I have a corpus of documents that I'm reading into a spark data frame.
I have tokeniked and vectorized the text and now I want to feed the vectorized data into an mllib LDA model. The LDA API docs seems to require the data to be:
rdd – RDD of documents, which are tuples of document IDs and term (word) count vectors. The term count vectors are “bags of words” with a fixed-size vocabulary (where the vocabulary size is the length of the vector). Document IDs must be unique and >= 0.
How can I get from my data frame to a suitable rdd?
from pyspark.mllib.clustering import LDA
from pyspark.ml.feature import Tokenizer
from pyspark.ml.feature import CountVectorizer
#read the data
tf = sc.wholeTextFiles("20_newsgroups/*")
#transform into a data frame
df = tf.toDF(schema=['file','text'])
#tokenize
tokenizer = Tokenizer(inputCol="text", outputCol="words")
tokenized = tokenizer.transform(df)
#vectorize
cv = CountVectorizer(inputCol="words", outputCol="vectors")
model = cv.fit(tokenized)
result = model.transform(tokenized)
#transform into a suitable rdd
myrdd = ?
#LDA
model = LDA.train(myrdd, k=2, seed=1)
PS : I'm using Apache Spark 1.6.3
Let's first organize imports, read the data, do some simple special characters removal and transform it into a DataFrame:
import re # needed to remove special character
from pyspark import Row
from pyspark.ml.feature import StopWordsRemover
from pyspark.ml.feature import Tokenizer, CountVectorizer
from pyspark.mllib.clustering import LDA
from pyspark.sql import functions as F
from pyspark.sql.types import StructType, StructField, LongType
pattern = re.compile('[\W_]+')
rdd = sc.wholeTextFiles("./data/20news-bydate/*/*/*") \
.mapValues(lambda x: pattern.sub(' ', x)).cache() # ref. https://stackoverflow.com/a/1277047/3415409
df = rdd.toDF(schema=['file', 'text'])
We will need to add an index to each Row. The following code snippet is inspired from this question about adding primary keys with Apache Spark :
row_with_index = Row(*["id"] + df.columns)
def make_row(columns):
def _make_row(row, uid):
row_dict = row.asDict()
return row_with_index(*[uid] + [row_dict.get(c) for c in columns])
return _make_row
f = make_row(df.columns)
indexed = (df.rdd
.zipWithUniqueId()
.map(lambda x: f(*x))
.toDF(StructType([StructField("id", LongType(), False)] + df.schema.fields)))
Once we have added the index, we can proceed to the features cleansing, extraction and transformation :
# tokenize
tokenizer = Tokenizer(inputCol="text", outputCol="tokens")
tokenized = tokenizer.transform(indexed)
# remove stop words
remover = StopWordsRemover(inputCol="tokens", outputCol="words")
cleaned = remover.transform(tokenized)
# vectorize
cv = CountVectorizer(inputCol="words", outputCol="vectors")
count_vectorizer_model = cv.fit(cleaned)
result = count_vectorizer_model.transform(cleaned)
Now, let's transform the results dataframe back to rdd
corpus = result.select(F.col('id').cast("long"), 'vectors').rdd \
.map(lambda x: [x[0], x[1]])
Our data is now ready to be trained :
# training data
lda_model = LDA.train(rdd=corpus, k=10, seed=12, maxIterations=50)
# extracting topics
topics = lda_model.describeTopics(maxTermsPerTopic=10)
# extraction vocabulary
vocabulary = count_vectorizer_model.vocabulary
We can print the topics descriptions now as followed :
for topic in range(len(topics)):
print("topic {} : ".format(topic))
words = topics[topic][0]
scores = topics[topic][1]
[print(vocabulary[words[word]], "->", scores[word]) for word in range(len(words))]
PS : This above code was tested with Spark 1.6.3.
Related
How do I return the individual auc-roc score for each fold/submodel when using crossValidator.
The documentation indicates that collectSubModels=True should save all models rather than just the best or average, but after inspecting model.subModels I can't find how to print them.
The below example works just missing the model.subModels.aucScore
Desired Result would be each fold with their corresponding score like [fold1:0.85, fold2:0.07, fold3:0.55]
from pyspark.ml.feature import VectorAssembler
from pyspark.ml.classification import RandomForestClassifier
from pyspark.ml.tuning import CrossValidator, ParamGridBuilder
from pyspark.ml.evaluation import BinaryClassificationEvaluator
#Creating test dataframe
training = spark.createDataFrame([
(1,0,1),
(1,0,0),
(0,1,1),
(0,1,0)], ["label", "feature1", "feature2"])
#Vectorizing features for modelling
assembler = VectorAssembler(inputCols=['feature1','feature2'],outputCol="features")
prepped = assembler.transform(training).select('label','features')
#setting variables and configuring CrossValidator
rf = RandomForestClassifier(labelCol="label", featuresCol="features")
params = ParamGridBuilder().build()
evaluator = BinaryClassificationEvaluator()
folds = 3
cv = CrossValidator(estimator=rf,
estimatorParamMaps=params,
evaluator=evaluator,
numFolds=folds,
collectSubModels=True
)
#Fitting model
model = cv.fit(prepped)
#Print Metrics
print(model)
print()
print(model.avgMetrics)
print()
print(model.subModels)
>>>>>Return:
>>>>>CrossValidatorModel_3a5c95c6d8d2
>>>>>()
>>>>>[0.8333333333333333]
>>>>>()
>>>>>[[RandomForestClassificationModel (uid=RandomForestClassifier_95da3a68af93) with 20 trees], >>>>>[RandomForestClassificationModel (uid=RandomForestClassifier_95da3a68af93) with 20 trees], >>>>>[RandomForestClassificationModel (uid=RandomForestClassifier_95da3a68af93) with 20 trees]]
I have a dataframe df. I have performed decisionTree classification algorithm on the dataframe. The two columns are label and features when algorithm is performed. The model is called dtc. How can I create a confusion matrix in pyspark?
dtc = DecisionTreeClassifier(featuresCol = 'features', labelCol = 'label')
dtcModel = dtc.fit(train)
predictions = dtcModel.transform(test)
from pyspark.mllib.linalg import Vectors
from pyspark.mllib.regression import LabeledPoint
from pyspark.mllib.evaluation import MulticlassMetrics
preds = df.select(['label', 'features']) \
.df.map(lambda line: (line[1], line[0]))
metrics = MulticlassMetrics(preds)
# Confusion Matrix
print(metrics.confusionMatrix().toArray())```
You need to cast to an rdd and map to tuple before calling metrics.confusionMatrix().toArray().
From the official documentation,
class pyspark.mllib.evaluation.MulticlassMetrics(predictionAndLabels)[source]
Evaluator for multiclass classification.
Parameters: predictionAndLabels – an RDD of (prediction, label) pairs.
Here is an example to guide you.
ML part
import pyspark.sql.functions as F
from pyspark.ml.feature import VectorAssembler
from pyspark.ml.classification import DecisionTreeClassifier
from pyspark.mllib.evaluation import MulticlassMetrics
from pyspark.sql.types import FloatType
#Note the differences between ml and mllib, they are two different libraries.
#create a sample data frame
data = [(1.54,3.45,2.56,0),(9.39,8.31,1.34,0),(1.25,3.31,9.87,1),(9.35,5.67,2.49,2),\
(1.23,4.67,8.91,1),(3.56,9.08,7.45,2),(6.43,2.23,1.19,1),(7.89,5.32,9.08,2)]
cols = ('a','b','c','d')
df = spark.createDataFrame(data, cols)
assembler = VectorAssembler(inputCols=['a','b','c'], outputCol='features')
df_features = assembler.transform(df)
#df.show()
train_data, test_data = df_features.randomSplit([0.6,0.4])
dtc = DecisionTreeClassifier(featuresCol='features',labelCol='d')
dtcModel = dtc.fit(train_data)
predictions = dtcModel.transform(test_data)
Evaluation part
#important: need to cast to float type, and order by prediction, else it won't work
preds_and_labels = predictions.select(['predictions','d']).withColumn('label', F.col('d').cast(FloatType())).orderBy('prediction')
#select only prediction and label columns
preds_and_labels = preds_and_labels.select(['prediction','label'])
metrics = MulticlassMetrics(preds_and_labels.rdd.map(tuple))
print(metrics.confusionMatrix().toArray())
Use this:
import sklearn
from pyspark.ml.classification import RandomForestClassifier
rf = RandomForestClassifier(featuresCol = 'features', labelCol = 'label', numTrees=500)
rfModel = rf.fit(train)
predictions_train = rfModel.transform(train)
y_true = predictions_train.select(['label']).collect()
y_pred = predictions_train.select(['prediction']).collect()
from sklearn.metrics import classification_report, confusion_matrix
print(classification_report(y_true, y_pred))
where train is your training data.
I am training an LDA model in pyspark (spark 2.1.1) on a customers review dataset. Now based on that model I want to predict the topics in the new unseen text.
I am using the following code to make the model
from pyspark import SparkConf, SparkContext
from pyspark.sql import SparkSession
from pyspark.sql import SQLContext, Row
from pyspark.ml.feature import CountVectorizer
from pyspark.ml.feature import HashingTF, IDF, Tokenizer, CountVectorizer, StopWordsRemover
from pyspark.mllib.clustering import LDA, LDAModel
from pyspark.ml.clustering import DistributedLDAModel, LocalLDAModel
from pyspark.mllib.linalg import Vector, Vectors
from pyspark.sql.functions import *
import pyspark.sql.functions as F
path = "D:/sparkdata/sample_text_LDA.txt"
sc = SparkContext("local[*]", "review")
spark = SparkSession.builder.appName('Basics').getOrCreate()
df = spark.read.csv("D:/sparkdata/customers_data.csv", header=True, inferSchema=True)
data = df.select("Reviews").rdd.map(list).map(lambda x: x[0]).zipWithIndex().map(lambda words: Row(idd= words[1], words = words[0].split(" "))).collect()
docDF = spark.createDataFrame(data)
remover = StopWordsRemover(inputCol="words",
outputCol="stopWordsRemoved")
stopWordsRemoved_df = remover.transform(docDF).cache()
Vector = CountVectorizer(inputCol="stopWordsRemoved", outputCol="vectors")
model = Vector.fit(stopWordsRemoved_df)
result = model.transform(stopWordsRemoved_df)
corpus = result.select("idd", "vectors").rdd.map(lambda x: [x[0],Vectors.fromML(x[1])]).cache()
# Cluster the documents topics using LDA
ldaModel = LDA.train(corpus, k=3,maxIterations=100,optimizer='online')
topics = ldaModel.topicsMatrix()
vocabArray = model.vocabulary
print(ldaModel.describeTopics())
wordNumbers = 10 # number of words per topic
topicIndices = sc.parallelize(ldaModel.describeTopics(maxTermsPerTopic = wordNumbers))
def topic_render(topic): # specify vector id of words to actual words
terms = topic[0]
result = []
for i in range(wordNumbers):
term = vocabArray[terms[i]]
result.append(term)
return result
topics_final = topicIndices.map(lambda topic: topic_render(topic)).collect()
for topic in range(len(topics_final)):
print("Topic" + str(topic) + ":")
for term in topics_final[topic]:
print (term)
print ('\n')
Now I have a dataframe with a column having new customer reviews and I want to predict that to which topic cluster they belong.
I have searched for answers, mostly the following way is recommended, as here Spark MLlib LDA, how to infer the topics distribution of a new unseen document?.
newDocuments: RDD[(Long, Vector)] = ...
topicDistributions = distLDA.toLocal.topicDistributions(newDocuments)
However, I get the following error:
'LDAModel' object has no attribute 'toLocal'.
Neither do it have topicDistribution attribute.
So are these attributes not supported in spark 2.1.1?
So any other way to infer topics from the unseen data?
You're going to need to pre-process the new data:
# import a new data set to be passed through the pre-trained LDA
data_new = pd.read_csv('YourNew.csv', encoding = "ISO-8859-1");
data_new = data_new.dropna()
data_text_new = data_new[['Your Target Column']]
data_text_new['index'] = data_text_new.index
documents_new = data_text_new
#documents_new = documents.dropna(subset=['Preprocessed Document'])
# process the new data set through the lemmatization, and stopwork functions
processed_docs_new = documents_new['Preprocessed Document'].map(preprocess)
# create a dictionary of individual words and filter the dictionary
dictionary_new = gensim.corpora.Dictionary(processed_docs_new[:])
dictionary_new.filter_extremes(no_below=15, no_above=0.5, keep_n=100000)
# define the bow_corpus
bow_corpus_new = [dictionary_new.doc2bow(doc) for doc in processed_docs_new]
Then you can just pass it through the trained LDA as a function. All you need is that bow_corpus:
ldamodel[bow_corpus_new[:len(bow_corpus_new)]]
If you want it out in a csv try this:
a = ldamodel[bow_corpus_new[:len(bow_corpus_new)]]
b = data_text_new
topic_0=[]
topic_1=[]
topic_2=[]
for i in a:
topic_0.append(i[0][1])
topic_1.append(i[1][1])
topic_2.append(i[2][1])
d = {'Your Target Column': b['Your Target Column'].tolist(),
'topic_0': topic_0,
'topic_1': topic_1,
'topic_2': topic_2}
df = pd.DataFrame(data=d)
df.to_csv("YourAllocated.csv", index=True, mode = 'a')
I hope this helps :)
I am calculating TF and IDF using spark mllib algorithm of a directory that contains csv files with the following code:
import argparse
from os import system
### args parsing
parser = argparse.ArgumentParser(description='runs TF/IDF on a directory of
text docs')
parser.add_argument("-i","--input", help="the input in HDFS",
required=True)
parser.add_argument("-o", '--output', help="the output in HDFS",
required=True )
parser.add_argument("-mdf", '--min_document_frequency', default=1 )
args = parser.parse_args()
docs_dir = args.input
d_out = "hdfs://master:54310/" + args.output
min_df = int(args.min_document_frequency)
# import spark-realated stuff
from pyspark import SparkContext
from pyspark.mllib.feature import HashingTF
from pyspark.mllib.feature import IDF
sc = SparkContext(appName="TF-IDF")
# Load documents (one per line).
documents = sc.textFile(docs_dir).map(lambda title_text:
title_text[1].split(" "))
hashingTF = HashingTF()
tf = hashingTF.transform(documents)
# IDF
idf = IDF().fit(tf)
tfidf = idf.transform(tf)
#print(tfidf.collect())
#save
tfidf.saveAsTextFile(d_out)
Using
print(tfidf.collect())
I get this output:
[SparseVector(1048576, {812399: 4.3307}), SparseVector(1048576, {411697:
0.0066}), SparseVector(1048576, {411697: 0.0066}), SparseVector(1048576,
{411697: 0.0066}), SparseVector(1048576, {411697: 0.0066}), ....
I have also tested the KMeans mllib algorithm :
from __future__ import print_function
import sys
import numpy as np
from pyspark import SparkContext
from pyspark.mllib.clustering import KMeans
runs=4
def parseVector(line):
return np.array([float(x) for x in line.split(' ')])
if __name__ == "__main__":
if len(sys.argv) != 3:
print("Usage: kmeans <file> <k>", file=sys.stderr)
exit(-1)
sc = SparkContext(appName="KMeans")
lines = sc.textFile(sys.argv[1])
data = lines.map(parseVector)
k = int(sys.argv[2])
model = KMeans.train(data, k, runs)
print("Final centers: " + str(model.clusterCenters))
print("Total Cost: " + str(model.computeCost(data)))
sc.stop()
with this sample test case
0.0 0.0 0.0
0.1 0.1 0.1
0.2 0.2 0.2
9.0 9.0 9.0
9.1 9.1 9.1
9.2 9.2 9.2
and it works fine.
Now I want to apply the rdd output from tfidf above in the KMeans algorithm but I don't know how is it possible to transform the rdd like the sample text above, or how to split properly the rdd in the KMeans algorithm to work properly.
I really need some help with this one.
UPDATE
My real question is how can i read the input to apply it to KMeans mllib from a text file like this
(1048576,[155412,857472,756332],[1.75642010278,2.41857747478,1.97365255252])
(1048576,[159196,323305,501636],[2.98856378408,1.63863706713,2.44956728334])
(1048576,[135312,847543,743411],[1.42412015238,1.58759872958,2.01237484818])
UPDATE2
I am not sure at all but i think i need to go from above vectors to the below array so as to apply it directly to KMeans mllib algorithm
1.75642010278 2.41857747478 1.97365255252
2.98856378408 1.63863706713 2.44956728334
1.42412015238 1.58759872958 2.01237484818
The output of IDF is a dataframe of SparseVector. KMeans takes a vector as input (sparse or dense), hence, there should be no need to make any transformations. You should be able to use the output column from IDF directly as input to KMeans.
If you need to save the data to disk in between running the TFIDF and KMeans, I would recommend saving it as a csv through the dataframe API.
First convert to a dataframe using Row:
from pyspark.sql import Row
row = Row("features") # column name
df = tfidf.map(row).toDF()
An alternative way to convert without import:
df = tfidf.map(lambda x: (x, )).toDF(["features"])
After the conversion save the dataframe as a parquet file:
df.write.parquet('/path/to/save/file')
To read the data, simply use:
from pyspark.sql import SQLContext
sqlContext = SQLContext(sc)
df = sqlContext.read.parquet('/path/to/file')
# converting from dataframe into an RDD[Vector]
data = df.rdd.map(list)
If you in any case need to convert from a vector saved as a string, that is also possible. Here is some example code:
from pyspark.mllib.linalg import Vectors, VectorUDT
from pyspark.sql.functions import udf
df = sc.parallelize(["(7,[1,2,4],[1,1,1])"]).toDF(["features"])
parse = udf(lambda s: Vectors.parse(s), VectorUDT())
df.select(parse("features"))
First an example dataframe is created with the same formatting. Then an UDF is used to parse the string into a vector. If you want an rdd instead of the dataframe, use the code above at the "reading from parquet" part to convert.
However, the output from IDF is very sparse. The vectors have a length of 1048576 and only one of these have a values over 1. KMeans would not give you any interesting results.
I would recommend you to look into word2vec instead. It will give you a more compact vector for each word and clustering these vectors would make more sense. Using this method you can receive a map of words to their vector representations which can be used for clustering.
when i try to feed df2 to kmeans i get the following error
clusters = KMeans.train(df2, 10, maxIterations=30,
runs=10, initializationMode="random")
The error i get:
Cannot convert type <class 'pyspark.sql.types.Row'> into Vector
df2 is a dataframe created as follow:
df = sqlContext.read.json("data/ALS3.json")
df2 = df.select('latitude','longitude')
df2.show()
latitude| longitude|
60.1643075| 24.9460844|
60.4686748| 22.2774728|
how can i convert this two columns to Vector and feed it to KMeans?
ML
The problem is that you missed the documentation's example, and it's pretty clear that the method train requires a DataFrame with a Vector as features.
To modify your current data's structure you can use a VectorAssembler. In your case it could be something like:
from pyspark.sql.functions import *
vectorAssembler = VectorAssembler(inputCols=["latitude", "longitude"],
outputCol="features")
# For your special case that has string instead of doubles you should cast them first.
expr = [col(c).cast("Double").alias(c)
for c in vectorAssembler.getInputCols()]
df2 = df2.select(*expr)
df = vectorAssembler.transform(df2)
Besides, you should also normalize your features using the class MinMaxScaler to obtain better results.
MLLib
In order to achieve this using MLLib you need to use a map function first, to convert all your string values into Double, and merge them together in a DenseVector.
rdd = df2.map(lambda data: Vectors.dense([float(c) for c in data]))
After this point you can train your MLlib's KMeans model using the rdd variable.
I got PySpark 2.3.1 to perform KMeans on a DataFrame as follows:
Write a list of the columns you want to include in the clustering analysis:
feat_cols = ['latitude','longitude']`
You need all of the columns to be numeric values:
expr = [col(c).cast("Double").alias(c) for c in feat_cols]
df2 = df2.select(*expr)
Create your features vector with mllib.linalg.Vectors:
from pyspark.ml.feature import VectorAssembler
assembler = VectorAssembler(inputCols=feat_cols, outputCol="features")
df3 = assembler.transform(df2).select('features')
You should normalize your features as normalization is not always required, but it rarely hurts (more about this here):
from pyspark.ml.feature import StandardScaler
scaler = StandardScaler(
inputCol="features",
outputCol="scaledFeatures",
withStd=True,
withMean=False)
scalerModel = scaler.fit(df3)
df4 = scalerModel.transform(df3).drop('features')\
.withColumnRenamed('scaledFeatures', 'features')
Turn your DataFrame object df4 into a dense vector RDD:
from pyspark.mllib.linalg import Vectors
data5 = df4.rdd.map(lambda row: Vectors.dense([x for x in row['features']]))
Use the obtained RDD object as input for KMeans training:
from pyspark.mllib.clustering import KMeans
model = KMeans.train(data5, k=3, maxIterations=10)
Example: classify a point p in your vector space:
prediction = model.predict(p)