cosine_similarity between 2 pandas df列得到余弦距离
cosine_similarity between 2 pandas df column to get cosine distance
我有一个如下所示的数据框:
vector_a vector_b
[1,2,3] [2,5,6]
[0,2,1] [2,9,1]
[4,7,1] [1,7,4]
我想在列 vector_a 和 vector_b 之间执行 sklearn 的 cosine_similarity
以获得同一数据框中名为 'cosine_distance' 的新列。请注意 vector_a 和 vector_b 是 list
.
的 pandas df
列
这是我尝试过的:
df['vector_a'] = df['vector_a'].apply(lambda x: np.asarray(x))
df['vector_b'] = df['vector_b'].apply(lambda x: np.asarray(x))
df['cosine_distance'] = cosine_similarity(df['vector_a'].apply(lambda x: np.transpose(x)),
df['vector_b'].apply(lambda x: np.transpose(x)))
我得到了这个错误:
---> 58 df['cosine_distance'] = cosine_similarity(df['vector_a'].apply(lambda x: np.transpose(x)), df['vector_b'].apply(lambda x: np.transpose(x)))
~\Anaconda3\lib\site-packages\sklearn\metrics\pairwise.py in cosine_similarity(X, Y, dense_output)
1025 # to avoid recursive import
1026
-> 1027 X, Y = check_pairwise_arrays(X, Y)
1028
1029 X_normalized = normalize(X, copy=True)
~\Anaconda3\lib\site-packages\sklearn\metrics\pairwise.py in check_pairwise_arrays(X, Y, precomputed, dtype)
110 else:
111 X = check_array(X, accept_sparse='csr', dtype=dtype,
--> 112 estimator=estimator)
113 Y = check_array(Y, accept_sparse='csr', dtype=dtype,
114 estimator=estimator)
~\Anaconda3\lib\site-packages\sklearn\utils\validation.py in check_array(array, accept_sparse, accept_large_sparse, dtype, order, copy, force_all_finite, ensure_2d, allow_nd, ensure_min_samples, ensure_min_features, warn_on_dtype, estimator)
494 try:
495 warnings.simplefilter('error', ComplexWarning)
--> 496 array = np.asarray(array, dtype=dtype, order=order)
497 except ComplexWarning:
498 raise ValueError("Complex data not supported\n"
~\Anaconda3\lib\site-packages\numpy\core\numeric.py in asarray(a, dtype, order)
536
537 """
--> 538 return array(a, dtype, copy=False, order=order)
539
540
ValueError: setting an array element with a sequence.
提前致谢!
TLDR:
df['cosine_similarity'] = df.apply(
lambda row: cosine_similarity([row['vector_a']], [row['vector_b']])[0][0],
axis=1)
解释:
cosine_similarity
需要 2D np.array 或列表列表。它不知道如何解释列表的 pd.Series。但是,即使我们确实将其转换为列表的列表,也会出现下一个问题:
cosine_similarity
returns 完全相似。所以,让我们限制成对比较,人为地创建第二个维度(注意 [row['vector_a']], [row['vector_b']]
中额外的方括号),然后取 1x1 数组的唯一元素(cosine_similarity(...)[0][0]
末尾的零)
我有一个如下所示的数据框:
vector_a vector_b
[1,2,3] [2,5,6]
[0,2,1] [2,9,1]
[4,7,1] [1,7,4]
我想在列 vector_a 和 vector_b 之间执行 sklearn 的 cosine_similarity
以获得同一数据框中名为 'cosine_distance' 的新列。请注意 vector_a 和 vector_b 是 list
.
df
列
这是我尝试过的:
df['vector_a'] = df['vector_a'].apply(lambda x: np.asarray(x))
df['vector_b'] = df['vector_b'].apply(lambda x: np.asarray(x))
df['cosine_distance'] = cosine_similarity(df['vector_a'].apply(lambda x: np.transpose(x)),
df['vector_b'].apply(lambda x: np.transpose(x)))
我得到了这个错误:
---> 58 df['cosine_distance'] = cosine_similarity(df['vector_a'].apply(lambda x: np.transpose(x)), df['vector_b'].apply(lambda x: np.transpose(x)))
~\Anaconda3\lib\site-packages\sklearn\metrics\pairwise.py in cosine_similarity(X, Y, dense_output)
1025 # to avoid recursive import
1026
-> 1027 X, Y = check_pairwise_arrays(X, Y)
1028
1029 X_normalized = normalize(X, copy=True)
~\Anaconda3\lib\site-packages\sklearn\metrics\pairwise.py in check_pairwise_arrays(X, Y, precomputed, dtype)
110 else:
111 X = check_array(X, accept_sparse='csr', dtype=dtype,
--> 112 estimator=estimator)
113 Y = check_array(Y, accept_sparse='csr', dtype=dtype,
114 estimator=estimator)
~\Anaconda3\lib\site-packages\sklearn\utils\validation.py in check_array(array, accept_sparse, accept_large_sparse, dtype, order, copy, force_all_finite, ensure_2d, allow_nd, ensure_min_samples, ensure_min_features, warn_on_dtype, estimator)
494 try:
495 warnings.simplefilter('error', ComplexWarning)
--> 496 array = np.asarray(array, dtype=dtype, order=order)
497 except ComplexWarning:
498 raise ValueError("Complex data not supported\n"
~\Anaconda3\lib\site-packages\numpy\core\numeric.py in asarray(a, dtype, order)
536
537 """
--> 538 return array(a, dtype, copy=False, order=order)
539
540
ValueError: setting an array element with a sequence.
提前致谢!
TLDR:
df['cosine_similarity'] = df.apply(
lambda row: cosine_similarity([row['vector_a']], [row['vector_b']])[0][0],
axis=1)
解释:
cosine_similarity
需要 2D np.array 或列表列表。它不知道如何解释列表的 pd.Series。但是,即使我们确实将其转换为列表的列表,也会出现下一个问题:cosine_similarity
returns 完全相似。所以,让我们限制成对比较,人为地创建第二个维度(注意[row['vector_a']], [row['vector_b']]
中额外的方括号),然后取 1x1 数组的唯一元素(cosine_similarity(...)[0][0]
末尾的零)