There’s a lot to know about search intent, from using deep learning to infer search intent by classifying text and breaking down SERP titles using Natural Language Processing (NLP) techniques, to clustering based on semantic relevance with the benefits explained.
Not only do we know the benefits of deciphering search intent — we have a number of techniques at our disposal for scale and automation, too.
But often, those involve building your own AI. What if you don’t have the time nor the knowledge for that?
In this column, you’ll learn a step-by-step process for automating keyword clustering by search intent using Python.
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SERPs Contain Insights for Search Intent
Some methods require that you get all of the copy from titles of the ranking content for a given keyword, then feed it into a neural network model (which you have to then build and test), or maybe you’re using NLP to cluster keywords.
There is another method that enables you to use Google’s very own AI to do the work for you, without having to scrape all the SERPs content and build an AI model.
Let’s assume that Google ranks site URLs by the likelihood of the content satisfying the user query in descending order. It follows that if the intent for two keywords is the same, then the SERPs are likely to be similar.
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For years, many SEO professionals compared SERP results for keywords to infer shared (or shared) search intent to stay on top of Core Updates, so this is nothing new.
The value-add here is the automation and scaling of this comparison, offering both speed and greater precision.
How to Cluster Keywords by Search Intent at Scale Using Python (With Code)
Begin with your SERPs results in a CSV download.
1. Import the list into your Python notebook.
import pandas as pd import numpy as np serps_input = pd.read_csv('data/sej_serps_input.csv') serps_input
Below is the SERPs file now imported into a Pandas dataframe.
2. Filter Data for Page 1
We want to compare the Page 1 results of each SERP between keywords.
We’ll split the dataframe into mini keyword dataframes to run the filtering function before recombining into a single dataframe, because we want to filter at keyword level:
# Split serps_grpby_keyword = serps_input.groupby("keyword") k_urls = 15 # Apply Combine def filter_k_urls(group_df): filtered_df = group_df.loc[group_df['url'].notnull()] filtered_df = filtered_df.loc[filtered_df['rank'] <= k_urls] return filtered_df filtered_serps = serps_grpby_keyword.apply(filter_k_urls) # Combine ## Add prefix to column names #normed = normed.add_prefix('normed_') # Concatenate with initial data frame filtered_serps_df = pd.concat([filtered_serps],axis=0) del filtered_serps_df['keyword'] filtered_serps_df = filtered_serps_df.reset_index() del filtered_serps_df['level_1'] filtered_serps_df
3. Convert Ranking URLs to a String
Because there are more SERP result URLs than keywords, we need to compress those URLs into a single line to represent the keyword’s SERP.
Here’s how:
# convert results to strings using Split Apply Combine filtserps_grpby_keyword = filtered_serps_df.groupby("keyword") def string_serps(df): df['serp_string'] = ''.join(df['url']) return df # Combine strung_serps = filtserps_grpby_keyword.apply(string_serps) # Concatenate with initial data frame and clean strung_serps = pd.concat([strung_serps],axis=0) strung_serps = strung_serps[['keyword', 'serp_string']]#.head(30) strung_serps = strung_serps.drop_duplicates() strung_serps
Below shows the SERP compressed into a single line for each keyword.
4. Compare SERP Similarity
To perform the comparison, we now need every combination of keyword SERP paired with other pairs:
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# align serps def serps_align(k, df): prime_df = df.loc[df.keyword == k] prime_df = prime_df.rename(columns = "serp_string" : "serp_string_a", 'keyword': 'keyword_a') comp_df = df.loc[df.keyword != k].reset_index(drop=True) prime_df = prime_df.loc[prime_df.index.repeat(len(comp_df.index))].reset_index(drop=True) prime_df = pd.concat([prime_df, comp_df], axis=1) prime_df = prime_df.rename(columns = "serp_string" : "serp_string_b", 'keyword': 'keyword_b', "serp_string_a" : "serp_string", 'keyword_a': 'keyword') return prime_df columns = ['keyword', 'serp_string', 'keyword_b', 'serp_string_b'] matched_serps = pd.DataFrame(columns=columns) matched_serps = matched_serps.fillna(0) queries = strung_serps.keyword.to_list() for q in queries: temp_df = serps_align(q, strung_serps) matched_serps = matched_serps.append(temp_df) matched_serps
The above shows all of the keyword SERP pair combinations, making it ready for SERP string comparison.
There is no open source library that compares list objects by order, so the function has been written for you below.
