Question

我尝试将此处的名称键值组合为dict值的键，并将源值计为所述父键的键，并将计数值作为其值。

override func viewWillDisappear(_ animated: Bool) {
        super.viewWillDisappear(animated)
        if isMovingFromParentViewController {
            if let viewControllers = self.navigationController?.viewControllers {
                if (viewControllers.count >= 1) {
                    let previousViewController = viewControllers[viewControllers.count-1] as! NameOfDestinationViewController
                    // whatever you want to do
                    previousViewController.callOrModifySomething()
                }
            }
        }
    }

如何实现以下输出？

data = [ {'name':'Gill', 'source':'foo'}, {'name':'Gill', 'source':'foo'}, {'name':'Gill', 'source':'foo'}, {'name':'Gill', 'source':'bar'}, {'name':'Gill', 'source':'bar'}, {'name':'Gill', 'source':'bar'}, {'name':'Gill', 'source':'bar'}, {'name':'Gill', 'source':'bar'}, {'name':'Dave', 'source':'foo'}, {'name':'Dave', 'source':'foo'}, {'name':'Dave', 'source':'foo'}, {'name':'Dave', 'source':'foo'}, {'name':'Dave', 'source':'egg'}, {'name':'Dave', 'source':'egg'}, {'name':'Dave', 'source':'egg'}, {'name':'Dave', 'source':'egg'}, {'name':'Dave', 'source':'egg'}, {'name':'Dave', 'source':'egg'}, {'name':'Dave', 'source':'egg'} ]

我认为有可能使用1班轮......

Answer 1

按名称使用itertools.groupby 组，然后collections.Counter 计算属于每个名称的来源类别：

from collections import Counter
from itertools import groupby

f = lambda x: x['name']
dct = {k: Counter(d['source'] for d in g) for k, g in groupby(data, f)}
print(dct)
# {'Gill': Counter({'bar': 5, 'foo': 3}), 'Dave': Counter({'egg': 7, 'foo': 4})}

Answer 2

这显然不是一个单行，但简单而且非常直接。适用于任何数量的值。

results = {}
key = 'name'
for line in data:
    tracked_key = line[key]
    results.setdefault(tracked_key, {})
    for k, v in line.iteritems():
        if k == key:
            continue
        results[tracked_key].setdefault(v, 0)
        results[tracked_key][v] += 1

Python - 按键值分组dicts列表，将单独的键值计为dict？

2 个答案: