Python：迭代两个列表作为其值的列表

时间：2012-07-31 00:11:48

标签： python dictionary loops

我构建了两个包含图像直方图值的字典。每个字典都有图像文件的文件名作为键，三个一维向量的列表作为其值组合在一起。

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

以下是我的一本词典的实际代表：

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']}

我的最终目标是将这些词典中的两个传递给一个方法并实际运行cosign值，

示例：因此每个字典都有列表作为其值，因此对于每个字典键，我想在dictionary1＆＃39; s key1之间进行向量乘法，velu1与dictionary2 key1，value1，

我有矢量乘法函数所以我试图弄清楚如何正确迭代，我正在考虑使用yield函数，但是当我尝试时它并没有真正起作用。这就是我到目前为止所做的：

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

仅供参考：我没有在cosign计算方面寻求帮助。

如果我可以将密钥从一个字典隔离到另一个字典，那么我的当前代码将数据吐出的方式然后我可以完成剩下的工作，例如计算余弦值。 

   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

显然，你可以看到我正在吐出许多重复相同值的方法

这些是我正在处理的实际词典：

词典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']}

词典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']}

我有lamba功能 

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

我想迭代字典1和字典2并获取dictionary1的fcolor值 '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'

和dictionary2的fcolor值 

'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'

将它们发送到我的lamba函数cosinLamba(valu1, value2) value1和值2是字符串，这就是为什么我将它们作为值保存在我的字典中。 我想为fcolor，纹理，边缘做我为每个字典中给定图片存储的所有向量。

2 个答案:

答案 0 :(得分：2)

您可以先将代表更改为：

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

或

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

例如，要获得第一种情况的相应列表：

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_])

将您的表示与字符串列表一起使用，如下所示：

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)

您不应将数字存储为ascii字符串列表。如果你需要节省内存，你可以使用numpy数组。 cosinLamba已经接受了它们。 

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))}

调用cosine_similarity（）的代码与您的表示完全相同。

答案 1 :(得分：0)

我不熟悉您想要执行的cosign计算，但除此之外，如果我对您的问题的理解是正确的，那么以下代码应该有效：

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)
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