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The function ‘serp_compare’ compares the overlap of sites and the order of those sites between SERPs.
import py_stringmatching as sm ws_tok = sm.WhitespaceTokenizer() # Only compare the top k_urls results def serps_similarity(serps_str1, serps_str2, k=15): denom = k+1 norm = sum([2*(1/i - 1.0/(denom)) for i in range(1, denom)]) ws_tok = sm.WhitespaceTokenizer() serps_1 = ws_tok.tokenize(serps_str1)[:k] serps_2 = ws_tok.tokenize(serps_str2)[:k] match = lambda a, b: [b.index(x)+1 if x in b else None for x in a] pos_intersections = [(i+1,j) for i,j in enumerate(match(serps_1, serps_2)) if j is not None] pos_in1_not_in2 = [i+1 for i,j in enumerate(match(serps_1, serps_2)) if j is None] pos_in2_not_in1 = [i+1 for i,j in enumerate(match(serps_2, serps_1)) if j is None] a_sum = sum([abs(1/i -1/j) for i,j in pos_intersections]) b_sum = sum([abs(1/i -1/denom) for i in pos_in1_not_in2]) c_sum = sum([abs(1/i -1/denom) for i in pos_in2_not_in1]) intent_prime = a_sum + b_sum + c_sum intent_dist = 1 - (intent_prime/norm) return intent_dist # Apply the function matched_serps['si_simi'] = matched_serps.apply(lambda x: serps_similarity(x.serp_string, x.serp_string_b), axis=1) serps_compared = matched_serps[['keyword', 'keyword_b', 'si_simi']] serps_compared
Now that the comparisons have been executed, we can start clustering keywords.
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We will be treating any keywords which have a weighted similarity of 40% or more.
# group keywords by search intent simi_lim = 0.4 # join search volume keysv_df = serps_input[['keyword', 'search_volume']].drop_duplicates() keysv_df.head() # append topic vols keywords_crossed_vols = serps_compared.merge(keysv_df, on = 'keyword', how = 'left') keywords_crossed_vols = keywords_crossed_vols.rename(columns = 'keyword': 'topic', 'keyword_b': 'keyword', 'search_volume': 'topic_volume') # sim si_simi keywords_crossed_vols.sort_values('topic_volume', ascending = False) # strip NANs keywords_filtered_nonnan = keywords_crossed_vols.dropna() keywords_filtered_nonnan
We now have the potential topic name, keywords SERP similarity, and search volumes of each.
You’ll note that keyword and keyword_b have been renamed to topic and keyword, respectively.
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Now we’re going to iterate over the columns in the dataframe using the lamdas technique.
The lamdas technique is an efficient way to iterate over rows in a Pandas dataframe because it converts rows to a list as opposed to the .iterrows() function.
Here goes:
queries_in_df = list(set(keywords_filtered_nonnan.topic.to_list())) topic_groups_numbered = topics_added = [] def find_topics(si, keyw, topc): i = 0 if (si >= simi_lim) and (not keyw in topics_added) and (not topc in topics_added): i += 1 topics_added.append(keyw) topics_added.append(topc) topic_groups_numbered[i] = [keyw, topc] elif si >= simi_lim and (keyw in topics_added) and (not topc in topics_added): j = [key for key, value in topic_groups_numbered.items() if keyw in value] topics_added.append(topc) topic_groups_numbered[j[0]].append(topc) elif si >= simi_lim and (not keyw in topics_added) and (topc in topics_added): j = [key for key, value in topic_groups_numbered.items() if topc in value] topics_added.append(keyw) topic_groups_numbered[j[0]].append(keyw) def apply_impl_ft(df): return df.apply( lambda row: find_topics(row.si_simi, row.keyword, row.topic), axis=1) apply_impl_ft(keywords_filtered_nonnan) topic_groups_numbered = k:list(set(v)) for k, v in topic_groups_numbered.items() topic_groups_numbered
Below shows a dictionary containing all the keywords clustered by search intent into numbered groups:
1: ['fixed rate isa', 'isa rates', 'isa interest rates', 'best isa rates', 'cash isa', 'cash isa rates'], 2: ['child savings account', 'kids savings account'], 3: ['savings account', 'savings account interest rate', 'savings rates', 'fixed rate savings', 'easy access savings', 'fixed rate bonds', 'online savings account', 'easy access savings account', 'savings accounts uk'], 4: ['isa account', 'isa', 'isa savings']
Let’s stick that into a dataframe:
topic_groups_lst = [] for k, l in topic_groups_numbered.items(): for v in l: topic_groups_lst.append([k, v]) topic_groups_dictdf = pd.DataFrame(topic_groups_lst, columns=['topic_group_no', 'keyword']) topic_groups_dictdf
The search intent groups above show a good approximation of the keywords inside them, something that an SEO expert would likely achieve.
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Although we only used a small set of keywords, the method can obviously be scaled to thousands (if not more).
Activating the Outputs to Make Your Search Better
Of course, the above could be taken further using neural networks processing the ranking content for more accurate clusters and cluster group naming, as some of the commercial products out there already do.
For now, with this output you can:
- Incorporate this into your own SEO dashboard systems to make your trends and SEO reporting more meaningful.
- Build better paid search campaigns by structuring your Google Ads accounts by search intent for a higher Quality Score.
- Merge redundant facet ecommerce search URLs.
- Structure a shopping site’s taxonomy according to search intent instead of a typical product catalog.
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I’m sure there are more applications that I haven’t mentioned — feel free to comment on any important ones that I’ve not already mentioned.
In any case, your SEO keyword research just got that little bit more scalable, accurate, and quicker!
More Resources:
Image Credits
Featured image: Astibuag/Shutterstock.com
All screenshots taken by author, July 2021
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