I have a QueryDyct object (request.data) in my method and want to do some modifications on it so I have copied it: data_dict = data.copy(). On creating an empty list in it, it creates an additional inner list:
data_dict['emails'] = []
data_dict['emails'].append('foo#mail.com')
data_dict['emails'].append('bar#mail.com')
Instead of crating an empty list and append data into it, it creates a list of list, and appends data into inner list:
On PyCharm watch it is shown as:
Why it behaves like this?
And for the further processing it is treated (by django validator) as an email with address ['foo#mail.com', 'bar#mail.com'], but I want to have two different emails with appended addresses.
How can I construct normal empty list in QueryDict ?
A QueryDict is a MultiValueDict to begin with, as you can repeat GET-parameters. I.e. it handles its values in lists by default. Good news is that you can use its features directly to handle lists:
from django.http.request import QueryDict
d = QueryDict(mutable=True)
d.setlist('emails', ['abc#abc.com', 'def#abc.com'])
d['emails']
# 'def#abc.com'
d.getlist('emails')
# ['abc#abc.com', 'def#abc.com']
d.appendlist('emails', 'foo#bar.baz')
d
# <QueryDict: {'emails': ['abc#abc.com', 'def#abc.com', 'foo#bar.baz']}>
d.urlencode()
# 'emails=abc%40abc.com&emails=def%40abc.com&emails=foo%40bar.baz'
Related
So my current code looks like this:
response = requests.get("https://pokeapi.co/api/v2/pokemon/palkia").json()
data = response["types"][1]
print(data)
Write now it only give the dictionary of the second typing, how can I get the just the name of the to typing's as the result
response["types"] is a list of dictionaries, and by indexing it with [1] you access the second element. To get the names for all elements in response["types"], you could e.g. do
data = [e["type"]["name"] for e in response["types"]]
I have a json code from API and I want to get new chat members with the code below but I only get the first two results and not the last (Tester). Why? It should itereate through the whole json file, shouldn't it?
r = requests.get("https://api.../getUpdates").json()
chat_members = []
a = 0
for i in r:
chat_members.append(r['result'][a]['message']['new_chat_members'][0]['last_name'])
a = a + 1
Json here:
{"ok":true,"result":[{"update_id":213849278,
"message":{"message_id":37731,"from":{"id":593029363,"is_bot":false,"first_name": "#tutu"},"chat":{"id":-1001272017174,"title":"tester account","username":"v_glob","type":"supergroup"},"date":1537470595,"new_chat_participant":{"id":593029363,"is_bot":false,"first_name":"tutu "},"new_chat_member":{"id":593029363,"is_bot":false,"first_name":"\u7535\u62a5\u589e\u7c89\uff0c\u4e2d\u82f1\u6587\u5ba2\u670d\uff0c\u62c9\u4eba\u6e05\u5783\u573e\u8f6f\u4ef6\uff0c\u5e7f\u544a\u63a8\u5e7f\uff0cKYC\u6750\u6599\u8ba4\u8bc1\uff0c","last_name":"#tutupeng"},"new_chat_members":[{"id":593029363,"is_bot":false,"first_name":"\u7535\u62a5\u589e\u7c89\uff0c\u4e2d\u82f1\u6587\u5ba2\u670d\uff0c\u62c9\u4eba\u6e05\u5783\u573e\u8f6f\u4ef6\uff0c\u5e7f\u544a\u63a8\u5e7f\uff0cKYC\u6750\u6599\u8ba4\u8bc1\uff0c","last_name":"#tutu"}]}},{"update_id":213849279,
"message":{"message_id":37732,"from":{"id":658150956,"is_bot":false,"first_name":"Rebecca","last_name":"Lawson"},"chat":{"id":-10012720,"title":"v glob OFFICIAL","username":"v_glob","type":"supergroup"},"date":1537484441,"new_chat_participant":{"id":65815,"is_bot":false,"first_name":"Rebecca","last_name":"Lawson"},"new_chat_member":{"id":65815,"is_bot":false,"first_name":"Rebecca","last_name":"Lawson"},"new_chat_members":[{"id":65815,"is_bot":false,"first_name":"Rebecca","last_name":"Lawson"}]}},{"update_id":213849280,
"message":{"message_id":12,"from":{"id":696749142,"is_bot":false,"first_name":"daniel","language_code":"cs-cz"},"chat":{"id":696749142,"first_name":"daniel","type":"private"},"date":1537537013,"text":"/stat","entities":[{"offset":0,"length":5,"type":"bot_command"}]}},{"update_id":213849281,
"message":{"message_id":37740,"from":{"id":669620,"is_bot":false,"first_name":"Ivan","last_name":"Tester"},"chat":{"id":-100127201,"title":"test account","username":"v_glob","type":"supergroup"},"date":1537537597,"new_chat_participant":{"id":669620191,"is_bot":false,"first_name":"Ivan","last_name":"Tester"},"new_chat_member":{"id":669620191,"is_bot":false,"first_name":"Ivan","last_name":"Tester"},"new_chat_members":[{"id":669620191,"is_bot":false,"first_name":"Ivan","last_name":"Tester"}]}}]}
Because you iterate over the entire response dict. The top level only has two items, so that's what you iterate over. Note that you don't actually use the iterator variable, and you have a completely unnecessary separate counter.
