Develop Reference

Adding selective elements of one dictionary to another

I have a list as:
my_list = ["10", "12", "32", "23"]
and a dictionary as:
my_dict = {
'one': {'index': 0, 'sec_key': 'AB', 'id': '10'},
'two': {'index': 0, 'sec_key': 'CD', 'id': '12'},
'three': {'index': 0, 'sec_key': 'EF', 'id': '32'}
}
I want to keep a dictionary say final_dict which will have content of my_dict only if the id is present in my_list. I tried to do that by:
sf_dict = dict()
for list_id in my_list:
for key, value in my_dict.items():
if list_id == value['id']:
print("Values : {} {} {}".format(value['index'], value['sec_key'], value['id']))
sf_dict[key] = key
sf_dict[key]['index'] = value['index']
sf_dict[key]['sec_key'] = value['sec_key']
sf_dict[key]['id'] = value['id']
print(sf_dict)
I'm able to print the values but the assigning of those values is failing due to the error as:
TypeError: 'str' object does not support item assignment
Where am I making mistake?

You can use a dictionary comprehension to loop over your my_dict and check if the value stored in id is in your list.
my_list = ["10", "12", "32", "23"]
my_dict = {
'one': {'index': 0, 'sec_key': 'AB', 'id': '10'},
'two': {'index': 0, 'sec_key': 'CD', 'id': '12'},
'three': {'index': 0, 'sec_key': 'EF', 'id': '32'}
}
result = {key:value for key, value in my_dict.items() if value['id'] in my_list}
#{'one': {'index': 0, 'sec_key': 'AB', 'id': '10'}, 'two': {'index': 0, 'sec_key': 'CD', 'id': '12'}, 'three': {'index': 0, 'sec_key': 'EF', 'id': '32'}}
my_list = ["10"]
#{'one': {'index': 0, 'sec_key': 'AB', 'id': '10'}}

The already given answers should do the trick, altough maybe it is important to get why your approach isn't right.
The problem with what you did is that when you do sf_dict[key] = key, you're setting the value of the dictionary for the key key with the value key. But key is a string, thus, when you later try sf_dict[key]['index'] = value['index'] it tells you that you cannot assign on top a string.
Try to remove these lines
sf_dict[key]['index'] = value['index']
sf_dict[key]['sec_key'] = value['sec_key']
sf_dict[key]['id'] = value['id']
and simply replace them with sf_dict[key] = value. This way, when you find the id you're searching for, you assign all the dictionary at once to the corresponding key.

Here is one solution:
final_dict={}
for i in my_dict:
if my_dict[i]['id'] in my_list:
final_dict[i]=my_dict[i]
Example:
my_list = ["10", "32", "23"]
Output:
print(final_dict)
{'one': {'index': 0, 'sec_key': 'AB', 'id': '10'}, 'three': {'index': 0, 'sec_key': 'EF', 'id': '32'}}

for list_id in my_list:
for key, value in my_dict.items():
if list_id == value['id']:
print("Values : {} {} {}".format(value['index'], value['sec_key'], value['id']))
sf_dict[key] = {'index':value['index'],'sec_key':value['sec_key'],'id':value['id']}
print(sf_dict)
the code should be like this.

#problem is here
sf_dict[key] = key
above line means sf_dict = {'one': 'one'} for example
so next lines mean 'one'['anything'] =
so using index to assign into str is not allowed.
better solution is already answered above, just pointing issue in exist code

cerberus: Validate an optional field occurs at least once

I'm using cerberus to validate data. One of my fields is optional - it doesn't need to be present for every item. However, the key must be populated at least once across the entire data array.
As an example, say I want to validate the key 'c' occurs in at least one dictionary in my data list:
from cerberus import Validator
has_c = {'data': [{'a': 1, 'b': 2}, {'b': 2}, {'c': 3}]}
no_c = {'data': [{'a': 1, 'b': 2}, {'a': 1}]}
schema = {'data':
{'type': 'list',
'schema': {
'type': 'dict',
'schema': {
'a': {'required': True},
'b': {'required': True},
'c': {'required': False, 'at_least_one': True}
}
}
}
}
class MyValidator(Validator) # Some fancy code...
....
v = MyValidator()
v.validate(has_c, schema) # Passes
v.validate(no_c, schema) # Fails
This seems doable outside of cerberus, but I'd like to keep the method in my validator if possible.

