Question

我正在使用Python 2.7。我试图比较词典中的价值项目。

我有两个问题。首先是长度为1的字典中值的迭代。我总是得到一个错误，因为python不迭代整数而单个项作为值是python的整数。我试图将项目更改为字符串。我试图将整个字典更改为字符串。在这些情况下，迭代比较逗号和括号，这不是目的。因此我复制了该项目。 key：1和key：4中的值是同一个项目，两次。这个解决方案有效，但它自然会改变一些结果。

第二个问题是匹配计数无法正常工作。是的，因为第一个问题，但还有另一个问题。我的代码将所有值项与字典中的所有其他值项进行比较。不比较相同的模数。如果比较中有一个匹配，则没有问题。但是如果比较有两个匹配，代码将输出结果两次作为一个匹配。但我想要一个结果，有两场比赛。代码包含很多注释，所以我希望你能解读它。

    #dicModul is a dictionary with the modulnames(keys of dictionary) and the components of the modul(value of dictionary).
# If a modul consists of one component, it is impossible to iterate the modul, therefore this component is two times in a modul.
# Any ideas how to iterate a single value item in a dictionary?
dicModul = {0:(0,1),1:(1,1), 2:(2,3,4,5,6,1,7,2), 
          3:(8,1),4:(9,9), 5:(10,10,5,11,0,12,13), 6:(10,11,9,7)}
#The list is needed for the iteration and to operate the different modul in the dictionary by following for loops.
ModulList = [0,1,2,3,4,5,6] 
#Counter is needed for counting same components in different moduls.
Counter = 0
for modKey in ModulList: #For-loop is needed for iteration of keys
    for modKey2 in ModulList: #2nd for-loop of Modulkeys.The components of moduls have to be checked for matches in other moduls.
                              #Therefore a second iteration of the modul list is needed.
        if modKey == modKey2: #Same moduls do not have to be checked.
            print "Skip same Moduls" 
            Counter  = 0     #The modkey2 have to changed by iteration. The counter have to be reset.
        elif modKey != modKey2: #Different moduls have to be checked.
                for modVal in dicModul[modKey]:  #Iteration of components for iterated modul.
                    if modVal in dicModul[modKey2]: #Checking iterated components in different modul
                        Counter = Counter + 1  #If component was in different moduls, counter will add +1
                        print "Modul: "+str(modKey)+" % Modul: "+str(modKey2)+ " Matches= "+str(Counter) 
                        #print function for both moduls and number of same components 
                        Counter =0

我认为，如果此时关于上一个检查键和已检查键之间的分离，则计数器不会每次都从0开始。我试图解决问题并寻找解决方案，但我没有找到这种情况。

loop on a dictionary (previous values) python

How to access previous/next element while for looping?

Count items in dictionary

我想解决方案必须看起来像下一个代码我用＃ - ＆gt;

标记了我的建议

#dicModul is a dictionary with the modulnames(keys of dictionary) and the components of the modul(value of dictionary).
# If a modul consists of one component, it is impossible to iterate the modul, therefore this component is two times in a modul.
# Any ideas how to iterate a single value item in a dictionary?
dicModul = {0:(0,1),1:(1,1), 2:(2,3,4,5,6,1,7,2), 
          3:(8,1),4:(9,9), 5:(10,10,5,11,0,12,13), 6:(10,11,9,7)}
#The list is needed for the iteration and to operate the different modul in the dictionary by following for loops.
ModulList = [0,1,2,3,4,5,6] 
#Counter is needed for counting same components in different moduls.
Counter = 0
for modKey in ModulList: #For-loop is needed for iteration of keys
    for modKey2 in ModulList: #2nd for-loop of Modulkeys.The components of moduls have to be checked for matches in other moduls.
                              #Therefore a second iteration of the modul list is needed.
        if modKey == modKey2: #Same moduls do not have to be checked.
            print "Skip same Moduls" 
            Counter  = 0     #The modkey2 have to changed by iteration. The counter have to be reset  
        elif modKey != modKey2: #Different moduls have to be checked.
                for modVal in dicModul[modKey]:  #Iteration of components for iterated modul.
                    if modVal in dicModul[modKey2]: #Checking iterated components in different modul
                #->     if modKey2 == previous modKey2: #Checking if is the previous modul, so counter is not reset.
                #->           Counter = Counter +1 
                #->           print "Modul: "+str(modKey)+" % Modul: "+str(modKey2)+ " Matches= "+str(Counter) 
                #->     else:
                #->        Counter = 1   #Counter is setted 1, because a same component is found in a different modul.
                    elif:
                        Counter = 0

