Question

假设我有两个数组：

arrayOne = [["james", 35], ["michael", 28], ["steven", 23], 
            ["jack", 18], ["robert", 12]]
arrayTwo = [["charles", 45], ["james",  36], ["trevor", 24], 
            ["michael", 17], ["steven", 4]]

我想合并它们，以便我有一个2D数组，其中每个内部数组的第一个元素是名称（james，charles等）。内部数组的第二个元素是它在arrayOne中的相应值，如果没有相应的值，则它将为0.相反，对于第三个元素。只要数字与名称匹配，订单就不重要了。换句话说，我会得到这样的东西

arrayResult = [["james", 35, 36], ["michael", 28, 17], ["steven", 23, 4],
               ["jack", 18, 0], ["robert", 12, 0], ["charles", 0, 45],
               ["trevor", 0, 4]]

此外，我正在努力让它为这个数组结果添加更多“列”，如果我要给另一个数组。

Answer 1

>>> dict1 = dict(arrayOne)
>>> dict2 = dict(arrayTwo)
>>> keyset = set(dict1.keys() + dict2.keys())
>>> [[key, dict1.get(key, 0), dict2.get(key, 0)] for key in keyset]
[['james', 35, 36], ['robert', 12, 0], ['charles', 0, 45], 
 ['michael', 28, 17], ['trevor', 0, 24], ['jack', 18, 0], 
 ['steven', 23, 4]]

如果要添加多个列，这会变得有点复杂;一本字典是最好的。但是在正确的位置使0成为挑战，因为当我们在“主词典”中添加名称时，我们必须确保它以正确长度的0列表开头。。我很想为此创建一个新类，但首先，这是一个基于函数的基本解决方案：

def add_column(masterdict, arr):
    mdlen = len(masterdict[masterdict.keys()[0]])
    newdict = dict(arr)
    keyset = set(masterdict.keys() + newdict.keys())
    for key in keyset:
        if key not in masterdict:
            masterdict[key] = [0] * mdlen
        masterdict[key].append(newdict.get(key, 0))

arrayOne =   [["james", 35],
              ["michael", 28],
              ["steven", 23],
              ["jack", 18],
              ["robert", 12]]
arrayTwo =   [["charles", 45],
              ["james",  36],
              ["trevor", 24],
              ["michael", 17],
              ["steven", 4]]
arrayThree = [["olliver", 11],
              ["james",  39],
              ["john", 22],
              ["michael", 13],
              ["steven", 6]]

masterdict = dict([(i[0], [i[1]]) for i in arrayOne])

add_column(masterdict, arrayTwo)
print masterdict
add_column(masterdict, arrayThree)
print masterdict

输出：

{'james': [35, 36], 'robert': [12, 0], 'charles': [0, 45], 
 'michael': [28, 17], 'trevor': [0, 24], 'jack': [18, 0], 
 'steven': [23, 4]}
{'james': [35, 36, 39], 'robert': [12, 0, 0], 'charles': [0, 45, 0], 
  'michael': [28, 17, 13], 'trevor': [0, 24, 0], 'olliver': [0, 0, 11], 
  'jack': [18, 0, 0], 'steven': [23, 4, 6], 'john': [0, 0, 22]}

Answer 2

看起来你真正需要的是词典，而不是数组。如果使用字典，这个问题就变得容易多了。转换为dicts并非易事：

dictOne = dict(arrayOne)
dictTwo = dict(arrayTwo)

从那里，你可以像这样把它们放在一起：

combined = dict()
for name in set(dictOne.keys() + dictTwo.keys()):
  combined[name] = [ dictOne.get(name, 0), dictTwo.get(name, 0) ]

这样做是创建一个名为combined的新词典，我们将把最终数据放入其中。然后，我们从两个原始词典中创建一组键。使用集合确保我们不做任何两次。最后，我们遍历这组键并将每对值添加到combined字典，如果没有值，则告诉调用.get方法提供0。如果你需要将组合字典切换回数组，那也很容易：

arrayResult = []
for name in combined:
  arrayResult.append([ name ] + combined[name])

假设您要在结果字典中添加另一列，您只需将中间代码更改为如下所示：

combined = dict()
for name in set(dictOne.keys() + dictTwo.keys() + dictThree.keys()):
  combined[name] = [ dictOne.get(name, 0), dictTwo.get(name, 0), dictThree.get(name, 0) ]

如果你想将所有这些逻辑封装在一个函数中（这是我推荐的），你可以这样做：

def combine(*args):
  # Create a list of dictionaries from the arrays we passed in, since we are
  # going to use dictionaries to solve the problem.
  dicts = [ dict(a) for a in args ]

  # Create a list of names by looping through all dictionaries, and through all
  # the names in each dictionary, adding to a master list of names
  names = []
  for d in dicts:
    for name in d.keys():
      names.append(name)

  # Remove duplicates in our list of names by making it a set
  names = set(names)

  # Create a result dict to store results in
  result = dict()

  # Loop through all the names, and add a row for each name, pulling data from
  # each dict we created in the beginning
  for name in names:
    result[name] = [ d.get(name, 0) for d in dicts ]

  # Return, secure in the knowledge of a job well done. :-)
  return result

# Use the function:
resultDict = combine(arrayOne, arrayTwo, arrayThree)

合并2d数组

2 个答案: