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Write a function to add key-value pairs

I'd like to write a function that will take one argument (a text file) to use its contents as keys and assign values to the keys. But I'd like the keys to go from 1 to n:
{'A': 1, 'B': 2, 'C': 3, 'D': 4... }.
I tried to write something like this:
Base code which kind of works:
filename = 'words.txt'
with open(filename, 'r') as f:
text = f.read()
ready_text = text.split()
def create_dict(lst):
""" go through the arg, stores items in it as keys in a dict"""
dictionary = dict()
for item in lst:
if item not in dictionary:
dictionary[item] = 1
else:
dictionary[item] += 1
return dictionary
print(create_dict(ready_text))
The output: {'A': 1, 'B': 1, 'C': 1, 'D': 1... }.
Attempt to make the thing work:
def create_dict(lst):
""" go through the arg, stores items in it as keys in a dict"""
dictionary = dict()
values = list(range(100)) # values
for item in lst:
if item not in dictionary:
for value in values:
dictionary[item] = values[value]
else:
dictionary[item] = values[value]
return dictionary
The output: {'A': 99, 'B': 99, 'C': 99, 'D': 99... }.
My attempt doesn't work. It gives all the keys 99 as their value.
Bonus question: How can I optimaze my code and make it look more elegant/cleaner?
Thank you in advance.

You can use dict comprehension with enumerate (note the start parameter):
words.txt:
colorless green ideas sleep furiously
Code:
with open('words.txt', 'r') as f:
words = f.read().split()
dct = {word: i for i, word in enumerate(words, start=1)}
print(dct)
# {'colorless': 1, 'green': 2, 'ideas': 3, 'sleep': 4, 'furiously': 5}
Note that "to be or not to be" will result in {'to': 5, 'be': 6, 'or': 3, 'not': 4}, perhaps what you don't want. Having only one entry out of two (same) words is not the result of the algorithm here. Rather, it is inevitable as long as you use a dict.

Your program sends a list of strings to create_dict. For each string in the list, if that string is not in the dictionary, then the dictionary value for that key is set to 1. If that string has been encountered before, then the value of that key is increased by 1. So, since every key is being set to 1, then that must mean there are no repeat keys anywhere, meaning you're sending a list of unique strings.
So, in order to have the numerical values increase with each new key, you just have to increment some number during your loop:
num = 0
for item in lst:
num += 1
dictionary[item] = num
There's an easier way to loop through both numbers and list items at the same time, via enumerate():
for num, item in enumerate(lst, start=1): # start at 1 and not 0
dictionary[item] = num

You can use this code. If an item has been in the lst more than once, the idx is considered one time in dictionary!
def create_dict(lst):
""" go through the arg, stores items in it as keys in a dict"""
dictionary = dict()
idx = 1
for item in lst:
if item not in dictionary:
dictionary[item]=idx
idx += 1
return dictionary

int values in dict appending as list

I have a for loop which is going through multiple dictionaries and adding the values under common keys. The input dictionary has keys that are strings and values that are int's. For some reason its adding the values as lists of one value (e.g. {"01":[12],[44]}). I want it to add the int on its own but cant get that working for some reason. I'm using the code below, is there something i am missing ?
dw = defaultdict()
dw = {}
for key, value in listb.items():
dw[key].append(value)

If you want to forgo all good practice and not use defaultdict(list), you can use setdefault and call it every single time you choose to add a value. This is inefficient and not idiomatic, but it will work.
In [1]: from collections import defaultdict
In [2]: a = defaultdict(list)
In [3]: b = {}
In [4]: a[1].append(1)
In [5]: b.setdefault(1, []).append(1)
In [6]: a
Out[6]: defaultdict(list, {1: [1]})
In [7]: b
Out[7]: {1: [1]}
In [8]:

As long as the values in the dicts are ints (not lists):
dw = {}
for key, value in listb.items():
try: # Key exists in dictionary and value is a list
dw[key].append(value)
except KeyError: # Key does not yet exist in dictionary
dw[key] = value
except AttributeError: # Key exist in dictionary and value is not a list
dw[key] = [dw[key], value]

If you mean to add key/value pairs to the dictionary (and not append to an array), it's:
for key, value in listb.items():
dw[key] = value
EDIT: or is it something like this you're after?
listb = {'1': 3, '2': 5}
dw = {'1': 5, '2': 9}
for key, value in listb.items():
if key not in dw.keys():
dw[key] = []
else:
dw[key] = [dw[key]]
dw[key].append(value)
which gives dw = {'2': [9, 5], '1': [5, 3]}

If you have a list like listb = [{'a': 1, 'b': 2}, {'a': 3, 'b': 4, 'c': 5}, {'b': 1}], you can try this:
dw = {}
for d in listb:
for k, v in d.items():
if k in dw:
if isinstance(dw[k], list):
dw[k].append(v)
elif isinstance(dw[k], int):
dw[k] = [dw[k], v]
else:
dw[k] = v
print(dw)
{'a': [1, 3], 'b': [2, 4, 1], 'c': 5}
>>>

keyerror 1 in my code

I am writing a function that take dictionary input and return list of keys which have unique values in that dictionary. Consider,
ip = {1: 1, 2: 1, 3: 3}
so output should be [3] as key 3 has unique value which is not present in dict.
Now there is problem in given fuction:
def uniqueValues(aDict):
dicta = aDict
dum = 0
for key in aDict.keys():
for key1 in aDict.keys():
if key == key1:
dum = 0
else:
if aDict[key] == aDict[key1]:
if key in dicta:
dicta.pop(key)
if key1 in dicta:
dicta.pop(key1)
listop = dicta.keys()
print listop
return listop
I am getting error like:
File "main.py", line 14, in uniqueValues
if aDict[key] == aDict[key1]: KeyError: 1
Where i am doing wrong?

Your main problem is this line:
dicta = aDict
You think you're making a copy of the dictionary, but actually you still have just one dictionary, so operations on dicta also change aDict (and so, you remove values from adict, they also get removed from aDict, and so you get your KeyError).
One solution would be
dicta = aDict.copy()
(You should also give your variables clearer names to make it more obvious to yourself what you're doing)
(edit) Also, an easier way of doing what you're doing:
def iter_unique_keys(d):
values = list(d.values())
for key, value in d.iteritems():
if values.count(value) == 1:
yield key
print list(iter_unique_keys({1: 1, 2: 1, 3: 3}))

Use Counter from collections library:
from collections import Counter
ip = {
1: 1,
2: 1,
3: 3,
4: 5,
5: 1,
6: 1,
7: 9
}
# Generate a dict with the amount of occurrences of each value in 'ip' dict
count = Counter([x for x in ip.values()])
# For each item (key,value) in ip dict, we check if the amount of occurrences of its value.
# We add it to the 'results' list only if the amount of occurrences equals to 1.
results = [x for x,y in ip.items() if count[y] == 1]
# Finally, print the results list
print results
Output:
[3, 4, 7]

Remove the smallest element(s) from a dictionary

I have a function such that there is a dictionary as parameters, with the value associated to be an integer. I'm trying to remove the minimum element(s) and return a set of the remaining keys.
I am programming in python. I cant seem to remove key value pairs with the same key or values. My code does not work for the 2nd and 3rd example
This is how it would work:
remaining({A: 1, B: 2, C: 2})
{B, C}
remaining({B: 2, C : 2})
{}
remaining({A: 1, B: 1, C: 1, D: 4})
{D}
This is what I have:
def remaining(d : {str:int}) -> {str}:
Remaining = set(d)
Remaining.remove(min(d, key=d.get))
return Remaining

One approach is to take the minimum value, then build a list of keys that are equal to it and utilise dict.viewkeys() which has set-like behaviour and remove the keys matching the minimum value from it.
d = {'A': 1, 'B': 1, 'C': 1, 'D': 4}
# Use .values() and .keys() and .items() for Python 3.x
min_val = min(d.itervalues())
remaining = d.viewkeys() - (k for k, v in d.iteritems() if v == min_val)
# set(['D'])
On a side note, I find it odd that {B: 2, C : 2} should be {} as there's not actually anything greater for those to be the minimum as it were.

That's because you're trying to map values to keys and map allows different keys to have the same values but not the other way! you should implement a map "reversal" as described here, remove the minimum key, and then reverse the map back to its original form.
from collections import defaultdict
# your example
l = {'A': 1, 'B': 1, 'C': 1, 'D': 4}
# reverse the dict
d1 = {}
for k, v in l.iteritems():
d1[v] = d1.get(v, []) + [k]
# remove the min element
del d1[min(d1, key=d1.get)]
#recover the rest to the original dict minus the min
res = {}
for k, v in d1.iteritems():
for e in v:
res[e] = k
print res
Comment:
#Jon Clements's solution is more elegant and should be accepted as the answer

Take the minimum value and construct a set with all the keys which are not associated to that value:
def remaining(d):
m = min(d.values())
return {k for k,v in d.items() if v != m}
If you don't like set comprehensions that's the same as:
def remaining(d):
m = min(d.values())
s = set()
for k,v in d.items():
if v != m:
s.add(k)
return s

This removes all the items with the minimum value.
import copy
def remaining(dic):
minimum = min([i for i in dic.values()])
for k, v in copy.copy(dic.items()):
if v == minimum: dic.pop(k)
return set(dic.keys())

An easier way would be to use pd.Series.idxmin() or pd.Series.min(). These functions allow you to find the index of the minimum value or the minimum value in a series, plus pandas allows you to create a named index.
import pandas as pd
import numpy as np
A = pd.Series(np.full(shape=5,fill_value=0))#create series of 0
A = A.reindex(['a','b','c','d','e'])#set index, similar to dictionary names
A['a'] = 2
print(A.max())
#output 2.0
print(A.idxmax())#you can also pop by index without changing other indices
#output a