If you want the method to be in the Validator subclass, then you will want to create a custom rule just like you were thinking.
from cerberus import Validator
test_with_c = {'data': [{'a': 1, 'b': 2}, {'b': 2}, {'c': 3}]}
test_with_no_c = {'data': [{'a': 1, 'b': 2}, {'a': 1}]}
class MyValidator(Validator):
def _validate_has_c(self, has_c, field, value):
seen_c = False
for v in value:
if "c" in v:
seen_c = True
if has_c and not seen_c:
self._error(field, "Must contain a 'c' key")
schema = {
"data": {
"type": "list",
"has_c": True
}
}
v = MyValidator(schema)
print(v(test_with_c), v.errors)
print(v(test_with_no_c), v.errors)
Running this will yield the results you want with respect to looking for a c key in one of the elements. Running that code yields
True {}
False {'data': ["Must contain a 'c' key"]}

KeyError: 'name' Why can't I use 'name'?

I wanna make a dictionary has name's key & data.In views.py I wrote
data_dict ={}
def try_to_int(arg):
try:
return int(arg)
except:
return arg
def main():
book4 = xlrd.open_workbook('./data/excel1.xlsx')
sheet4 = book4.sheet_by_index(0)
data_dict_origin = OrderedDict()
tag_list = sheet4.row_values(0)[1:]
for row_index in range(1, sheet4.nrows):
row = sheet4.row_values(row_index)[1:]
row = list(map(try_to_int, row))
data_dict_origin[row_index] = dict(zip(tag_list, row))
if data_dict_origin['name'] in data_dict:
data_dict[data_dict_origin['name']].update(data_dict_origin)
else:
data_dict[data_dict_origin['name']] = data_dict_origin
main()
When I printed out data_dict,it is
OrderedDict([(1, {'user_id': '100', 'group': 'A', 'name': 'Tom', 'dormitory': 'C'}), (2, {'user_id': '50', 'group': 'B', 'name': 'Blear', 'dormitory': 'E'})])
My ideal dictionary is
dicts = {
Tom: {
'user_id': '100',
'group': 'A',
'name': 'Tom',
'dormitory': 'C'
},
Blear: {
},
}
How should I fix this?What should I write it?

The code is using the wrong key in the dictionary. The keys are 1, 2, and do not have the name key. You can use this code instead:
for value in data_dict.values():
if value['name'] in data_dict:
data_dict[value['name']].update(value)
else:
data_dict[value['name']] = value

Your data_dict_origin has numbers as keys and dicts as values (which technically makes it a sparse array of dicts). The "name" key exists in those dicts, not in your data_dict.

Python Function: Given Scrabble Tile, What is the value?

I'm trying to write a simple function that given a letter, you are returned the value of the scrabble tile. Here is what I have:
def letterPoint(letter):
letter = letter.upper()
lettersWorthOne =(['A','E','I','N','O','R','S','T'])
lettersWorthTwo = (['D','G'])
lettersWorthThree = (['B','C','M','P'])
lettersWorthFour = (['F','H','U','V','W','Y'])
lettersWorthFive = (['K'])
lettersWorthEight = (['J','X'])
lettersWorthTen = (['Q','Z'])
if letterWorthOne:
print '1'
if letterWorthTwo:
print '2'
if letterWorthThree:
print '3'
if letterWorthFour:
print '4'
if letterWorthFive:
print '5'
if letterWorthEight:
print '8'
if letterWorthTen:
print '10'

Use a dictionary. Rather than
lettersWorthTwo = (['D','G']), etc.
You would have a data structure along the lines of:
letterValues = {'D':2, 'G':2, ... }
Then a lookup for value is simply:
letterValues['D'] # returns 2 for the value of the tile

