I have a folder with hundreds of txt files I need to analyse for similarity. Below is an example of a script I use to run similarity analysis. In the end I get an array or a matrix I can plot etc.
I would like to see how many pairs there are with cos_similarity > 0.5 (or any other threshold I decide to use), removing cos_similarity == 1 when I compare the same files, of course.
Secondly, I need a list of these pairs based on file names.
So the output for the example below would look like:
1
and
["doc1", "doc4"]
Will really appreciate your help as I feel a bit lost not knowing which direction to go.
This is an example of my script to get the matrix:
doc1 = "Amazon's promise of next-day deliveries could be investigated amid customer complaints that it is failing to meet that pledge."
doc2 = "The BBC has been inundated with comments from Amazon Prime customers. Most reported problems with deliveries."
doc3 = "An Amazon spokesman told the BBC the ASA had confirmed to it there was no investigation at this time."
doc4 = "Amazon's promise of next-day deliveries could be investigated amid customer complaints..."
documents = [doc1, doc2, doc3, doc4]
# In my real script I iterate through a folder (path) with txt files like this:
#def read_text(path):
# documents = []
# for filename in glob.iglob(path+'*.txt'):
# _file = open(filename, 'r')
# text = _file.read()
# documents.append(text)
# return documents
import nltk, string, numpy
nltk.download('punkt') # first-time use only
stemmer = nltk.stem.porter.PorterStemmer()
def StemTokens(tokens):
return [stemmer.stem(token) for token in tokens]
remove_punct_dict = dict((ord(punct), None) for punct in string.punctuation)
def StemNormalize(text):
return StemTokens(nltk.word_tokenize(text.lower().translate(remove_punct_dict)))
nltk.download('wordnet') # first-time use only
lemmer = nltk.stem.WordNetLemmatizer()
def LemTokens(tokens):
return [lemmer.lemmatize(token) for token in tokens]
remove_punct_dict = dict((ord(punct), None) for punct in string.punctuation)
def LemNormalize(text):
return LemTokens(nltk.word_tokenize(text.lower().translate(remove_punct_dict)))
from sklearn.feature_extraction.text import CountVectorizer
LemVectorizer = CountVectorizer(tokenizer=LemNormalize, stop_words='english')
LemVectorizer.fit_transform(documents)
tf_matrix = LemVectorizer.transform(documents).toarray()
from sklearn.feature_extraction.text import TfidfTransformer
tfidfTran = TfidfTransformer(norm="l2")
tfidfTran.fit(tf_matrix)
tfidf_matrix = tfidfTran.transform(tf_matrix)
cos_similarity_matrix = (tfidf_matrix * tfidf_matrix.T).toarray()
from sklearn.feature_extraction.text import TfidfVectorizer
TfidfVec = TfidfVectorizer(tokenizer=LemNormalize, stop_words='english')
def cos_similarity(textlist):
tfidf = TfidfVec.fit_transform(textlist)
return (tfidf * tfidf.T).toarray()
cos_similarity(documents)
Out:
array([[ 1. , 0.1459739 , 0.03613371, 0.76357693],
[ 0.1459739 , 1. , 0.11459266, 0.19117117],
[ 0.03613371, 0.11459266, 1. , 0.04732164],
[ 0.76357693, 0.19117117, 0.04732164, 1. ]])
As I understood your question, you want to create a function that reads the output numpy array and a certain value (threshold) in order to return two things:
how many docs are bigger than or equal the given threshold
the names of these docs.
So, here I've made the following function which takes three arguments:
the output numpy array from cos_similarity() function.
list of document names.
a certain number (threshold).
And here it's:
def get_docs(arr, docs_names, threshold):
output_tuples = []
for row in range(len(arr)):
lst = [row+1+idx for idx, num in \
enumerate(arr[row, row+1:]) if num >= threshold]
for item in lst:
output_tuples.append( (docs_names[row], docs_names[item]) )
return len(output_tuples), output_tuples
Let's see it in action:
>>> docs_names = ["doc1", "doc2", "doc3", "doc4"]
>>> arr = cos_similarity(documents)
>>> arr
array([[ 1. , 0.1459739 , 0.03613371, 0.76357693],
[ 0.1459739 , 1. , 0.11459266, 0.19117117],
[ 0.03613371, 0.11459266, 1. , 0.04732164],
[ 0.76357693, 0.19117117, 0.04732164, 1. ]])
>>> threshold = 0.5
>>> get_docs(arr, docs_names, threshold)
(1, [('doc1', 'doc4')])
>>> get_docs(arr, docs_names, 1)
(0, [])
>>> get_docs(lst, docs_names, 0.13)
(3, [('doc1', 'doc2'), ('doc1', 'doc4'), ('doc2', 'doc4')])
Let's see how this function works:
first, I iterate over every row of the numpy array.
Second, I iterate over every item in the row whose index is bigger than the row's index. So, we are iterating in a traingular shape like so:
and that's because each pair of documents is mentioned twice in the whole array. We can see that the two values arr[0][1] and arr[1][0] are the same. You also should notice that the diagonal items arn't included because we knew for sure that they are 1 as evey document is very similar to itself :).
Finally, we get the items whose values are bigger than or equal the given threshold, and return their indices. These indices are used later to get the documents names.
Related
OK so let's say I have a situation where I have a bunch of objects in different classifications and I need to know the total possible combinations of these objects so I end up with an input that looks like this
{'raw':[{'AH':['P','C','R','Q','L']},
{'BG':['M','A','S','B','F']},
{'KH':['E','V','G','N','Y']},
{'KH':['E','V','G','N','Y']},
{'NM':['1','2','3','4','5']}]}
Where the keys AH, BG, KH, NM constitute groups, the values are list that hold individual objects and a finished group would constitute one member of each list, in this example KH is listed twice so each finished group would have 2 members of KH in it. I build something that handles this, it looks like this.
class Builder():
def __init__(self, data):
self.raw = data['raw']
node = []
for item in self.raw:
for k in item.keys():
node.append({k:0})
logger.debug('node: %s' % node)
#Parse out groups#
self.groups = []
increment = -2
while True:
try:
assert self.raw[increment].values()[0][node[increment][node[increment].keys()[0]]]
increment = -2
for x in self.raw[-1].values()[0]:
group = []
for k in range(0,len(node[:-1])):
position = node[k].keys()[0]
player = self.raw[k].values()[0][node[k][node[k].keys()[0]]]
group.append({position:player})
group.append({self.raw[-1].keys()[0]:x})
if self.repeatRemovals(group):
self.groups.append(group)
node[increment][node[increment].keys()[0]]+=1
except IndexError:
node[increment][node[increment].keys()[0]] = 0
increment-=1
try:
node[increment][node[increment].keys()[0]]+=1
except IndexError:
break
for group in self.groups:
logger.debug(group)
def repeatRemovals(self, group):
for x in range(0, len(group)):
for y in range(0, len(group)):
if group[x].values()[0] == group[y].values()[0] and x != y:
return False
return True
if __name__ == '__main__':
groups = Builder({'raw':[{'AH':['P','C','R','Q','L']},
{'BG':['M','A','S','B','F']},
{'KH':['E','V','G','N','Y']},
{'KH':['E','V','G','N','Y']},
{'NM':['1','2','3','4','5']}]})
logger.debug("Total groups: %d" % len(groups.groups))
The output of running this should clearly state my intended goal, if I have failed to do so in text. My concern is the time it takes to handle large classification of objects, when a classification has some 40 something objects in it, it is in the matrix three times and there are 7 other classifications with comparable object sizes. I think the numpy library could help me, but I am new to scientific programming and am not sure how to go about it, or if it would be worth it, could anyone provide some insight? Thank you.
Try this:
Remove duplicated values
Calculate all possibilities using permutation and factorial
Like that:
https://www.youtube.com/watch?v=Oc50d2GqXx0
I would like to sort a list or an array using python to achive the following:
Say my initial list is:
example_list = ["retg_1_gertg","fsvs_1_vs","vrtv_2_srtv","srtv_2_bzt","wft_3_btb","tvsrt_3_rtbbrz"]
I would like to get all the elements that have 1 behind the first underscore together in one list and the ones that have 2 together in one list and so on. So the result should be:
sorted_list = [["retg_1_gertg","fsvs_1_vs"],["vrtv_2_srtv","srtv_2_bzt"],["wft_3_btb","tvsrt_3_rtbbrz"]]
My code:
import numpy as np
import string
example_list = ["retg_1_gertg","fsvs_1_vs","vrtv_2_srtv","srtv_2_bzt","wft_3_btb","tvsrt_3_rtbbrz"]
def sort_list(imagelist):
# get number of wafers
waferlist = []
for image in imagelist:
wafer_id = string.split(image,"_")[1]
waferlist.append(wafer_id)
waferlist = set(waferlist)
waferlist = list(waferlist)
number_of_wafers = len(waferlist)
# create list
sorted_list = []
for i in range(number_of_wafers):
sorted_list.append([])
for i in range(number_of_wafers):
wafer_id = waferlist[i]
for image in imagelist:
if string.split(image,"_")[1] == wafer_id:
sorted_list[i].append(image)
return sorted_list
sorted_list = sort_list(example_list)
works but it is really awkward and it involves many for loops that slow down everything if the lists are large.
Is there any more elegant way using numpy or anything?
Help is appreciated. Thanks.
I'm not sure how much more elegant this solution is; it is a bit more efficient. You could first sort the list and then go through and filter into final set of sorted lists:
example_list = ["retg_1_gertg","fsvs_1_vs","vrtv_2_srtv","srtv_2_bzt","wft_3_btb","tvsrt_3_rtbbrz"]
sorted_list = sorted(example_list, key=lambda x: x[x.index('_')+1])
result = [[]]
current_num = sorted_list[0][sorted_list[0].index('_')+1]
index = 0
for i in example_list:
if current_num != i[i.index('_')+1]:
current_num = i[i.index('_')+1]
index += 1
result.append([])
result[index].append(i)
print result
If you can make assumptions about the values after the first underscore character, you could clean it up a bit (for example, if you knew that they would always be sequential numbers starting at 1).
I'm newbie in Python and I'm struggling in create a list of sums generated by a for loop.
I got an school assignment where my program have to simulate the scores of a class of blind students in a multiple choice test.
def blindwalk(): # Generates the blind answers in a test with 21 questions
import random
resp = []
gab = ["a","b","c","d"]
for n in range(0,21):
resp.append(random.choice(gab))
return(resp)
def gabarite(): # Generates the official answer key of the tests
import random
answ_gab = []
gab = ["a","b","c","d"]
for n in range(0,21):
answ_gab.append(random.choice(gab))
return(answ_gab)
def class_tests(A): # A is the number of students
alumni = []
A = int(A)
for a in range(0,A):
alumni.append(blindwalk())
return alumni
def class_total(A): # A is the number of students
A = int(A)
official_gab = gabarite()
tests = class_tests(A)
total_score = []*0
for a in range(0,A):
for n in range(0,21):
if tests[a][n] == official_gab[n]:
total_score[a].add(1)
return total_score
When I run the class_total() function, I get this error:
total_score[a].add(1)
IndexError: list index out of range
Question is: How I valuate the scores of each student and create a list with them, because this is what I want to do with the class_total() function.
I also tried
if tests[a][n] == official_gab[n]:
total_score[a] += 1
But I got the same error, so I think I don't fully understand how lists work in Python yet.
Thanks!
(Also, I'm not a English native-speaker, so please tell me if I couldn't be clear enough)
This line:
total_score = []*0
And in fact, any of the following lines:
total_score = []*30
total_score = []*3000
total_score = []*300000000
Cause total_score to be instantiated as an empty list. It doesn't even have a 0th index, in this case! If you'd like to initiate every value to x in a list of length l , the syntax would look more like:
my_list = [x]*l
Alternatively, instead of thinking about the size before-hand, you can use .append instead of trying to access a particular index, as in:
my_list = []
my_list.append(200)
# my_list is now [200], my_list[0] is now 200
my_list.append(300)
# my_list is now [200,300], my_list[0] is still 200 and my_list[1] is now 300
I have created code in Python 2.7 which saves sales data for various products into a text file using the write() method. My limited Python skills have hit the wall with the next step - I need code which can read this data from the text file and then calculate and display the mean average number of sales of each item. The data is stored in the text file like the data shown below (but I am able to format it differently if that would help).
Product A,30
Product B,26
Product C,4
Product A,40
Product B,18
Product A,31
Product B,13
Product C,3
After far too long Googling around this to no avail, any pointers on the best way to manage this would be greatly appreciated. Thanks in advance.
You can read from the file, then split each line by a space (' '). Then, it is just a matter of creating a dictionary, and appending each new item to a list which is the value for each letter key, then using sum and len to get the average.
Example
products = {}
with open("myfile.txt") as product_info:
data = product_info.read().split('\n') #Split by line
for item in data:
_temp = item.split(' ')[1].split(',')
if _temp[0] not in products.keys():
products[_temp[0]] = [_temp[1]]
else:
products[_temp[0]] = products[_temp[0]]+[_temp[1]]
product_list = [[item, float(sum(key))/len(key)] for item, key in d.items()]
product_list.sort(key=lambda x:x[0])
for item in product_list:
print 'The average of {} is {}'.format(item[0], item[1])
from __future__ import division
dict1 = {}
dict2 = {}
file1 = open("input.txt",'r')
for line in file1:
if len(line)>2:
data = line.split(",")
a,b = data[0].strip(),data[1].strip()
if a in dict1:
dict1[a] = dict1[a] + int(b)
else:
dict1[a] = int(b)
if a in dict2:
dict2[a] = dict2[a] + 1
else:
dict2[a] = 1
for k,v in dict1.items():
for m,n in dict2.items():
if k == m:
avg = float(v/n)
print "%s Average is: %0.6f"%(k,float(avg))
Output:
Product A Average is: 33.666667
Product B Average is: 19.000000
Product C Average is: 3.500000
I've been trying to add the number of 2 list inside a dictionnary. The thing is, I need to verify if the value in the selected row and column is already in the dictionnary, if so I want to add the double entry list to the value (another double entry list) already existing in the dictionnary. I'm using a excel spreadsheet + xlrd so i can read it up. I' pretty new to this.
For exemple, the code is checking the account (a number) in the specified row and columns, let's say the value is 10, then if it's not in the dictionnary, it add the 2 values corresponding to this count, let's say [100, 0] as a value to this key. This is working as intended.
Now, the hard part is when the account number is already in the dictionnary. Let's say its the second entry for the account number 10. and it's [50, 20]. I want the value associated to the key "10" to be [150, 20].
I've tried the zip method but it seems to return radomn result, Sometimes it adds up, sometime it doesn't.
import xlrd
book = xlrd.open_workbook("Entry.xls")
print ("The number of worksheets is", book.nsheets)
print ("Worksheet name(s):", book.sheet_names())
sh = book.sheet_by_index(0)
print (sh.name,"Number of rows", sh.nrows,"Number of cols", sh.ncols)
liste_compte = {}
for rx in range(4, 10):
if (sh.cell_value(rowx=rx, colx=4)) not in liste_compte:
liste_compte[((sh.cell_value(rowx=rx, colx=4)))] = [sh.cell_value(rowx=rx, colx=6), sh.cell_value(rowx=rx, colx=7)]
elif (sh.cell_value(rowx=rx, colx=4)) in liste_compte:
three = [x + y for x, y in zip(liste_compte[sh.cell_value(rowx=rx, colx=4)],[sh.cell_value(rowx=rx, colx=6), sh.cell_value(rowx=rx, colx=7)])]
liste_compte[(sh.cell_value(rowx=rx, colx=4))] = three
print (liste_compte)
I'm not going to directly untangle your code, but just help you with a general example that does what you want:
def update_balance(existing_balance, new_balance):
for column in range(len(existing_balance)):
existing_balance[column] += new_balance[column]
def update_account(accounts, account_number, new_balance):
if account_number in accounts:
update_balance(existing_balance = accounts[account_number], new_balance = new_balance)
else:
accounts[account_number] = new_balance
And finally you'd do something like (assuming your xls looks like [account_number, balance 1, balance 2]:
accounts = dict()
for row in xls:
update_account(accounts = accounts,
account_number = row[0],
new_balance = row[1:2])