2

画像のヒストグラム値を含む2つの辞書を作成しました。各辞書には、キーとして画像ファイルのファイル名があり、値としてまとめられた3つの1次元ベクトルのリストがあります。

例： {'someFileName.jpg' : ['forecolor=2,3,5,5,6','edge=2,4,5','texture=5,4,3']}

これが私の辞書の1つの実際の表現です：

Dictionary1

{'/Users/images/Transcend-8GB-Class-10-SDHC-Flash-Memory-Card.jpg': ['fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3', 'edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63', 'texture=1,78,27,37,13,6,6,7,78']}

Dictionary2

{'/Users/images/kodax-camera-M531.jpg': ['fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0', 'edge=1,4,1,88,128,22,8,39,25,142,230,226,31,60,64,255,252,12,76', 'texture=1,15,32,31,28,19,16,12,98']}

私の最終目標は、これらのディクショナリのうち2つをメソッドに渡し、実際にcosign値を実行することです。

例：各ディクショナリには値としてリストがあるので、各ディクショナリキーについて、dictionary1のkey1、velu1とdictionary2 key1、value1の間でベクトル乗算を実行します。

私はベクトル乗算関数を持っているので、適切に反復する方法を理解しようとしているので、yield関数を使用することを考えていましたが、実際に試しても機能しませんでした。これは私がこれまでに持っているものです：

def cosignSimilarity(image1VectorDict, image2VectorDict):
    for image1Key, image2Value in image1VectorDict.iteritems():
        print image1Key
        for aValue in image1Value:
            print aValue
            for image2Key, image2Value in image2VectorDict.iteritems():
                for eValue in image2Value:
                    print aValue
                    print "\n"
                    print eValue

参考までに：私はコサインの計算について助けを求めていません。

これは、ある辞書から別の辞書にキーを分離できれば、現在のコードがデータを吐き出している方法です。その後、正弦値の計算などの残りの作業を行うことができます。

   First  Dictionary
    {'/Users/test/Transcend-8GB-Class-10-SDHC-Flash-Memory-Card.jpg': ['fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3', 'edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63', 'texture=1,78,27,37,13,6,6,7,78']}
    ------------------
Second Dictionary
    {'/Users/test/kodax-camera-M531.jpg': ['fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0', 'edge=1,4,1,88,128,22,8,39,25,142,230,226,31,60,64,255,252,12,76', 'texture=1,15,32,31,28,19,16,12,98']}
    ++++++++++++++++++
    /Users/test/Transcend-8GB-Class-10-SDHC-Flash-Memory-Card.jpg
    fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3
    fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3


    fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0
    fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3


    edge=1,4,1,88,128,22,8,39,25,142,230,226,31,60,64,255,252,12,76
    fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3


    texture=1,15,32,31,28,19,16,12,98
    edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63
    edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63


    fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0
    edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63


    edge=1,4,1,88,128,22,8,39,25,142,230,226,31,60,64,255,252,12,76
    edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63


    texture=1,15,32,31,28,19,16,12,98
    texture=1,78,27,37,13,6,6,7,78
    texture=1,78,27,37,13,6,6,7,78


    fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0
    texture=1,78,27,37,13,6,6,7,78


    edge=1,4,1,88,128,22,8,39,25,142,230,226,31,60,64,255,252,12,76
    texture=1,78,27,37,13,6,6,7,78


    texture=1,15,32,31,28,19,16,12,98

明らかにあなたが見ることができるように私は同じ価値の多くの繰り返しへの道を吐き出している

これらは私が扱っている実際の辞書です：

辞書1：

{'/Users/test/Transcend-8GB-Class-10-SDHC-Flash-Memory-Card.jpg': ['fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3', 'edge=1,252,1,32,124,194,63,252,67,15,240,1,7,244,66,47,0,192,63', 'texture=1,78,27,37,13,6,6,7,78']}

辞書2：

{'/Users/test/kodax-camera-M531.jpg': ['fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0', 'edge=1,4,1,88,128,22,8,39,25,142,230,226,31,60,64,255,252,12,76', 'texture=1,15,32,31,28,19,16,12,98']}

ランバ機能があります

cosinLamba = lambda a, b : round(NP.inner(a, b)/(LA.norm(a)*LA.norm(b)), 3)

辞書1と辞書2を繰り返し処理して、dictionary1のfcolor値を取得したい 'fcolor=2,4,14,5,0,0,0,0,0,0,0,0,0,0,12,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,15,6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,20,9,0,0,0,2,2,0,0,0,0,0,0,0,0,0,13,6,0,0,0,1,0,0,0,0,0,0,0,0,0,0,10,8,0,0,0,1,2,0,0,0,0,0,0,0,0,0,17,17,0,0,0,3,6,0,0,0,0,0,0,0,0,0,7,5,0,0,0,2,0,0,0,0,0,0,0,0,0,0,4,3,0,0,0,1,1,0,0,0,0,0,0,0,0,0,6,6,0,0,0,2,3'

およびdictionary2のfcolor値

'fcolor=2,74,6,20,30,1,2,0,1,0,0,0,1,3,2,0,0,0,0,0,1,1,1,0,0,2,0,0,0,2,2,0,0,0,0,0,2,2,1,0,0,5,0,0,0,1,4,0,0,0,0,0,2,2,1,0,0,1,0,0,0,3,1,0,0,0,0,0,1,1,0,0,0,3,0,0,0,1,2,0,0,0,0,0,2,2,1,0,0,4,0,0,0,0,5,0,0,0,0,0,2,1,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0'

それらをランバ関数に送信しますcosinLamba(valu1, value2) 。value1とvalue2は文字列であるため、これらを値として辞書に保存しました。そして、fcolor、texture、edgeに対して、各辞書の特定の画像に保存したすべてのベクトルを実行したいと思います。

4

2 に答える 2

2

表現を次のように変更することから始めることができます。

{'someFileName.jpg' : {'forecolor': [2,3,5,5,6],'edge': [2,4,5],'texture':[5,4,3]}}

または

{('someFileName.jpg', 'forecolor'): [2,3,5,5,6],
 ('someFileName.jpg', 'edge'): [2,4,5],
 ('someFileName.jpg', 'texture'):[5,4,3]}

たとえば、最初のケースに対応するリストを取得するには、次のようにします。

from itertools import product

# pair info for each image with info of every image from another dictionary
for (fn1, d1), (fn2,d2) in product(dict1.iteritems(), dict2.iteritems()):
    for property_, list_value in d1.iteritems():
        compute_cosine_similarity(list_value, d2[property_])

文字列のリストで表現を使用すると、次のようになります。

from itertools import product

# pair info for each image with info of every image from another dictionary
for (fn1,lst1), (fn2,lst2) in product(dict1.iteritems(), dict2.iteritems()):
    # assume all lists has the same order of elements
    for string_value1, string_value2 in zip(lst1, lst2):
        compute(string_value1, string_value2)

数値をASCII文字列のリストとして保存しないでください。メモリを節約する必要がある場合は、numpy配列を使用できます。cosinLambaすでにそれらを受け入れます。

from collections import namedtuple
import numpy as np

Info = namedtuple('Info', 'forecolor edge texture')

dict1 = {'someFileName.jpg': Info(np.array([...], dtype=np.uint8),
                                  np.array([...], dtype=np.uint8),
                                  np.array([...], dtype=np.uint8))}

cosine_similarity（）を呼び出すコードは、表現の場合とまったく同じです。

于 2012-07-31T00:30:55.257 に答える
0

私はあなたが実行したいコサイン計算に慣れていませんが、それを除けば、あなたの質問の私の理解が正しければ、次のコードが機能するはずです：

for key1, vals1 in dict1.iteritems():
    vals2 = dict2[key1]
    for val1, val2 in zip(vals1, vals2):
        # you now have the corresponding values for each image file
        compute_cosign(val1, val2)
于 2012-07-31T00:28:23.593 に答える