I've been programming for less than four weeks and have run into a problem that I cannot figure out. I'm trying to append a string value to an existing key with an existing string stored in it but if any value already exists in the key I get "str object has no attribute 'append'.
I've tried turning the value to list but this also does not work. I need to use the .append() attribute because update simply replaces the value in clientKey instead of appending to whatever value is already stored. After doing some more research, I understand now that I need to somehow split the value stored in clientKey.
Any help would be greatly appreciated.
data = {}
while True:
clientKey = input().upper()
refDate = strftime("%Y%m%d%H%M%S", gmtime())
refDate = refDate[2 : ]
ref = clientKey + refDate
if clientKey not in data:
data[clientKey] = ref
elif ref in data[clientKey]:
print("That invoice already exists")
else:
data[clientKey].append(ref)
break
You can't .append() to a string because a string is not mutable. If you want your dictionary value to be able to contain multiple items, it should be a container type such as a list. The easiest way to do this is just to add the single item as a list in the first place.
if clientKey not in data:
data[clientKey] = [ref] # single-item list
Now you can data[clientkey].append() all day long.
A simpler approach for this problem is to use collections.defaultdict. This automatically creates the item when it's not there, making your code much simpler.
from collections import defaultdict
data = defaultdict(list)
# ... same as before up to your if
if clientkey in data and ref in data[clientkey]:
print("That invoice already exists")
else:
data[clientKey].append(ref)
You started with a string value, and you cannot call .append() on a string. Start with a list value instead:
if clientKey not in data:
data[clientKey] = [ref]
Now data[clientKey] references a list object with one string in it. List objects do have an append() method.
If you want to keep appending to the string you can use data[clientKey]+= ref
Related
I have this problem where I would like to add a value to a dictionary but the key is duplicate.
I would like the key to to hold a list with multiple values
this is what I have tried
def storingPassword():
username=("bob")#key,
password=("PASSWROD1")#value
allpasswords={
"jeff":"jeff 123 ",
"bob" : "bob 123"
}
if username not in allpasswords:
allpasswords[username]=password
else:
allpasswords[username].append(password)
return allpasswords
but i keep getting this error
"AttributeError: 'str' object has no attribute 'append'"
I expect a output something like this;
"jeff":"jeff 123 ",
"bob" : ["bob 123","PASSWORD1"]
That's because the value in your allpasswords dict is a string and you are trying to treat it like a list. Why are you trying to make your data structure complex with few values as list and few as string? I recommend to convert everything to list for a simpler logic.
Hence your code should be like this:
allpasswords={
"jeff": ["jeff 123 "],
"bob" : ["bob 123"]
}
allpasswords[username].append(password)
Instead of using dict object, you can use collections.defaultdict. It will let you define a dict with default value as list. So you don't need to even explicitly initialise value of new key as list. For example:
from collections import defaultdict
my_dict = defaultdict(list)
my_dict['new_key'].append('new_value')
# dictionary will hold the value:
# {'new_key': ['new_value']})
Initiate your dictionary entry with a list instead of just a string.
allpasswords[username] = [password] # List containing a single password
You will then be able to append to it.
(Having some entries contain a string while others contain a list of strings is best avoided - when it is time to look them up or print them, you would have to check each time whether it is a list or string.)
I am asking an ElasticSearch database to provide me with a list of indices and their creation dates using Python 2.7 and the Requests package. The idea is to quickly calculate which indices have exceeded the retention policy and need to be put to sleep.
The request works perfectly and the results are exactly what I want. However, when I run the code below, when I try to convert the json result to a dict, the type of theDict is correct but it reports a size of 1, when there should be at least a couple dozen entries. What am I doing wrong? I have a feeling it's something really dumb but I just can't snag it! :)
import json
import requests
esEndPoint = "https://localhost:9200"
retrieveString = "/_cat/indices?h=index,creation.date.string&format=json&s=creation.date"
# Gets the current indices and their creation dates
def retrieveIndicesAndDates():
try:
theResult = requests.get(esEndPoint+retrieveString)
print (theResult.content)
except Exception as e:
print("Unable to retrieve list of indices with creation dates.")
print("Error: "+e)
exit(3)
return theResult.content
def main():
theDict = dict(json.loads(retrieveIndicesAndDates()))
print(type(theDict)) # Reports correct type
print(len(theDict)) # Always outputs "1" ??
for index, creationdate in theDict.items():
print("Index: ",index,", Creation date: ",theDict[index])
return
The json the call returns:
[{"index":".kibana","creation.date.string":"2017-09-14T15:01:38.611Z"},{"index":"logstash-2018.07.23","creation.date.string":"2018-07-23T00:00:01.024Z"},{"index":"cwl-2018.07.23","creation.date.string":"2018-07-23T00:00:03.877Z"},{"index":"k8s-testing-internet-2018.07.23","creation.date.string":"2018-07-23T14:19:10.024Z"},{"index":"logstash-2018.07.24","creation.date.string":"2018-07-24T00:00:01.023Z"},{"index":"k8s-testing-internet-2018.07.24","creation.date.string":"2018-07-24T00:00:01.275Z"},{"index":"cwl-2018.07.24","creation.date.string":"2018-07-24T00:00:02.157Z"},{"index":"k8s-testing-internet-2018.07.25","creation.date.string":"2018-07-25T00:00:01.022Z"},{"index":"logstash-2018.07.25","creation.date.string":"2018-07-25T00:00:01.186Z"},{"index":"cwl-2018.07.25","creation.date.string":"2018-07-25T00:00:04.012Z"},{"index":"logstash-2018.07.26","creation.date.string":"2018-07-26T00:00:01.026Z"},{"index":"k8s-testing-internet-2018.07.26","creation.date.string":"2018-07-26T00:00:01.185Z"},{"index":"cwl-2018.07.26","creation.date.string":"2018-07-26T00:00:02.587Z"},{"index":"k8s-testing-internet-2018.07.27","creation.date.string":"2018-07-27T00:00:01.027Z"},{"index":"logstash-2018.07.27","creation.date.string":"2018-07-27T00:00:01.144Z"},{"index":"cwl-2018.07.27","creation.date.string":"2018-07-27T00:00:04.485Z"},{"index":"ctl-2018.07.27","creation.date.string":"2018-07-27T09:02:09.854Z"},{"index":"cfl-2018.07.27","creation.date.string":"2018-07-27T11:12:44.681Z"},{"index":"elb-2018.07.27","creation.date.string":"2018-07-27T11:13:51.340Z"},{"index":"cfl-2018.07.24","creation.date.string":"2018-07-27T11:45:23.697Z"},{"index":"cfl-2018.07.23","creation.date.string":"2018-07-27T11:45:24.646Z"},{"index":"cfl-2018.07.25","creation.date.string":"2018-07-27T11:45:25.700Z"},{"index":"cfl-2018.07.26","creation.date.string":"2018-07-27T11:45:26.341Z"},{"index":"elb-2018.07.24","creation.date.string":"2018-07-27T11:45:27.440Z"},{"index":"elb-2018.07.25","creation.date.string":"2018-07-27T11:45:29.572Z"},{"index":"elb-2018.07.26","creation.date.string":"2018-07-27T11:45:36.170Z"},{"index":"logstash-2018.07.28","creation.date.string":"2018-07-28T00:00:01.023Z"},{"index":"k8s-testing-internet-2018.07.28","creation.date.string":"2018-07-28T00:00:01.316Z"},{"index":"cwl-2018.07.28","creation.date.string":"2018-07-28T00:00:03.945Z"},{"index":"elb-2018.07.28","creation.date.string":"2018-07-28T00:00:53.992Z"},{"index":"ctl-2018.07.28","creation.date.string":"2018-07-28T00:07:19.543Z"},{"index":"k8s-testing-internet-2018.07.29","creation.date.string":"2018-07-29T00:00:01.026Z"},{"index":"logstash-2018.07.29","creation.date.string":"2018-07-29T00:00:01.378Z"},{"index":"cwl-2018.07.29","creation.date.string":"2018-07-29T00:00:04.100Z"},{"index":"elb-2018.07.29","creation.date.string":"2018-07-29T00:00:59.241Z"},{"index":"ctl-2018.07.29","creation.date.string":"2018-07-29T00:06:44.199Z"},{"index":"logstash-2018.07.30","creation.date.string":"2018-07-30T00:00:01.024Z"},{"index":"k8s-testing-internet-2018.07.30","creation.date.string":"2018-07-30T00:00:01.179Z"},{"index":"cwl-2018.07.30","creation.date.string":"2018-07-30T00:00:04.417Z"},{"index":"elb-2018.07.30","creation.date.string":"2018-07-30T00:01:01.442Z"},{"index":"ctl-2018.07.30","creation.date.string":"2018-07-30T00:08:28.936Z"},{"index":"cfl-2018.07.30","creation.date.string":"2018-07-30T06:52:16.739Z"}]
Your error is trying to convert a list of dicts to a dict:
theDict = dict(json.loads(retrieveIndicesAndDates()))
# ^^^^^ ^
That would only work for a dict of lists. It would be redundant, though.
Just use the reply directly. Each entry is a dict with the appropriate keys:
data = json.loads(retrieveIndicesAndDates())
for entry in data:
print("Index: ", entry["index"], ", Creation date: ", entry["creation.date.string"])
So what happens when you do convert that list to a dict? Why is there just one entry?
The dict understands three initialisation methods: keywords, mappings and iterables. A list fits the last one.
Initialisation from an iterable goes through it and expects key-value iterables as elements. If one were to do it manually, it would look like this:
def sequence2dict(sequence):
map = {}
for element in sequence:
key, value = element
map[key] = value
return map
Notice how each element is unpacked via iteration? In the reply each element is a dict with two entries. Iteration on that yields the two keys but ignores the values.
key, value = {"index":".kibana","creation.date.string":"2017-09-14T15:01:38.611Z"}
print(key, '=>', value) # prints "index => creation.date.string"
To the dict constructor, every element in the reply has the same key-value pair: "index" and "creation.date.string". Since keys in a dict are unique, all elements collapse to the same entry: {"index": "creation.date.string"}.
This has taken me over a day of trial and error. I am trying to keep a dictionary of queries and their respective matches in a search. My problem is that there can be one or more matches. My current solution is:
match5[query_site] will already have the first match but if it finds another match it will append it using the code below.
temp5=[] #temporary variable to create array
if isinstance(match5[query_site],list): #check if already a list
temp5.extend(match5[query_site])
temp5.append(match_site)
else:
temp5.append(match5[query_site])
match5[query_site]=temp5 #add new location
That if statement is literally to prevent extend converting my str element into an array of letters. If I try to initialize the first match as a single element array I get None if I try to directly append. I feel like there should be a more pythonic method to achieve this without a temporary variable and conditional statement.
Update: Here is an example of my output when it works
5'flank: ['8_73793824', '6_133347883', '4_167491131', '18_535703', '14_48370386']
3'flank: X_11731384
There's 5 matches for my "5'flank" and only 1 match for my "3'flank".
So what about this:
if query_site not in match5: # here for the first time
match5[query_site] = [match_site]
elif isinstance(match5[query_site], str): # was already here, a single occurrence
match5[query_site] = [match5[query_site], match_site] # make it a list of strings
else: # already a list, so just append
match5[query_site].append(match_site)
I like using setdefault() for cases like this.
temp5 = match5.setdefault(query_site, [])
temp5.append(match_site)
It's sort of like get() in that it returns an existing value if the key exists but you can provide a default value. The difference is that if the key doesn't exist already setdefault inserts the default value into the dict.
This is all you need to do
if query_site not in match5:
match5[query_site] = []
temp5 = match5[query_site]
temp5.append(match_site)
You could also do
temp5 = match5.setdefault(query_site, [])
temp5.append(match_site)
Assuming match5 is a dictionary, what about this:
if query_site not in match5: # first match ever
match5[query_site] = [match_site]
else: # entry already there, just append
match5[query_site].append(temp5)
Make the entries of the dictionary to be always a list, and just append to it.
I've got a problem. I would like to call to a dictionary and add an item, but I want to call to it by a variable.
if zacetek in predmeti:
ocena = str(raw_input("Vpisite oceno: "))
procent = str(raw_input("Vpisite procent: "))
zacetek[ocena] = procent
Zacetek is a string inside a list named predmeti. Dictionaries are named the same as the strings in that list. I basicly check if the input(zacetek) is in those list of strings and then call to a dictionary named like that. But I can't call because it says that zacetek is a string and not an object. How could I get a dictionary like this?