To point out why your code doesn't work, because you're not comparing your letter to the list.
#Change from this:
if letterWorthOne:
print '1'
#to this, should work
if letter in letterWorthOne:
print '1'
.....
Use a python dictionary is the way to go.
Further to the solution someone has already posted. You can also construct a more content dictionary like this:
Letters = {
'a': { 'quantity' : 9, 'value': 1},
'b': { 'quantity' : 2, 'value': 3},
'c': { 'quantity' : 2, 'value': 3},
'd': { 'quantity' : 4, 'value': 2},
'e': { 'quantity' : 12, 'value': 1},
'f': { 'quantity' : 2, 'value': 4},
'g': { 'quantity' : 3, 'value': 2},
'h': { 'quantity' : 2, 'value': 4},
'i': { 'quantity' : 9, 'value': 1},
'j': { 'quantity' : 1, 'value': 8},
'k': { 'quantity' : 1, 'value': 5},
'l': { 'quantity' : 4, 'value': 1},
'm': { 'quantity' : 2, 'value': 3},
'n': { 'quantity' : 6, 'value': 1},
'o': { 'quantity' : 8, 'value': 1},
'p': { 'quantity' : 2, 'value': 3},
'q': { 'quantity' : 1, 'value': 10},
'r': { 'quantity' : 6, 'value': 1},
's': { 'quantity' : 4, 'value': 1},
't': { 'quantity' : 6, 'value': 1},
'u': { 'quantity' : 4, 'value': 1},
'v': { 'quantity' : 2, 'value': 4},
'w': { 'quantity' : 2, 'value': 4},
'x': { 'quantity' : 1, 'value': 8},
'y': { 'quantity' : 2, 'value': 4},
'z': { 'quantity' : 1, 'value': 10},
'*': { 'quantity' : 2, 'value': 0}
}
# to get to it's "content", like this:
Letters['a']
{'quantity': 9, 'value': 1}
# you can then get its 'value' or 'quantity' in a tile bag
Letters['a']['value']
1
# if you MUST use a function, do this with above dictionary, although it's quite pointless
def letter_point(letter):
return Letters[letter.upper()]['value']

In letterPoint(), letterWorthOne and lettersWorthOne are separate variables. Each lettersWorth* variable holds a list, and you appear to want letterWorthOne to contain a boolean value (True or False) specifying whether or not letter is in the lettersWorthOne list. To determine whether a value is in a collection, use operator in.
def letterPoint(letter):
letter = letter.upper()
lettersWorthOne =(['A','E','I','N','O','R','S','T'])
lettersWorthTwo = (['D','G'])
letterWorthOne = letter in lettersWorthOne
if letterWorthOne:
print '1'
letterWorthTwo = letter in lettersWorthTwo
if letterWorthTwo:
print '2'
# rest of values omitted for brevity
print 'E worth'
letterPoint('E')
print 'D worth'
letterPoint('D')
This program produces the following output:
E worth
1
D worth
2
This explains why your existing function doesn't work. But in the long run, I'd recommend using a dictionary to hold the value and quantity of each letter, and store the values as numbers instead of strings so that you can add the values of all letters in a word.

To use the code you began with, you could consider changing the lines to:
if letter in lettersWorthOne print 1
...
This is because the data structure you're using is a list (it is surrounded by [] brackets). The way to use these lists in your function is to see if they contain the letter in the list using the code:
if <variable> in <list> print <value>
The () brackets aren't doing anything, as mentioned in another answer so you can get rid of them.
This is just to explain to you why you don't see any result in the function you've written. The other answers which suggest using a dict (dictionary) are a better approach in practice.

Python defaultdict get original dictionary

Let's say we have:
from collections import defaultdict
original_dict = { 'somegroupofelements':{'name':1, 'group':1 ,'results':[1,2,3,4]}, 'somegroupofelements2':{'name':2, 'group': 2 ,'results':[1,2,3,4]}, 'somegroupofelements3':{'name':3, 'group':3 ,'results':[1,2,3,4]} }
new_dict = defaultdict(list)
for key, value in original_dict.iteritems():
# i need to organize things grouped for making the right latex tables
# and for updating some values...
value['key']=key
new_dict[value['group']].append(value)
I want that new_dict, after I've done my work, to be organized again just like the original_dict? Like reconstruct the original_dict from the new_dict.

So you end up with a dictionary in the form:
>>> d = { 100 : [{'name':1, 'group':100, 'key':'group1'},
... {'name':2, 'group':100, 'key':'group2'}],
... 200 : [{'name':3, 'group':200, 'key':'group3'}] }
Which can be transformed back into a dictionary using a dict comprehension:
>>> orig = { x['key']:x for v in d.values() for x in v }
>>> orig
{'group1': {'group': 100, 'name': 1, 'key': 'group1'},
'group3': {'group': 200, 'name': 3, 'key': 'group3'},
'group2': {'group': 100, 'name': 2, 'key': 'group2'}}
If you want, you can then delete the superflous key field of the items:
>>> for v in orig.values(): del v['key']
...
>>> orig
{'group1': {'group': 100, 'name': 1},
'group3': {'group': 200, 'name': 3},
'group2': {'group': 100, 'name': 2}}