Abh*_*ury 6 python dataframe cosine-similarity pyspark
我有一个数据集,其中包含工人的人口统计信息,如年龄性别,地址等及其工作地点.我从数据集创建了一个RDD并将其转换为DataFrame.
每个ID有多个条目.因此,我创建了一个DataFrame,其中只包含工人的ID和他/她工作过的各个办公地点.
|----------|----------------|
| **ID** **Office_Loc** |
|----------|----------------|
| 1 |Delhi, Mumbai, |
| | Gandhinagar |
|---------------------------|
| 2 | Delhi, Mandi |
|---------------------------|
| 3 |Hyderbad, Jaipur|
-----------------------------
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我想根据办公地点计算每个工人与其他工人之间的余弦相似度.
所以,我遍历了DataFrame的行,从DataFrame中检索了一行:
myIndex = 1
values = (ID_place_df.rdd.zipWithIndex()
.filter(lambda ((l, v), i): i == myIndex)
.map(lambda ((l,v), i): (l, v))
.collect())
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然后使用地图
cos_weight = ID_place_df.select("ID","office_location").rdd\
.map(lambda x: get_cosine(values,x[0],x[1]))
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计算提取的行和整个DataFrame之间的余弦相似度.
我不认为我的方法是好的,因为我在迭代DataFrame的行,它失败了使用spark的整个目的.在pyspark有更好的方法吗?好心提醒.
MaF*_*aFF 16
您可以使用该mllib包来计算L2每行的TF-IDF 的范数.然后将表与自身相乘以得到余弦相似度,作为两乘两个L2范数的点积:
1. RDD
rdd = sc.parallelize([[1, "Delhi, Mumbai, Gandhinagar"],[2, " Delhi, Mandi"], [3, "Hyderbad, Jaipur"]])
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计算TF-IDF:
documents = rdd.map(lambda l: l[1].replace(" ", "").split(","))
from pyspark.mllib.feature import HashingTF, IDF
hashingTF = HashingTF()
tf = hashingTF.transform(documents)
Run Code Online (Sandbox Code Playgroud)您可以指定HashingTF要素数,以使要素矩阵更小(列数更少).
tf.cache()
idf = IDF().fit(tf)
tfidf = idf.transform(tf)
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计算L2规范:
from pyspark.mllib.feature import Normalizer
labels = rdd.map(lambda l: l[0])
features = tfidf
normalizer = Normalizer()
data = labels.zip(normalizer.transform(features))
Run Code Online (Sandbox Code Playgroud)通过将矩阵与自身相乘来计算余弦相似度:
from pyspark.mllib.linalg.distributed import IndexedRowMatrix
mat = IndexedRowMatrix(data).toBlockMatrix()
dot = mat.multiply(mat.transpose())
dot.toLocalMatrix().toArray()
array([[ 0. , 0. , 0. , 0. ],
[ 0. , 1. , 0.10794634, 0. ],
[ 0. , 0.10794634, 1. , 0. ],
[ 0. , 0. , 0. , 1. ]])
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或者:使用笛卡尔积和dotnumpy数组上的函数:
data.cartesian(data)\
.map(lambda l: ((l[0][0], l[1][0]), l[0][1].dot(l[1][1])))\
.sortByKey()\
.collect()
[((1, 1), 1.0),
((1, 2), 0.10794633570596117),
((1, 3), 0.0),
((2, 1), 0.10794633570596117),
((2, 2), 1.0),
((2, 3), 0.0),
((3, 1), 0.0),
((3, 2), 0.0),
((3, 3), 1.0)]
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由于您似乎使用的是数据帧,因此您可以使用该spark ml包:
import pyspark.sql.functions as psf
df = rdd.toDF(["ID", "Office_Loc"])\
.withColumn("Office_Loc", psf.split(psf.regexp_replace("Office_Loc", " ", ""), ','))
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计算TF-IDF:
from pyspark.ml.feature import HashingTF, IDF
hashingTF = HashingTF(inputCol="Office_Loc", outputCol="tf")
tf = hashingTF.transform(df)
idf = IDF(inputCol="tf", outputCol="feature").fit(tf)
tfidf = idf.transform(tf)
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from pyspark.ml.feature import Normalizer
normalizer = Normalizer(inputCol="feature", outputCol="norm")
data = normalizer.transform(tfidf)
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from pyspark.mllib.linalg.distributed import IndexedRow, IndexedRowMatrix
mat = IndexedRowMatrix(
data.select("ID", "norm")\
.rdd.map(lambda row: IndexedRow(row.ID, row.norm.toArray()))).toBlockMatrix()
dot = mat.multiply(mat.transpose())
dot.toLocalMatrix().toArray()
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或者:使用连接和UDFfor函数dot:
dot_udf = psf.udf(lambda x,y: float(x.dot(y)), DoubleType())
data.alias("i").join(data.alias("j"), psf.col("i.ID") < psf.col("j.ID"))\
.select(
psf.col("i.ID").alias("i"),
psf.col("j.ID").alias("j"),
dot_udf("i.norm", "j.norm").alias("dot"))\
.sort("i", "j")\
.show()
+---+---+-------------------+
| i| j| dot|
+---+---+-------------------+
| 1| 2|0.10794633570596117|
| 1| 3| 0.0|
| 2| 3| 0.0|
+---+---+-------------------+
Run Code Online (Sandbox Code Playgroud)本教程列出了乘以大规模矩阵的不同方法:https://labs.yodas.com/large-scale-matrix-multiplication-with-pyspark-or-how-to-match-two-large-datasets-of-company -1be4b1b2871e
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