这是我期望的解决方案

Modul   Modul   Matches
skip same modul     
0   1   1
0   2   1
0   3   1
0   4   0
0   5   1
skip same modul     
1   2   1
1   3   1
1   4   0
1   5   0
skip same modul     
2   3   1
2   4   0
2   5   1
2   6   1
skip same modul     
3   4   0
3   5   0
skip same modul     
4   5   0
4   6   1
skip same modul     
5   6   3

Answer 1

我建议你看一下for循环，以及它们是如何工作的。 Using domain name with WAMP是一个很棒的（免费）互动资源。我不是100％肯定我明白你在问什么，所以我主要是关注你的问题。我想我明白你要做的是什么，但如果我这样做，你就会让自己变得非常困难。

迭代包含单个项目的字典：

dicty = {'a': 1}    
for key in dicty:
    print key

迭代字典中的值和键：

for key, val in dicty.iteritems():
    print key
    print val

相当于：

for key in stuff:
    val = stuff[key]
    print key
    print val

计数（键）匹配两个词：

count = 0
keys_counted = []

dicty_a = {'key1': 'val1', 'key2': 'val2'}
dicty_b = {'key1': 'val1', 'keyB': 'valB'}
for keyA, valA in dicty_a.iteritems():
    for keyB, valB in dicty_b.iteritems():
        if keyB == keyA and keyA not in keys_counted:
            count += 1
            keys_counted.append(valA)

确保不重复检查的最常见解决方案是构建列表，检查它，然后在最后将其丢弃。例如：

list_printed = []
for x in range(5):
    for y in range(5):
        if [x, y] not in list_printed:
            print x, y
            list_printed.append([y,x])

输出：

Enumerate()允许您获取字典中项目的键和值，或者如果您在列表中循环，则获取迭代次数和值。

关于迭代单个项目的问题。 Codeacademy，特别是单项。你需要尾随逗号或者它不是一个元组，它是一个int。老实说，这对我来说是一个错误，而不是一个功能，但乐观地说，它可能对某些计算机技术原因很重要，这远远超出了我的薪水。此外，区分列表[]和元组()的重要一点是元组是不可变的。在这种情况下，您可能不需要使用元组。话虽如此，你得到的错误只是因为，你说，你不能迭代一个整数：

loopy = (1,)
for x in loopy:
    print x # this will _not_ raise an error

loopy = (1)
for x in loopy:
    print x # this _will_ raise an error

loopy = [1]
for x in loopy:
    print x # this will _not_ raise an error

loopy = 1
for x in loopy:
    print x # this _will_ raise an error

将此应用于您要执行的操作：

dicModul = {0:(0,1),1:(1,1), 2:(2,3,4,5,6,1,7,2), 
          3:(8,1),4:(9,9), 5:(10,10,5,11,0,12,13), 6:(10,11,9,7)}

keys_checked = []

def get_tuple_matches(t1, t2):
    counter = 0
    matched_list = []
    for x in t1:
        for y in t2:
            stuff = [x, y]
            if stuff not in matched_list and x == y:
                counter +=1
                matched_list.append(stuff)
    return counter

for key_outerloop, val_outerloop in dicModul.iteritems():
    for key_innerloop, val_innerloop in dicModul.iteritems():
        if key_outerloop == key_innerloop:
            print "skip..."
        elif [key_innerloop, key_outerloop] not in keys_checked:
            matches = get_tuple_matches(val_outerloop, val_innerloop)
            keys_checked.append([key_outerloop, key_innerloop])
            print "Modul: " + str(key_outerloop) + " | Modul: " + str(key_innerloop) + " | Matches= "+ str(matches)

输出：

skip...
Modul: 0 | Modul: 1 | Matches= 1
Modul: 0 | Modul: 2 | Matches= 1
Modul: 0 | Modul: 3 | Matches= 1
Modul: 0 | Modul: 4 | Matches= 0
Modul: 0 | Modul: 5 | Matches= 1
Modul: 0 | Modul: 6 | Matches= 0
skip...
Modul: 1 | Modul: 2 | Matches= 1
Modul: 1 | Modul: 3 | Matches= 1
Modul: 1 | Modul: 4 | Matches= 0
Modul: 1 | Modul: 5 | Matches= 0
Modul: 1 | Modul: 6 | Matches= 0
skip...
Modul: 2 | Modul: 3 | Matches= 1
Modul: 2 | Modul: 4 | Matches= 0
Modul: 2 | Modul: 5 | Matches= 1
Modul: 2 | Modul: 6 | Matches= 1
skip...
Modul: 3 | Modul: 4 | Matches= 0
Modul: 3 | Modul: 5 | Matches= 0
Modul: 3 | Modul: 6 | Matches= 0
skip...
Modul: 4 | Modul: 5 | Matches= 0
Modul: 4 | Modul: 6 | Matches= 1
skip...
Modul: 5 | Modul: 6 | Matches= 2
skip...

Tuples are weird

比较字典中的价值项和计算匹配

1 个答案: