {str(i + 1): {t: data[j][i] for j, t in enumerate(title)}
for i in range(len(data[0]))}
It gives as expected:
{'1': {'Age': 5, 'Date': '2015-10-08', 'Name': 'Alex'},
'2': {'Age': 6, 'Date': '2016-07-17', 'Name': 'Jerry'}}
How would I generate a list of values of a certain field of objects in a list?
Given the list of objects:
[ {name: "Joe", group: 1}, {name: "Kirk", group: 2}, {name: "Bob", group: 1}]
I want to generate list of the name field values:
["Joe", "Kirk", "Bob"]
The built-in filter() function seems to come close, but it will return the entire objects themselves.
I'd like a clean, one line solution such as:
filterLikeFunc(function(obj){return obj.name}, mylist)
Sorry, I know that's c syntax.
Just replace filter built-in function with map built-in function.
And use get function which will not give you key error in the absence of that particular key to get value for name key.
data = [{'name': "Joe", 'group': 1}, {'name': "Kirk", 'group': 2}, {'name': "Bob", 'group': 1}]
print map(lambda x: x.get('name'), data)
In Python 3.x
print(list(map(lambda x: x.get('name'), data)))
Results:
['Joe', 'Kirk', 'Bob']
Using List Comprehension:
print [each.get('name') for each in data]
Using a list comprehension approach you get:
objects = [{'group': 1, 'name': 'Joe'}, {'group': 2, 'name': 'Kirk'}, {'group': 1, 'name': 'Bob'}]
names = [i["name"] for i in objects]
For a good intro to list comprehensions, see https://docs.python.org/2/tutorial/datastructures.html
Just iterate over your list of dicts and pick out the name value and put them in a list.
x = [ {'name': "Joe", 'group': 1}, {'name': "Kirk", 'group': 2}, {'name': "Bob", 'group': 1}]
y = [y['name'] for y in x]
print(y)
i have a result tuple of dictionaries.
result = ({'name': 'xxx', 'score': 120L }, {'name': 'xxx', 'score': 100L}, {'name': 'yyy', 'score': 10L})
I want to uniqify it. After uniqify operation result = ({'name': 'xxx', 'score': 120L }, {'name': 'yyy', 'score': 10L})
The result contain only one dictionary of each name and the dict should have maximum score. The final result should be in the same format ie tuple of dictionary.
from operator import itemgetter
names = set(d['name'] for d in result)
uniq = []
for name in names:
scores = [res for res in result if res['name'] == name]
uniq.append(max(scores, key=itemgetter('score')))
I'm sure there is a shorter solution, but you won't be able to avoid filtering the scores by name in some way first, then find the maximum for each name.
Storing scores in a dictionary with names as keys would definitely be preferable here.
I would create an intermediate dictionary mapping each name to the maximum score for that name, then turn it back to a tuple of dicts afterwards:
>>> result = ({'name': 'xxx', 'score': 120L }, {'name': 'xxx', 'score': 100L}, {'name': 'xxx', 'score': 10L}, {'name':'yyy', 'score':20})
>>> from collections import defaultdict
>>> max_scores = defaultdict(int)
>>> for d in result:
... max_scores[d['name']] = max(d['score'], max_scores[d['name']])
...
>>> max_scores
defaultdict(<type 'int'>, {'xxx': 120L, 'yyy': 20})
>>> tuple({name: score} for (name, score) in max_scores.iteritems())
({'xxx': 120L}, {'yyy': 20})
Notes:
1) I have added {'name': 'yyy', 'score': 20} to your example data to show it working with a tuple with more than one name.
2)I use a defaultdict that assumes the minimum value for score is zero. If the score can be negative you will need to change the int parameter of defaultdict(int) to a function that returns a number smaller than the minimum possible score.
Incidentally I suspect that having a tuple of dictionaries is not the best data structure for what you want to do. Have you considered alternatives, such as having a single dict, perhaps with a list of scores for each name?
I would reconsider the data structure to fit your needs better (for example dict hashed with name with list of scores as value), but I would do like this:
import operator as op
import itertools as it
result = ({'name': 'xxx', 'score': 120L },
{'name': 'xxx', 'score': 100L},
{'name': 'xxx', 'score': 10L},
{'name':'yyy', 'score':20})
# groupby
highscores = tuple(max(namegroup, key=op.itemgetter('score'))
for name,namegroup in it.groupby(result,
key=op.itemgetter('name'))
)
print highscores
How about...
inp = ({'name': 'xxx', 'score': 120L }, {'name': 'xxx', 'score': 100L}, {'name': 'yyy', 'score': 10L})
temp = {}
for dct in inp:
if dct['score'] > temp.get(dct['name']): temp[dct['name']] = dct['score']
result = tuple({'name': name, 'score': score} for name, score in temp.iteritems())