Instead, you should be iterating over the result dict:
for result in r['result']:
if "new_chat_members" in result['message']:
chat_members.append(result['message']['new_chat_members'][0]['last_name'])
A colleague of mine has come up with a solution:
for i in l['result']:
chat_members.append(i['message']['new_chat_member']['first_name'])
To sum up: Iterate through 'result' with no positional arguments
I'm building a small Django app. The user is to build up a list of pairs of strings, by repeatedly submitting a pair of forms. The list is stored as value of a session variable.
But there is a problem. When the list is loaded by the view, the strings which were to be elements of the pairs in the list wind up getting prefixed by the letter u. So for example as elements of the list instead of
['a','b']
['c','d']
we get
[u'a',u'b']
[u'c',u'd']
My impression is that Django treats values of session variables as strings. So maybe the problem has something to do with the conversion of lists?
Here is the source of the view:
def plisting(request):
if 'plist' not in request.session:
request.session['plist']=[]
plist = request.session['plist']
if 'entry' in request.POST:
entry = str(request.POST['entry'])
key = str(request.POST['key'])
plist = plist+[[entry,key]]
request.session['plist'] = plist
return render(request,'evaluator/plisting.html',{'plist':plist})
The u at the beginning of the string returned to your view function just means that the strings are encoded in unicode. You need not handle it specially in your code. Accessing the list just normally will return you the expected strings.
This is what I mean:
>>> a = [u'a',u'b']
>>> b = [u'c',u'd']
>>> a[1]
'b'
>>> b[0]
'c'
As evident from the output, when you access the elements, you do not see the 'u' as part of the strings.
Hope this helps.
I have a list of dictionaries called lod. All dictionaries have the same keys but different values. I am trying to update one specific value in the list of values for the same key in all the dictionaries.
I am attempting to do it with the following for loop:
for i in range(len(lod)):
a=lod[i][key][:]
a[p]=a[p]+lov[i]
lod[i][key]=a
What's happening is each is each dictionary is getting updated len(lod) times so lod[0][key][p] is supposed to have lov[0] added to it but instead it is getting lov[0]+lov[1]+.... added to it.
What am I doing wrong?
Here is how I declared the list of dicts:
lod = [{} for _ in range(len(dataul))]
for j in range(len(dataul)):
for i in datakl:
rrdict[str.split(i,',')[0]]=list(str.split(i,',')[1:len(str.split(i,','))])
lod[j]=rrdict
The problem is in how you created the list of dictionaries. You probably did something like this:
list_of_dicts = [{}] * 20
That's actually the same dict 20 times. Try doing something like this:
list_of_dicts = [{} for _ in range(20)]
Without seeing how you actually created it, this is only an example solution to an example problem.
To know for sure, print this:
[id(x) for x in list_of_dicts]
If you defined it in the * 20 method, the id is the same for each dict. In the list comprehension method, the id is unique.
This it where the trouble starts: lod[j] = rrdict. lod itself is created properly with different dictionaries. Unfortunately, afterwards any references to the original dictionaries in the list get overwritten with a reference to rrdict. So in the end, the list contains only references to one single dictionary. Here is some more pythonic and readable way to solve your problem:
lod = [{} for _ in range(len(dataul))]
for rrdict in lod:
for line in datakl:
splt = line.split(',')
rrdict[splt[0]] = splt[1:]
You created the list of dictionaries correctly, as per other answer.
However, when you are updating individual dictionaries, you completely overwrite the list.
Removing noise from your code snippet:
lod = [{} for _ in range(whatever)]
for j in range(whatever):
# rrdict = lod[j] # Uncomment this as a possible fix.
for i in range(whatever):
rrdict[somekey] = somevalue
lod[j] = rrdict
Assignment on the last line throws away the empty dict that was in lod[j] and inserts a reference to the object represented by rrdict.
Not sure what your code does, but see a commented-out line - it might be the fix you are looking for.
I have a dict like this:
(100002: 'APPLE', 100004: 'BANANA', 100005: 'CARROT')
I am trying to make my dict have ints for the keys (as it does now) but have sets for the values (rather than strings as it is now.) My goal is to be able to read from a .csv file with one column for the key (an int which is the item id number) and then columns for things like size, shape, and color. I want to add this information into my dict so that only the information for keys already in dict are added.
My goal dict might look like this:
(100002: set(['APPLE','MEDIUM','ROUND','RED']), 100004: set(['Banana','MEDIUM','LONG','YELLOW']), 100005: set(['CARROT','MEDIUM','LONG','ORANGE'])
Starting with my dict of just key + string for item name, I tried code like this to read the extra information in from a .csv file:
infile = open('FileWithTheData.csv', 'r')
for line in infile.readlines():
spl_line = line.split(',')
if int(spl_line[0]) in MyDict.keys():
MyDict[int(spl_line[0])].update(spl_line[1:])
Unfortunately this errors out saying AttributeError: 'str' object has no attribute 'update'. My attempts to change my dictionary's values into sets so that I can then .update them have yielded things like this: (100002: set(['A','P','L','E']), 100004: set(['B','A','N']), 100005: set(['C','A','R','O','T']))
I want to convert the values to a set so that the string that is currently the value will be the first string in the set rather than breaking up the string into letters and making a set of those letters.
I also tried making the values a set when I create the dict by zipping two lists together but it didn't seem to make any difference. Something like this
MyDict = dict(zip(listofkeys, set(listofnames)))
still makes the whole listofnames list into a set but it doesn't achieve my goal of making each value in MyDict into a set with the corresponding string from listofnames as the first string in the set.
How can I make the values in MyDict into a set so that I can add additional strings to that set without turning the string that is currently the value in the dict into a set of individual letters?
EDIT:
I currently make MyDict by using one function to generate a list of item ids (which are the keys) and another function which looks up those item ids to generate a list of corresponding item names (using a two column .csv file as the data source) and then I zip them together.
ANSWER:
Using the suggestions here I came up with this solution. I found that the section that has set()).update can easily be changed to list()).append to yield a list rather than a set (so that the order is preserved.) I also found it easier to update by .csv data input files by adding the column containing names to the FileWithTheData.csv so that I didn't have to mess with making the dict, converting the values to sets, and then adding in more data. My code for this section now looks like this:
MyDict = {}
infile = open('FileWithTheData.csv', 'r')
for line in infile.readlines():
spl_line = line.split(',')
if int(spl_line[0]) in itemidlist: #note that this is the list I was formerly zipping together with a corresponding list of names to make my dict
MyDict.setdefault(int(spl_line[0]), list()).append(spl_line[1:])
print MyDict
Your error is because originally your MyDict variable maps an integer to a string. When you are trying to update it you are treating the value like a set, when it is a string.
You can use a defaultdict for this:
combined_dict = defaultdict(set)
# first add all the values from MyDict
for key, value in MyDict.iteritems():
combined_dict[int(key)].add(value)
# then add the values from the file
infile = open('FileWithTheData.csv', 'r')
for line in infile.readlines():
spl_line = line.split(',')
combined_dict[int(sp_line[0])].update(spl_line[1:])
Your issue is with how you are initializing MyDict, try changing it to the following:
MyDict = dict(zip(listofkeys, [set([name]) for name in listofnames]))
Here is a quick example of the difference:
>>> listofkeys = [100002, 100004, 100005]
>>> listofnames = ['APPLE', 'BANANA', 'CARROT']
>>> dict(zip(listofkeys, set(listofnames)))
{100002: 'CARROT', 100004: 'APPLE', 100005: 'BANANA'}
>>> dict(zip(listofkeys, [set([name]) for name in listofnames]))
{100002: set(['APPLE']), 100004: set(['BANANA']), 100005: set(['CARROT'])}
set(listofnames) is just going to turn your list into a set, and the only effect that might have is to reorder the values as seen above. You actually want to take each string value in your list, and convert it to a one-element set, which is what the list comprehension does.
After you make this change, your current code should work fine, although you can just do the contains check directly on the dictionary instead of explicitly checking the keys (key in MyDict is the same as key in MyDict.keys()).