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Seeking help on Homework
I am given a list and asked to find the most occurring value in a list and returns the amount of times it is occurred. This question is fairly big and i have managed to get through the other parts by myself but this one stumped me.I should add that this is for an assignment any guidance would be appreciated.
Question Statement : Maximum (word) Frequency
For example in a book with the following words ['big', 'big', 'bat', 'bob', 'book'] the maximum frequency is 2, i.e., big is the most frequently occurring word, therefore 2 is the maximum frequency.
def maximum_frequency(new_list):
word_counter = {}
for word in new_list:
if word in word_counter:
word_counter[word] += 1
else:
word_counter[word] = 1
I have gotten this far but I am not sure if its right/where to go from here
Try this:
from collections import Counter
c = Counter(['big', 'big', 'bat', 'bob', 'book'])
max(c.items(), key=lambda x:x[1])
the max will returns the most one by its count, you can do:
key,rate = max(c.items(), key=lambda x:x[1])
the key will be big and the rate will be 2.
also, you can access all of the items count by c.items(). and the output will be
{'big': 2, 'bat': 1, 'bob': 1, 'book': 1}
Edit:
as schwobaseggl said the best practice to find from a counter is to use most_common.
c.most_common(1)[0]
You just need to count the occurrence of all the unique elements and compare the frequency with the previously computed frequency.
sample is a list of words.
def maxfreq(sample):
m=0
frequency=0
word=''
set_sample=list(set(sample))
for i in range(len(set_sample)):
c=sample.count(set_sample[i])
if c>m:
m=c
frequency=m
word=set_sample[i]
return (frequency,word)
Since it sounds like this is some kind of challenge and/or homework you're supposed to be working on, instead of directly providing a code sample let me give you some concepts.
First off, the best way to know if you've seen a word or not is to use a map, in Python -- the term is "dict" and the syntax is simple {}, you can store values like this: my_dict['value'] = true or whatever key/value you need.
So if you're going to read your words, one by one, and store them into this dict, the what should the value be? You know you want to know the maximum frequency, right? Well, so let's use that as our value. By default, if we add a word, we should make sure to set it's initial value to 1 (we've seen it once). And if we see a word a second time, we then increment our frequency.
Now that you have a dict full of words and their frequencies, perhaps you might be able to figure out how to find the one with the largest frequency?
So that being said, things you should look into are:
How to determine if a key exists in a dict
How to modify the value of a key in a dict
How to iterate a dict's key/value pairs
After that, your answer should be pretty easy to figure out.
try this :
>>> MyList = ["above", "big", "above", "cat", "cat", "above", "cat"]
>>> my_dict = {i:MyList.count(i) for i in MyList}
>>> my_dict
{'above': 3, 'big': 1, 'cat': 3}
It can also be accomplish using collections.Counter which is compatible with Python 2.7 or 3.x !
>>> from collections import Counter
>>> MyList = ['big', 'big', 'bat', 'bob', 'book']
>>> dict(Counter(MyList))
{'big': 2, 'bat': 1, 'bob': 1, 'book': 1}
If you are open to Pandas then it can be done as follows:
>>> import pandas as pd
>>> pd.Series(MyList).value_counts()
big 2
book 1
bob 1
bat 1
dtype: int64
#Answer to the OP's next Question in the comment section what if i wanted to get just the maximum value instead of the word .
>>> pd.Series(MyList).value_counts().max()
2
How about this:
def maximum_frequency(new_list):
word_counter = {}
for word in new_list:
if word in word_counter:
word_counter[word] += 1
else:
word_counter[word] = 1
max_freq = max(word_counter.items(), key=(lambda x: x[1]))
return max_freq
if __name__ == '__main__':
test_data = ['big', 'big', 'bat', 'bob', 'book']
print(maximum_frequency(test_data))
Output:
('big', 2)
Works fine with Python 2 and 3 and returns result as a tuple of most frequent word and occurrences count.
EDIT:
If you don't care at all which word has the highest count and you want only the frequency number you can simplify it a bit to:
def maximum_frequency(new_list):
word_counter = {}
for word in new_list:
if word in word_counter:
word_counter[word] += 1
else:
word_counter[word] = 1
return max(word_counter.values())
if __name__ == '__main__':
test_data = ['big', 'big', 'bat', 'bob', 'book']
print(maximum_frequency(test_data))
for the class data structures and algorithms at Tilburg University i got a question in an in class test:
build a dictionary from testfile.txt, with only unique values, where if a value appears again, it should be added to the total sum of that productclass.
the text file looked like this, it was not a .csv file:
apples,1
pears,15
oranges,777
apples,-4
oranges,222
pears,1
bananas,3
so apples will be -3 and the output would be {"apples": -3, "oranges": 999...}
in the exams i am not allowed to import any external packages besides the normal: pcinput, math, etc. i am also not allowed to use the internet.
I have no idea how to accomplish this, and this seems to be a big problem in my development of python skills, because this is a question that is not given in a 'dictionaries in python' video on youtube (would be to hard maybe), but also not given in a expert course because there this question would be to simple.
hope you guys can help!
enter code here
from collections import Counter
from sys import exit
from os.path import exists, isfile
##i did not finish it, but wat i wanted to achieve was build a list of the
strings and their belonging integers. then use the counter method to add
them together
## by splitting the string by marking the comma as the split point.
filename = input("filename voor input: ")
if not isfile(filename):
print(filename, "bestaat niet")
exit()
keys = []
values = []
with open(filename) as f:
xs = f.read().split()
for i in xs:
keys.append([i])
print(keys)
my_dict = {}
for i in range(len(xs)):
my_dict[xs[i]] = xs.count(xs[i])
print(my_dict)
word_and_integers_dict = dict(zip(keys, values))
print(word_and_integers_dict)
values2 = my_dict.split(",")
for j in values2:
print( value2 )
the output becomes is this:
[['schijndel,-3'], ['amsterdam,0'], ['tokyo,5'], ['tilburg,777'], ['zaandam,5']]
{'zaandam,5': 1, 'tilburg,777': 1, 'amsterdam,0': 1, 'tokyo,5': 1, 'schijndel,-3': 1}
{}
so i got the dictionary from it, but i did not separate the values.
the error message is this:
28 values2 = my_dict.split(",") <-- here was the error
29 for j in values2:
30 print( value2 )
AttributeError: 'dict' object has no attribute 'split'
I don't understand what your code is actually doing, I think you don't know what your variables are containing, but this is an easy problem to solve in Python. Split into a list, split each item again, and count:
>>> input = "apples,1 pears,15 oranges,777 apples,-4 oranges,222 pears,1 bananas,3"
>>> parts = input.split()
>>> parts
['apples,1', 'pears,15', 'oranges,777', 'apples,-4', 'oranges,222', 'pears,1', 'bananas,3']
Then split again. Behold the list comprehension. This is an idiomatic way to transform a list to another in python. Note that the numbers are strings, not ints yet.
>>> strings = [s.split(',') for s in strings]
>>> strings
[['apples', '1'], ['pears', '15'], ['oranges', '777'], ['apples', '-4'], ['oranges', '222'], ['pears', '1'], ['bananas', '3']]
Now you want to iterate over pairs, and sum all the same fruits. This calls for a dict:
>>> result = {}
>>> for fruit, countstr in pairs:
... if fruit not in result:
... result[fruit] = 0
... result[fruit] += int(countstr)
>>> result
{'pears': 16, 'apples': -3, 'oranges': 999, 'bananas': 3}
This pattern of adding an element if it doesn't exist comes up frequently. You should checkout defaultdict in the collections module. If you use that, you don't even need the if.
Let's walk through what you need to do to. First, check if the file exists and read the contents to a variable. Second, parse each line - you need to split the line on the comma, convert the number from a string to an integer, and then pass the values to a dictionary. In this case I would recommend using defaultdict from collections, but we can also do it with a standard dictionary.
from os.path import exists, isfile
from collections import defaultdict
filename = input("filename voor input: ")
if not isfile(filename):
print(filename, "bestaat niet")
exit()
# this reads the file to a list, removing newline characters
with open(filename) as f:
line_list = [x.strip() for x in f]
# create a dictionary
my_dict = {}
# update the value in the dictionary if it already exists,
# otherwise add it to the dictionary
for line in line_list:
k, v_str = line.split(',')
if k in my_dict:
my_dict[k] += int(v_str)
else:
my_dict[k] = int(v_str)
# print the dictionary
table_str = '{:<30}{}'
print(table_str.format('Item','Count'))
print('='*35)
for k,v in sorted(my_dict.item()):
print(table_str.format(k,v))
I wrote this function:
def make_upper(words):
for word in words:
ind = words.index(word)
words[ind] = word.upper()
I also wrote a function that counts the frequency of occurrences of each letter:
def letter_cnt(word,freq):
for let in word:
if let == 'A': freq[0]+=1
elif let == 'B': freq[1]+=1
elif let == 'C': freq[2]+=1
elif let == 'D': freq[3]+=1
elif let == 'E': freq[4]+=1
Counting letter frequency would be much more efficient with a dictionary, yes. Note that you are manually lining up each letter with a number ("A" with 0, et cetera). Wouldn't it be easier if we could have a data type that directly associated a letter with the number of times it occurs, without adding an extra set of numbers in between?
Consider the code:
freq = {"A":0, "B":0, "C":0, "D":0, ... ..., "Z":0}
for letter in text:
freq[letter] += 1
This dictionary is used to count frequencies much more efficiently than your current code does. You just add one to an entry for a given letter each time you see it.
I will also mention that you can count frequencies effectively with certain libraries. If you are interested in analyzing frequencies, look into collections.Counter() and possibly the collections.Counter.most_common() method.
Whether or not you decide to just use collections.Counter(), I would attempt to learn why dictionaries are useful in this context.
One final note: I personally found typing out the values for the "freq" dictionary to be tedious. If you want you could construct an empty dictionary of alphabet letters on-the-fly with this code:
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
freq = {letter:0 for letter in alphabet}
If you want to convert strings in the list to upper case using lambda, you may use it with map() as:
>>> words = ["Hello", "World"]
>>> map(lambda word: word.upper(), words) # In Python 2
['HELLO', 'WORLD']
# In Python 3, use it as: list(map(...))
As per the map() document:
map(function, iterable, ...)
Apply function to every item of iterable and return a list of the results.
For finding the frequency of each character in word, you may use collections.Counter() (sub class dict type) as:
>>> from collections import Counter
>>> my_word = "hello world"
>>> c = Counter(my_word)
# where c holds dictionary as:
# {'l': 3,
# 'o': 2,
# ' ': 1,
# 'e': 1,
# 'd': 1,
# 'h': 1,
# 'r': 1,
# 'w': 1}
As per Counter Document:
A Counter is a dict subclass for counting hashable objects. It is an unordered collection where elements are stored as dictionary keys and their counts are stored as dictionary values.
for the letter counting, don't reinvent the wheel collections.Counter
A Counter is a dict subclass for counting hashable objects. It is an unordered collection where elements are stored as dictionary keys and their counts are stored as dictionary values. Counts are allowed to be any integer value including zero or negative counts. The Counter class is similar to bags or multisets in other languages.
def punc_remove(words):
for word in words:
if word.isalnum() == False:
charl = []
for char in word:
if char.isalnum()==True:
charl.append(char)
ind = words.index(word)
delimeter = ""
words[ind] = delimeter.join(charl)
def letter_cnt_dic(word,freq_d):
for let in word:
freq_d[let] += 1
import string
def letter_freq(fname):
fhand = open(fname)
freqs = dict()
alpha = list(string.uppercase[:26])
for let in alpha: freqs[let] = freqs.get(let,0)
for line in fhand:
line = line.rstrip()
words = line.split()
punc_remove(words)
#map(lambda word: word.upper(),words)
words = [word.upper() for word in words]
for word in words:
letter_cnt_dic(word,freqs)
fhand.close()
return freqs.values()
You can read the docs about the Counter and the List Comprehensions or run this as a small demo:
from collections import Counter
words = ["acdefg","abcdefg","abcdfg"]
#list comprehension no need for lambda or map
new_words = [word.upper() for word in words]
print(new_words)
# Lets create a dict and a counter
letters = {}
letters_counter = Counter()
for word in words:
# The counter count and add the deltas.
letters_counter += Counter(word)
# We can do it to
for letter in word:
letters[letter] = letters.get(letter,0) + 1
print(letters_counter)
print(letters)
Super new to to python here, I've been struggling with this code for a while now. Basically the function returns a dictionary with the integers as keys and the values are all the words where the length of the word corresponds with each key.
So far I'm able to create a dictionary where the values are the total number of each word but not the actual words themselves.
So passing the following text
"the faith that he had had had had an affect on his life"
to the function
def get_word_len_dict(text):
result_dict = {'1':0, '2':0, '3':0, '4':0, '5':0, '6' :0}
for word in text.split():
if str(len(word)) in result_dict:
result_dict[str(len(word))] += 1
return result_dict
returns
1 - 0
2 - 3
3 - 6
4 - 2
5 - 1
6 - 1
Where I need the output to be:
2 - ['an', 'he', 'on']
3 - ['had', 'his', 'the']
4 - ['life', 'that']
5 - ['faith']
6 - ['affect']
I think I need to have to return the values as a list. But I'm not sure how to approach it.
I think that what you want is a dic of lists.
result_dict = {'1':[], '2':[], '3':[], '4':[], '5':[], '6' :[]}
for word in text.split():
if str(len(word)) in result_dict:
result_dict[str(len(word))].append(word)
return result_dict
Fixing Sabian's answer so that duplicates aren't added to the list:
def get_word_len_dict(text):
result_dict = {1:[], 2:[], 3:[], 4:[], 5:[], 6 :[]}
for word in text.split():
n = len(word)
if n in result_dict and word not in result_dict[n]:
result_dict[n].append(word)
return result_dict
Check out list comprehensions
Integers are legal dictionaries keys so there is no need to make the numbers strings unless you want it that way for some other reason.
if statement in the for loop controls flow to add word only once. You could get this effect more automatically if you use set() type instead of list() as your value data structure. See more in the docs. I believe the following does the job:
def get_word_len_dict(text):
result_dict = {len(word) : [] for word in text.split()}
for word in text.split():
if word not in result_dict[len(word)]:
result_dict[len(word)].append(word)
return result_dict
try to make it better ;)
Instead of defining the default value as 0, assign it as set() and within if condition do, result_dict[str(len(word))].add(word).
Also, instead of preassigning result_dict, you should use collections.defaultdict.
Since you need non-repetitive words, I am using set as value instead of list.
Hence, your final code should be:
from collections import defaultdict
def get_word_len_dict(text):
result_dict = defaultdict(set)
for word in text.split():
result_dict[str(len(word))].add(word)
return result_dict
In case it is must that you want list as values (I think set should suffice your requirement), you need to further iterate it as:
for key, value in result_dict.items():
result_dict[key] = list(value)
What you need is a map to list-construct (if not many words, otherwise a 'Counter' would be fine):
Each list stands for a word class (number of characters). Map is checked whether word class ('3') found before. List is checked whether word ('had') found before.
def get_word_len_dict(text):
result_dict = {}
for word in text.split():
if not result_dict.get(str(len(word))): # add list to map?
result_dict[str(len(word))] = []
if not word in result_dict[str(len(word))]: # add word to list?
result_dict[str(len(word))].append(word)
return result_dict
-->
3 ['the', 'had', 'his']
2 ['he', 'an', 'on']
5 ['faith']
4 ['that', 'life']
6 ['affect']
the problem here is you are counting the word by length, instead you want to group them. You can achieve this by storing a list instead of a int:
def get_word_len_dict(text):
result_dict = {}
for word in text.split():
if len(word) in result_dict:
result_dict[len(word)].add(word)
else:
result_dict[len(word)] = {word} #using a set instead of list to avoid duplicates
return result_dict
Other improvements:
don't hardcode the key in the initialized dict but let it empty instead. Let the code add the new keys dynamically when necessary
you can use int as keys instead of strings, it will save you the conversion
use sets to avoid repetitions
Using groupby
Well, I'll try to propose something different: you can group by length using groupby from the python standard library
import itertools
def get_word_len_dict(text):
# split and group by length (you get a list if tuple(key, list of values)
groups = itertools.groupby(sorted(text.split(), key=lambda x: len(x)), lambda x: len(x))
# convert to a dictionary with sets
return {l: set(words) for l, words in groups}
You say you want the keys to be integers but then you convert them to strings before storing them as a key. There is no need to do this in Python; integers can be dictionary keys.