Thank you!
It looks like you are trying to use the string as a dictionary. Try setting a variable to a dictionary, and setting it's keys.
predmeti = [] # strings here
zacetek_dict = {}
zacetek = raw_input('Input: ')
if zacetek in predmeti:
ocena = str(raw_input("Vpisite oceno: "))
procent = str(raw_input("Vpisite procent: "))
zacetek_dict[ocena] = procent
Since zacetek is already a string, you can't treat it like a dictionary.
If I'm understanding you correctly, you are trying to create a dictionary. You'll want a new variable for that. If this is inside a loop, instantiate the new variable outside the loop like this:
new_dict = {}
# or
new_dict = dict()
Otherwise, if you only need it to have one item in it, you can just set it up as you have:
new_dict = {ocena: procent}
Couple of things:
raw_input already gives you a string. You don't have to attempt to convert to a string (str(raw_input([prompt])), raw_input([prompt]) is good enough.
Also, Zacetek is a string. You can only index keys and values if the object is a dictionary. Since, Predmeti is a data type (list or dict), you can append to that, or even convert Zacetek to a string and append to that object.
predmeti = {zacetek: {}} # Zacetek is a dictionary
predmeti[zacetek][ocena] = procent # Get's the value of zacetek (Which is a
# dictionary),then adds a key and a
# value to it.
I am writing code in python.
My input line is "all/DT remaining/VBG all/NNS of/IN "
I want to create a dictionary with one key and multiple values
For example - all:[DT,NNS]
groupPairsByKey={}
Code:
for line in fileIn:
lineLength=len(line)
words=line[0:lineLength-1].split(' ')
for word in words:
wordPair=word.split('/')
if wordPair[0] in groupPairsByKey:
groupPairsByKey[wordPair[0]].append(wordPair[1])
<getting error here>
else:
groupPairsByKey[wordPair[0]] = [wordPair[1]]
Your problem is that groupPairsByKey[wordPair[0]] is not a list, but a string!
Before appending value to groupPairsByKey['all'], you need to make the value a list.
Your solution is already correct, it works perfectly in my case. Try to make sure that groupPairsByKey is a completely empty dictionary.
By the way, this is what i tried:
>>> words = "all/DT remaining/VBG all/NNS of/IN".split
>>> for word in words:
wordPair = word.split('/')
if wordPair[0] in groupPairsByKey:
groupPairsByKey[wordPair[0]].append(wordPair[1])
else:
groupPairsByKey[wordPair[0]] = [wordPair[1]]
>>> groupPairsByKey
{'of': ['IN'], 'remaining': ['VBG'], 'all': ['DT', 'NNS']}
>>>
Also, if your code is formatted like the one you posted here, you'll get an indentationError.
Hope this helps!
Although it looks to me like you should be getting an IndentationError, if you are getting the message
str object has no attribute append
then it means
groupPairsByKey[wordPair[0]]
is a str, and strs do not have an append method.
The code you posted does not show how
groupPairsByKey[wordPair[0]]
could have a str value. Perhaps put
if wordPair[0] in groupPairsByKey:
if isinstance(groupPairsByKey[wordPair[0]], basestring):
print('{}: {}'.format(*wordPair))
raise Hell
into your code to help track down the culprit.
You could also simplify your code by using a collections.defaultdict:
import collections
groupPairsByKey = collections.defaultdict(list)
for line in fileIn:
lineLength=len(line)
words=line[0:lineLength-1].split(' ')
for word in words:
wordPair=word.split('/')
groupPairsByKey[wordPair[0]].append(wordPair[1])
When you access a defaultdict with a missing key, the factory function -- in this case list -- is called and the returned value is used as the associated value in the defaultdict. Thus, a new key-value pair is automatically inserted into the defaultdict whenever it encounters a missing key. Since the default value is always a list, you won't run into the error
str object has no attribute append anymore -- unless you have
code which reassigns an old key-value pair to have a new value which is a str.
You can do:
my_dict["all"] = my_string.split('/')
in Python,