Appending values to dictionary in Python

I have a dictionary to which I want to append to each drug, a list of numbers. Like this:
append(0), append(1234), append(123), etc.
def make_drug_dictionary(data):
drug_dictionary={'MORPHINE':[],
'OXYCODONE':[],
'OXYMORPHONE':[],
'METHADONE':[],
'BUPRENORPHINE':[],
'HYDROMORPHONE':[],
'CODEINE':[],
'HYDROCODONE':[]}
prev = None
for row in data:
if prev is None or prev==row[11]:
drug_dictionary.append[row[11][]
return drug_dictionary
I later want to be able to access the entirr set of entries in, for example, 'MORPHINE'.
How do I append a number into the drug_dictionary?
How do I later traverse through each entry?

Just use append:
list1 = [1, 2, 3, 4, 5]
list2 = [123, 234, 456]
d = {'a': [], 'b': []}
d['a'].append(list1)
d['a'].append(list2)
print d['a']

You should use append to add to the list. But also here are few code tips:
I would use dict.setdefault or defaultdict to avoid having to specify the empty list in the dictionary definition.
If you use prev to to filter out duplicated values you can simplfy the code using groupby from itertools
Your code with the amendments looks as follows:
import itertools
def make_drug_dictionary(data):
drug_dictionary = {}
for key, row in itertools.groupby(data, lambda x: x[11]):
drug_dictionary.setdefault(key,[]).append(row[?])
return drug_dictionary
If you don't know how groupby works just check this example:
>>> list(key for key, val in itertools.groupby('aaabbccddeefaa'))
['a', 'b', 'c', 'd', 'e', 'f', 'a']

It sounds as if you are trying to setup a list of lists as each value in the dictionary. Your initial value for each drug in the dict is []. So assuming that you have list1 that you want to append to the list for 'MORPHINE' you should do:
drug_dictionary['MORPHINE'].append(list1)
You can then access the various lists in the way that you want as drug_dictionary['MORPHINE'][0] etc.
To traverse the lists stored against key you would do:
for listx in drug_dictionary['MORPHINE'] :
do stuff on listx

To append entries to the table:
for row in data:
name = ??? # figure out the name of the drug
number = ??? # figure out the number you want to append
drug_dictionary[name].append(number)
To loop through the data:
for name, numbers in drug_dictionary.items():
print name, numbers

If you want to append to the lists of each key inside a dictionary, you can append new values to them using + operator (tested in Python 3.7):
mydict = {'a':[], 'b':[]}
print(mydict)
mydict['a'] += [1,3]
mydict['b'] += [4,6]
print(mydict)
mydict['a'] += [2,8]
print(mydict)
and the output:
{'a': [], 'b': []}
{'a': [1, 3], 'b': [4, 6]}
{'a': [1, 3, 2, 8], 'b': [4, 6]}
mydict['a'].extend([1,3]) will do the job same as + without creating a new list (efficient way).

You can use the update() method as well
d = {"a": 2}
d.update{"b": 4}
print(d) # {"a": 2, "b": 4}

how do i append a number into the drug_dictionary?
Do you wish to add "a number" or a set of values?
I use dictionaries to build associative arrays and lookup tables quite a bit.
Since python is so good at handling strings,
I often use a string and add the values into a dict as a comma separated string
drug_dictionary = {}
drug_dictionary={'MORPHINE':'',
'OXYCODONE':'',
'OXYMORPHONE':'',
'METHADONE':'',
'BUPRENORPHINE':'',
'HYDROMORPHONE':'',
'CODEINE':'',
'HYDROCODONE':''}
drug_to_update = 'MORPHINE'
try:
oldvalue = drug_dictionary[drug_to_update]
except:
oldvalue = ''
# to increment a value
try:
newval = int(oldval)
newval += 1
except:
newval = 1
drug_dictionary[drug_to_update] = "%s" % newval
# to append a value
try:
newval = int(oldval)
newval += 1
except:
newval = 1
drug_dictionary[drug_to_update] = "%s,%s" % (oldval,newval)
The Append method allows for storing a list of values but leaves you will a trailing comma
which you can remove with
drug_dictionary[drug_to_update][:-1]
the result of the appending the values as a string means that you can append lists of values as you need too and
print "'%s':'%s'" % ( drug_to_update, drug_dictionary[drug_to_update])
can return
'MORPHINE':'10,5,7,42,12,'

vowels = ("a","e","i","o","u") #create a list of vowels
my_str = ("this is my dog and a cat") # sample string to get the vowel count
count = {}.fromkeys(vowels,0) #create dict initializing the count to each vowel to 0
for char in my_str :
if char in count:
count[char] += 1
print(count)