Regarding your question, simply initialize the values of the keys to empty lists instead of the number 0. Then, in the loop, append the word to the list stored under the appropriate key (the length of the word), like this:
string = "the faith that he had had had had an affect on his life"
def get_word_len_dict(text):
result_dict = {i : [] for i in range(1, 7)}
for word in text.split():
length = len(word)
if length in result_dict:
result_dict[length].append(word)
return result_dict
This results in the following:
>>> get_word_len_dict(string)
{1: [], 2: ['he', 'an', 'on'], 3: ['the', 'had', 'had', 'had', 'had', 'his'], 4: ['that', 'life'], 5: ['faith'], 6: ['affect']}
If you, as you mentioned, wish to remove the duplicate words when collecting your input string, it seems elegant to use a set and convert to a list as a final processing step, if this is needed. Also note the use of defaultdict so you don't have to manually initialize the dictionary keys and values as a default value set() (i.e. the empty set) gets inserted for each key that we try to access but not others:
from collections import defaultdict
string = "the faith that he had had had had an affect on his life"
def get_word_len_dict(text):
result_dict = defaultdict(set)
for word in text.split():
length = len(word)
result_dict[length].add(word)
return {k : list(v) for k, v in result_dict.items()}
This gives the following output:
>>> get_word_len_dict(string)
{2: ['he', 'on', 'an'], 3: ['his', 'had', 'the'], 4: ['life', 'that'], 5: ['faith'], 6: ['affect']}
Your code is counting the occurrence of each word length - but not storing the words themselves.
In addition to capturing each word into a list of words with the same size, you also appear to want:
If a word length is not represented, do not return an empty list for that length - just don't have a key for that length.
No duplicates in each word list
Each word list is sorted
A set container is ideal for accumulating the words - sets naturally eliminate any duplicates added to them.
Using defaultdict(sets) will setup an empty dictionary of sets -- a dictionary key will only be created if it is referenced in our loop that examines each word.
from collections import defaultdict
def get_word_len_dict(text):
#create empty dictionary of sets
d = defaultdict(set)
# the key is the length of each word
# The value is a growing set of words
# sets automatically eliminate duplicates
for word in text.split():
d[len(word)].add(word)
# the sets in the dictionary are unordered
# so sort them into a new dictionary, which is returned
# as a dictionary of lists
return {i:sorted(d[i]) for i in d.keys()}
In your example string of
a="the faith that he had had had had an affect on his life"
Calling the function like this:
z=get_word_len_dict(a)
Returns the following list:
print(z)
{2: ['an', 'he', 'on'], 3: ['had', 'his', 'the'], 4: ['life', 'that'], 5: ['faith'], 6: ['affect']}
The type of each value in the dictionary is "list".
print(type(z[2]))
<class 'list'>
In direct, my code so far is this :
from glob import glob
pattern = "D:\\report\\shakeall\\*.txt"
filelist = glob(pattern)
def countwords(fp):
with open(fp) as fh:
return len(fh.read().split())
print "There are" ,sum(map(countwords, filelist)), "words in the files. " "From directory",pattern
I want to add a code that counts unique words from pattern(42 txt files in this path) but I don't know how. Can anybody help me?
The best way to count objects in Python is to use collections.Counter class, which was created for that purposes. It acts like a Python dict but is a bit easier in use when counting. You can just pass a list of objects and it counts them for you automatically.
>>> from collections import Counter
>>> c = Counter(['hello', 'hello', 1])
>>> print c
Counter({'hello': 2, 1: 1})
Also Counter has some useful methods like most_common, visit documentation to learn more.
One method of Counter class that can also be very useful is update method. After you've instantiated Counter by passing a list of objects, you can do the same using update method and it will continue counting without dropping old counters for objects:
>>> from collections import Counter
>>> c = Counter(['hello', 'hello', 1])
>>> print c
Counter({'hello': 2, 1: 1})
>>> c.update(['hello'])
>>> print c
Counter({'hello': 3, 1: 1})
print len(set(w.lower() for w in open('filename.dat').read().split()))
Reads the entire file into memory, splits it into words using
whitespace, converts
each word to lower case, creates a (unique) set from the lowercase words, counts them
and prints the output
If you want to get count of each unique word, then use dicts:
words = ['Hello', 'world', 'world']
count = {}
for word in words :
if word in count :
count[word] += 1
else:
count[word] = 1
And you will get dict
{'Hello': 1, 'world': 2}