如何高效遍历字典？

Question

我是Python和编程新手。在

我的课本上说我必须做以下习题：

Create a second purchase summary that which accumulates total investment by ticker symbol. In the above sample data, there are two blocks of CAT.

These can easily be combined by creating a dict where the key is the ticker and the value is the list of blocks purchased. The program makes one pass through the data to create the dict. A pass through the dict can then create a report showing each ticker symbol and all blocks of stock.

除了硬编码之外，我想不出一种方法来添加“猫”股票的两个条目。在

## Stock Reports

stockDict = {"GM":"General Motors", "CAT":"Caterpillar", "EK":"Eastman Kodak",
             "FB":"Facebook"}
# symbol,prices,dates,shares
purchases = [("GM",100,"10-sep-2001",48), ("CAT",100,"01-apr-1999",24),
             ("FB",200,"01-jul-2013",56), ("CAT", 200,"02-may-1999",53)]

# purchase history:
print "Company", "\t\tPrice", "\tDate\n"
for stock in purchases:
    price = stock[1] * stock[3]
    name = stockDict[stock[0]]
    print name, "\t\t", price, "\t", stock[2]
print "\n"

# THIS IS THE PROBLEM SET I NEED HELP WITH:
# accumulate total investment by ticker symbol
byTicker = {}
# create dict
for stock in purchases:
    ticker = stock[0]
    block = [stock]
    if ticker in byTicker:
        byTicker[ticker] += block
    else:
        byTicker[ticker] = block

for i in byTicker.values():
    shares = i[0][3]
    price = i[0][1]
    investment = shares * price
    print investment

现在，输出是：

^{pr2}$

这不好，因为它没有计算这两只猫的存量。现在它只计算一个。代码应该足够灵活，我可以添加更多的猫股票。在

Answer 1

也许是这样的：

## Stock Reports

stockDict = {"GM":"General Motors", "CAT":"Caterpillar", "EK":"Eastman Kodak",
             "FB":"Facebook"}
# symbol,prices,dates,shares
purchases = [("GM",100,"10-sep-2001",48), ("CAT",100,"01-apr-1999",24),
             ("FB",200,"01-jul-2013",56), ("CAT", 200,"02-may-1999",53)]

# purchase history:
print "Company", "\t\tPrice", "\tDate\n"
for stock in purchases:
    price = stock[1] * stock[3]
    name = stockDict[stock[0]]
    print name, "\t\t", price, "\t", stock[2]
print "\n"

# THIS IS THE PROBLEM SET I NEED HELP WITH:
# accumulate total investment by ticker symbol
byTicker = {}
# create dict
for stock in purchases:
    ticker = stock[0]
    price = stock[1] * stock[3]
    if ticker in byTicker:
        byTicker[ticker] += price
    else:
        byTicker[ticker] = price

for ticker, price in byTicker.iteritems():
    print ticker, price

我得到的输出是：

^{pr2}$

这似乎是正确的。在

测试股票代码是否在byTickerdict中，可以告诉您是否已经记录了该股票的购买记录。如果有的话，你就加进去，如果没有，你就重新开始。这基本上就是你在做的，除了出于某种原因你收集了那张dict中给定股票的所有购买记录，当时你真正关心的只是购买的价格。在

您可以按照原来的方式构建dict，然后迭代存储在每个键下的项，并将它们相加。像这样：

totals = []
for ticker in byTicker:
    total = 0
    for purchase in byTicker[ticker]:
        total += purchase[1] * purchase[3]
    totals.append((ticker, total))

for ticker, total in totals:
    print ticker, total

对于kicks，您可以使用generator语句将其压缩为一行：

 print "\n".join("%s: %d" % (ticker, sum(purchase[1]*purchase[3] for purchase in byTicker[ticker])) for ticker in byTicker)

不过，后两种方法中的任何一种都是完全不需要做的，因为您已经在遍历每一次购买，所以您也可以像我在第一个示例中所展示的那样，累积每只股票的总价格。在

Answer 2

代码的问题在于，您没有迭代purchaes，而只是从每个ticker值中获取第一个元素。也就是说，byTicker看起来像：

byTicker: {
  "GM": [("GM",100,"10-sep-2001",48)],
  "CAT": [("CAT",100,"01-apr-1999",24), ("CAT", 200,"02-may-1999",53)],
  "FB": [("FB",200,"01-jul-2013",56)]
}

所以当你迭代这些值时，你实际上得到了三个列表。但当您处理这些列表时，实际上您只访问其中的第一个列表：

^{pr2}$

对应于“CAT”的值，i[0]为（“CAT”，100，“01-apr-1999”，24）。你也应该调查我！考虑迭代不同的购买：

for company, purchases in byTicker.items():
  investment = 0
  for purchase in purchases:
      investment += purchase[1] * purchase[3]
  print(company, investment)

Answer 3

您的问题是在代码的最后一部分，倒数第二位创建了针对每个股票代码的所有股票的列表，这很好：

for i in byTicker.values():
    shares = i[0][3]
    price = i[0][1]
    investment = shares * price
    print investment

这里的股票只供你使用。相反，请尝试：

^{pr2}$

这将把每个股票代码的所有股票加起来，对于你的例子，我给出了：

CAT 13000
FB 11200
GM 4800