In order to optimize a code in one single line, I am trying to write a determinate statement in my code without calling any function or method. While I was thinking about this I wondered if this is even possible in my case. I was searching some information about this but it seems to be very rarely, but in my current work I must be able to keep the code intact except that optimize section.
Hope you could give me a hand. Any help is welcome.
This is my current progress.
def count_chars(s):
'''(str) -> dict of {str: int}
Return a dictionary where the keys are the characters in s and the values
are how many times those characters appear in s.
>>> count_chars('abracadabra')
{'a': 5, 'r': 2, 'b': 2, 'c': 1, 'd': 1}
'''
d = {}
for c in s:
if not (c in d):
# This is the line it is assumed to be modified without calling function or method
else:
d[c] = d[c] + 1
return d
How about this, as mentioned in the comments, it does implicitly use functions, but I think it may be the sort of thing you are looking for?
s='abcab'
chars={}
for char in s:
if char not in chars:
chars[char]=0
chars[char]+=1
Result
{'a': 2, 'b': 2, 'c': 1}
Related
Here's a function that is supposed to swap dictionary keys and values. {'a': 3} is supposed to become {3: 'a'}.
def change_keys_values(d):
for key in d:
value = d[key]
del d[key]
d[value] = key
return d
I've realized that this function shouldn't work because I'm changing dictionary keys during iteration. This is the error I get: "dictionary keys changed during iteration". However, I don't get this error on a three key-value pair dictionary. So, while {'a': 3, 't': 8, 'r': 2, 'z': 44, 'u': 1, 'b': 4} results in the above mentioned error, {'a': 3, 't': 8, 'r': 2} gets solved without any issues. I'm using python 3. What is causing this?
You must never modify a dictionary inside a loop. The reason is the way the dictionaries are often implemented.
Hash Tables
Basically, when you create a dictionary, each item is indexed using the hash value of the key.
Dictionaries are implemented sparsely
Another implementation detail involves the fact that dictionaries are implemented in a sparse manner. Namely, when you create a dictionary, there are empty places in memory (called buckets). When you add or remove elements from a dictionary, it may hit a threshold where the dictionary key hashes are re-evaluated and as a consequence, the indexes are changed.
Roughly speaking, these two points are the reason behind the problem you are observing.
Moral Point: Never modify a dictionary inside a loop of any kind.
Here's a simple code to do what you want:
def change_keys_values(d):
new_dict = {value: key for key, value in d.items()}
return new_dict
You need to verify that the values are unique, after that, no problem :)
But be sure not to change a dictionary while parsing it. Otherwise, you could encounter an already changed index that get's interpreted twice or even more. I suggest making a new variable (a copy):
def invert(dict_: dict) -> dict:
if list(set(dict_.values())) == list(dict_.values()): # evaluates if "inverting key:value" is possible (if keys are unique)
return {b: a for a, b in dict_.items()}
else:
raise ValueError("Dictionary values contain duplicates. Inversion not possible!")
print(invert({"a": 1, "b": 2, "c": 3, "d": 4})) # works
print(invert({"a": 1, "b": 2, "c": 3, "d": 3})) # fails
To fix your issue, just iterate over copy, not the original dict:
import copy
def change_keys_values(d):
for key in copy.deepcopy(d):
value = d[key]
del d[key]
d[value] = key
return d
Then the good alternative using zip would be:
def change_keys_values(d):
a, b = zip(*d.items())
d = dict(list(zip(b,a)))
return d
So I was wondering if anybody wants to help me with this. I don't even understand where to begin? Any help would be appreciated.
Write a function called count_bases that counts the number of times each letter occurs in a given string. The results should be returned as a dictionary, with letters in upper case as keys and the number of occurrences as (integer) values
For example when the function is called with the string 'ATGATAGG', it should return {'A': 3, 'T': 2, 'G': 3, 'C': 0}. Please ensure your function uses return, not print(). The order of the keys in the dictionary does not need to follow this order (2 marks).
Make sure that your function works when passed any lower and/or uppercase DNA characters in the sequence string. (2 marks)
DNA sequences sometimes contain letters other than A, C, G to T to indicate degenerate nucleotides. For example, R can represent A or G (the purine bases). If the program encounters any letter other than A, C, G or T, it should also count the frequency of that letter and return within the dictionary object. (2 marks).
Use following code:
def count_bases(input_str):
result = {}
for s in input_str:
try:
result[s]+=1
except:
result[s] = 1
return result
print(count_bases('ATGATAGG'))
Output:
{'A': 3, 'T': 2, 'G': 3}
Try it:
def f(input):
d = {}
for s in input:
d[s] = d.get(s,0)+1
return d
from collections import Counter
def count_bases(sequence):
# since you want to count both lower and upper case letters,
# it'd be better if you convert the input sequence to either upper or lower.
sequence = sequence.upper()
# Counter (from collections) does the counting for you. It takes list as input.
# So, list(sequence) will separate letters from your sequence into a list of letters ('abc' => ['a', 'b', 'c'])
# It returns you a Counter object. Since you want a dictionary, cast it to dict.
return dict(Counter(list(sequence)))
count_bases('ATGATAGGaatdga')
{'A': 6, 'T': 3, 'G': 4, 'D': 1}
This question already has answers here:
Count the number of occurrences of a character in a string
(26 answers)
Closed 8 years ago.
I want a string such as 'ddxxx' to be returned as ('d': 2, 'x': 3). So far I've attempted
result = {}
for i in s:
if i in s:
result[i] += 1
else:
result[i] = 1
return result
where s is the string, however I keep getting a KeyError. E.g. if I put s as 'hello' the error returned is:
result[i] += 1
KeyError: 'h'
The problem is with your second condition. if i in s is checking for the character in the string itself and not in the dictionary. It should instead be if i in result.keys() or as Neil mentioned It can just be if i in result
Example:
def fun(s):
result = {}
for i in s:
if i in result:
result[i] += 1
else:
result[i] = 1
return result
print (fun('hello'))
This would print
{'h': 1, 'e': 1, 'l': 2, 'o': 1}
You can solve this easily by using collections.Counter. Counter is a subtype of the standard dict that is made to count things. It will automatically make sure that indexes are created when you try to increment something that hasn’t been in the dictionary before, so you don’t need to check it yourself.
You can also pass any iterable to the constructor to make it automatically count the occurrences of the items in that iterable. Since a string is an iterable of characters, you can just pass your string to it, to count all characters:
>>> import collections
>>> s = 'ddxxx'
>>> result = collections.Counter(s)
>>> result
Counter({'x': 3, 'd': 2})
>>> result['x']
3
>>> result['d']
2
Of course, doing it the manual way is fine too, and your code almost works fine for that. Since you get a KeyError, you are trying to access a key in the dictionary that does not exist. This happens when you happen to come accross a new character that you haven’t counted before. You already tried to handle that with your if i in s check but you are checking the containment in the wrong thing. s is your string, and since you are iterating the character i of the string, i in s will always be true. What you want to check instead is whether i already exists as a key in the dictionary result. Because if it doesn’t you add it as a new key with a count of 1:
if i in result:
result[i] += 1
else:
result[i] = 1
Using collections.Counter is the sensible solution. But if you do want to reinvent the wheel, you can use the dict.get() method, which allows you to supply a default value for missing keys:
s = 'hello'
result = {}
for c in s:
result[c] = result.get(c, 0) + 1
print result
output
{'h': 1, 'e': 1, 'l': 2, 'o': 1}
Here is a simple way of doing this if you don't want to use collections module:
>>> st = 'ddxxx'
>>> {i:st.count(i) for i in set(st)}
{'x': 3, 'd': 2}
Does anone have a fair algorithm for unifying (almost) arbitrary dicts? That is, given the dicts
a = {1: 1, 2: 2, 3: [1,2,3]}
b = {4: 4, 3: [5], 5: {'a': 0, 'b': {}}
c = {3: [{'A': '0'}], 5: {'b': {'B': 1}}}
unify (a, b, c)
yields
{1: 1,
2: 2,
3: [1, 2, 3, 5, {'A': '0'}],
4: 4,
5: {'a': 0, 'b': {'B': 1}}
}
I keep wanting a generic solution. I wind up searching for a generic solution a couple of times a year and not finding one (no Google, unify from unification and unify from union is not the same word!), and I keep putting off writing one myself. I very well know that programming Prolog leads to an odd pespective on life, but hey, how can one have a recursive dict/key/value-store and not have unification?
I have in the past needed ordering, hence lists, and back then I wound up not going for a generic version but hardcoding. This time around I don't actually need unification of sets/lists at all, and the fall back is to once again hardcode as I know what the keys can be ahead of time. But: If there already were a generic solution out there, I wouldn't have to reinvent the wheel again and again. It's just wrong to have to do that.
The really pythonic solution would probably start with a __unify__-method on all things that can be unified, it's that basic.
If you are stuck as to how to iterate through a dictionary, using a for loop iterates through the keys:
>>> for i in {1: "abc"}: print i
1
As the comments say, please specify what problems you're facing rather than asking SO to write the code for you.
Like zodiac mentioned, it's difficult to answer with out a direct problem; however, I'll try for a solution.
#Merge Lists of Dictionaries Functions
def merge_lists(l1, l2, key):
merged = {}
for item in l1+l2:
if item[key] not in merged:
merged[item[key]] = item
return [val for (_, val) in merged.items()]
tell me how this works
I have been trying to build a function to get letter frequencies from a string and store them in a dictionary.
I have done something like that :
s="today the weather was really nice"
def get_letter_freq(s):
for letter in(s):
x=letter.split()
f=dict()
for each_letter in x:
if f.has_key(x):
f[x]+=1
else:
f[x]=1
print f
Could you help me put things into order and find my mistakes?
Why I get an error that my 'f' is not defined?
In your code, your first for loop, where you have your letter.split() statement seems useless. Why you want to split a single character, you get in your loop?
Secondly, you have defined your f = dict() inside your function and
using it ouside. It will not be visible outside.
Third, your should not use f.has_key. Just do, key in my_dict to
check for availability of key in dict.
And at last, you can pass your dictionary as parameter to your
function. Then modify it there, and finally return it. (Although you can also do it without passing the dict in your function. just create a new one there, and return it).
So, in your code, almost everything is fine. You just need to remove your first for loop in function, and move f = dict() outside the function, before invoking it. And pass it as a paramter.
Way 1:
So, you can rather try the following modified code of yours: -
def get_letter_freq(my_dict, s):
for letter in s:
if letter in my_dict:
my_dict[letter] += 1
else:
my_dict[letter] = 1
return my_dict
my_dict = dict()
my_str = "today the weather was really nice"
print get_letter_freq(my_dict, my_str)
Way 2: -
Alternatively, you can also use a pre-defined library function Counter from collections, which does exactly what you want.
WAY 3: -
As suggested by #thebjorn in comment, you can also use defaultdict, which will make your task easier, in that, you won't have to check for the availability of key in dictionary before adding it. The count will automatically default to 0: -
from collections import defaultdict
def get_letter_freq(s):
my_dict = defaultdict(int)
for letter in s:
my_dict[letter] += 1
return my_dict
my_str = "today the weather was really nice"
print list(get_letter_freq(my_str).items())
Besides that indentation error your program has many other problems, like:
s = "today the weather was really nice"
def get_letter_freq(s):
f = dict()
for each_letter in s: #you can directly iterate over a string, so no need of split()
if each_letter in f: #has_key() has been deprecated
f[each_letter]+=1
else:
f[each_letter]=1
return f #better return the output from function
print get_letter_freq(s)
By the way collections.Counter() is good for this purpose:
In [61]: from collections import Counter
In [62]: strs = "today the weather was really nice"
In [63]: Counter(strs)
Out[63]: Counter({' ': 5, 'e': 5, 'a': 4, 't': 3, 'h': 2, 'l': 2, 'r': 2, 'w': 2, 'y': 2, 'c': 1, 'd': 1, 'i': 1, 'o': 1, 'n': 1, 's': 1})
f is defined inside get_letter_freq, you can't access it from outside.
Your function should return the constructed dictionary.
You should actually call the function.
What do you expect from splitting a single letter? Just leave that part out, and you don't need the inner loop.
print f needs to be indented, if it has to be part of get_letter_freq.
& f does not exist outside get_letter_freq. Hence the error.
import string
s="today the weather was really nice"
print dict([ ( letter, s.count(letter)) for letter in string.lowercase[:25]])
If case sensitivity is important use s.lower().count(letter) instead.