Question

我有一个字典，其中包含一个扩展类名列表作为键，每个键指向一个列表，其中包含该类出现在不同罐子中的次数以及它出现的罐子。

例如：

classToJars = {
  'com.sun.xml.ws.policy.PolicyMapKey.class' :  [ 1, 'policy-2.3.1.jar'],
  'com.sun.xml.ws.policy.PolicyMerger.class' :  [ 1, 'policy-2.3.1.jar'],
  'com.sun.xml.ws.policy.PolicyAssertion.class' : [ 1,  'policy-2.3.1.jar' ],

  'com.sun.xml.bind.AccessorFactory.class'     : [1, 'jaxb-impl-2.2.6.jar'],
  'com.sun.xml.bind.AccessorFactoryImpl.class' : [1, 'jaxb-impl-2.2.6.jar'],
  'com.sun.xml.bind.AnyTypeAdapter.class'      : [1, 'jaxb-impl-2.2.6.jar' ],

  'org.apache.mina.integration.jmx.IoSessionManager.class' :      [1, 'mina-integration-jmx-1.1.7.jar'],
  'org.apache.mina.integration.jmx.IoServiceManager.class' :      [1, 'mina-integration-jmx-1.1.7.jar'],

  'org.apache.log4j.Appender.class' : [2, 'log4j-1.2.14.jar', 'log4j-1.2.15.jar'],
  'org.apache.log4j.AppenderSkeleton.class' : [2, 'log4j-1.2.14.jar', 'log4j-1.2.15.jar'],
  
  'com.sun.activation.registries.LineTokenizer.class' : [1, 'activation-1.1.jar'],
  'com.sun.activation.registries.LogSupport.class' : [1, 'activation-1.1.jar'],

  'com.sun.istack.Builder.class' : [2, 'jaxb-impl-2.2.6.jar istack-commons-runtime-2.4.jar'],
  'com.sun.istack.ByteArrayDataSource.class' : [2, 'jaxb-impl-2.2.6.jar istack-commons-runtime-2.4.jar'],
  
  'com.reuters.rfa.ansipage.Page.class' : [1, 'rfa-7.2.0.E2.jar'],
  'com.reuters.rfa.ansipage.PageUpdate.class' : [1, 'rfa-7.2.0.E2.jar'],

  'org.apache.http.impl.io.AbstractMessageWriter.class' : [1, 'rfa-7.2.0.E2.jar'],
  'org.apache.http.impl.io.ChunkedOutputStream.class' : [1, 'rfa-7.2.0.E2.jar']
}

这是一个大字典，有数千个键和值环绕着大量的罐子。我们的想法是能够折叠dict，如果值相同，则将其折叠到最大的公共子串。

例如：当我运行折叠函数时，上面的哈希应该减少到4行，如下所示：

'com.sun.xml.ws.policy' 			:  [ 1, 'policy-2.3.1.jar'],
'com.sun.xml.bind'      			:  [1, 'jaxb-impl-2.2.6.jar'],
'org.apache.mina.integration.jmx' 	:  [1, 'mina-integration-jmx-1.1.7.jar'],
'org.apache.log4j' 					:  [2, 'log4j-1.2.14.jar', 'log4j-1.2.15.jar'],
'com.sun.activation.registries' 	:  [1, 'activation-1.1.jar'],
'com.sun.istack' 					:  [2, 'jaxb-impl-2.2.6.jar istack-commons-runtime-2.4.jar'],
'com.reuters.rfa.ansipage' 			:  [1, 'rfa-7.2.0.E2.jar'],
'org.apache.http.impl.io' 			:  [1, 'rfa-7.2.0.E2.jar'],

等等。

因为com.reuters.rfa和org.apache.http之间没有任何共同之处，如果你选择最大的公共子字符串，它将返回一个空键。

在这种情况下，它应该分别粘贴com.reuters.rfa和org.apache.http。

关于如何实现这一目标的任何想法？

Answer 1

这是你想要的吗？

import os

classToJars = {
  'com.sun.xml.ws.policy.PolicyMapKey.class' :  [ 1, 'policy-2.3.1.jar'],
  'com.sun.xml.ws.policy.PolicyMerger.class' :  [ 1, 'policy-2.3.1.jar'],
  'com.sun.xml.ws.policy.PolicyAssertion.class' : [ 1,  'policy-2.3.1.jar' ],

  'com.sun.xml.bind.AccessorFactory.class'     : [1, 'jaxb-impl-2.2.6.jar'],
  'com.sun.xml.bind.AccessorFactoryImpl.class' : [1, 'jaxb-impl-2.2.6.jar'],
  'com.sun.xml.bind.AnyTypeAdapter.class'      : [1, 'jaxb-impl-2.2.6.jar' ],

  'org.apache.mina.integration.jmx.IoSessionManager.class' :
      [1, 'mina-integration-jmx-1.1.7.jar'],
  'org.apache.mina.integration.jmx.IoServiceManager.class' :
      [1, 'mina-integration-jmx-1.1.7.jar'],
  'org.apache.mina.integration.jmx.IoSessionManagerMBean.class' :
      [1, 'mina-integration-jmx-1.1.7.jar' ],

  'org.apache.log4j.Appender.class' : [2, 'log4j-1.2.14.jar', 'log4j-1.2.15.jar'],
  'org.apache.log4j.AppenderSkeleton.class' : [2, 'log4j-1.2.14.jar', 'log4j-1.2.15.jar'],
  'org.apache.log4j.AsyncAppender.class' : [2, 'log4j-1.2.14.jar log4j-1.2.15.jar'],
  # ...
  }

#
# from http://stackoverflow.com/a/21419164/866915
#
def common_prefix(names):
  prefix = os.path.commonprefix( [ n.split('.') for n in names ] )
  return '.'.join(prefix)

# return the first 3 components of a class name
def min_prefix(name):
  return '.'.join( name.split('.')[0:3] )

jarsForKey = {}
keyForClass = {}

for c in classToJars:
  jars = classToJars[c]
  s = '|'.join(jars[1:])
  jarsForKey[s] = classToJars[c]
  keyForClass[c] = s

# group together classes based on their key

sameKey = {}
for c in classToJars:
  s = keyForClass[c]
  sameKey.setdefault(s,[]).append(c)

# for each group of classes with the same key, find the largest common substring

for s in sameKey:
  cls = sameKey[s]  # all of the classes with the same key
  jars = jarsForKey[s]

  # partition cls into groups having at least 3 components in common

  group = {}
  for c in cls:
    m = min_prefix(c)
    group.setdefault(m, []).append(c)

  # find the common prefix for each group

  for m in group:
    cls = group[m]
    prefix = common_prefix(cls)
    print prefix, "==>", jars

需要折叠的Jars中的类的哈希或词典

1 个答案: