Question

我正在尝试打印一个字典（实际上是defaultdict），其中键是版本号（格式为6.0或6.1.2），值为浮点数。< / p>

# Parse file contents
firmware_percents = defaultdict(float)
with open('test.csv', 'r') as file:
    reader = csv.DictReader(file)

    for row in reader:
        # Don't care about iphone vs ipad
        # Must be a string - StrictVersion is apparently unhashable
        rev = row["Firmware Version"].split(" ")[2]

        # Dump extra spaces, the less than (assume .1%), and % sign
        percent = float(row["% of Sessions"].strip().lstrip("<").strip("%"))
        firmware_percents[rev] += percent

def pretty_print(d):
    for k in sorted(d, key=d.get, reverse=True):
        print("{0}: {1:.1f}".format(k, d[k]))

print("All versions:")
pretty_print(firmware_percents)

但是，当我这样做时，某些版本会无序打印：

All versions:
7.0: 44.2
7.0.2: 25.7
6.1.3: 14.2
6.1.4: 5.5
7.0.1: 3.2
6.1: 2.7
# You get the point

使用此输入文件：

Firmware Version,Sessions,% of Sessions
" iPhone 5.0.1 ",20," <0.1% "
" iPhone 6.0 ",26," 0.1% "
" iPhone 5.1.1 ",69," 0.3% "
" iPhone 5.1 ",2," <0.1% "
" iPhone 7.0 ",7401," 31.5% "
" iPhone 6.1 ",337," 1.4% "
" iPhone 6.1.3 ",2193," 9.3% "
" iPhone 6.1.2 ",84," 0.4% "
" iPhone 7.0.1 ",747," 3.2% "
" iPhone 7.0.2 ",4619," 19.7% "
" iPhone 6.0.1 ",37," 0.2% "
" iPhone 6.0.2 ",1," <0.1% "
" iPhone 6.1.4 ",1281," 5.5% "
" iPad 5.0 ",4," <0.1% "
" iPad 5.1 ",100," 0.4% "
" iPad 5.1.1 ",545," 2.3% "
" iPad 6.0 ",16," <0.1% "
" iPad 6.1 ",305," 1.3% "
" iPhone 7.0.3 ",1," <0.1% "
" iPhone 6.1.1 ",1," <0.1% "
" iPad 7.0 ",2979," 12.7% "
" iPad 6.0.1 ",100," 0.4% "
" iPad 6.1.3 ",1139," 4.9% "
" iPad 6.0.2 ",5," <0.1% "
" iPad 6.1.2 ",65," 0.3% "
" iPad 7.0.2 ",1404," 6.0% "

我已经尝试了pprint，虽然它按照正确的顺序排序，但它不会格式化浮点数（所以我最终会得到像14.5999999996这样的数字）。当我尝试仅使用主要版本时，它偶尔会打印像defaultdict(<class 'float'>, {'5': 3.3, '6': 24.2, '7': 73.2})这样的陌生感。

如何确保以格式化百分比的版本顺序打印这些内容？

按顺序我的意思是按主要分类然后构建/超级小调（7.0.2＆gt; 7.0.1＆gt; 6.1.4＆gt; 6.1等等）？

Answer 1

我认为@dornhege首先发现了主要问题：key=d.get意味着你要按值排序，而不是键。

我会利用stdlib包distutils来自动处理版本内容：

>>> sorted(firmware_percents, key=distutils.version.StrictVersion, reverse=True)
['7.0.3', '7.0.2', '7.0.1', '7.0', '6.1.4', '6.1.3', '6.1.2', '6.1.1', '6.1', '6.0.2', '6.0.1', '6.0', '5.1.1', '5.1', '5.0.1', '5.0']
>>> sorted(firmware_percents, key=distutils.version.LooseVersion, reverse=True)
['7.0.3', '7.0.2', '7.0.1', '7.0', '6.1.4', '6.1.3', '6.1.2', '6.1.1', '6.1', '6.0.2', '6.0.1', '6.0', '5.1.1', '5.1', '5.0.1', '5.0']

这比.拆分更有优势，因为处理7.0.2rc1之类的事情会更聪明。

Answer 2

将版本号转换为整数列表以进行排序：

def pretty_print(d):
    for k in sorted(d, key=lambda x: x.split("."), reverse=True):
        print("{0}: {1:.1f}".format(k, d[k]))

但使用operator.methodcaller可能会更有效率。

from operator import methodcaller
def pretty_print(d):
    for k in sorted(d, key=methodcaller('split', '.'), reverse=True):
        print("{0}: {1:.1f}".format(k, d[k]))

Answer 3

在Python中，字典不可排序 - 而使用字典“get”方法作为排序键的食谱是按字典值排序 - 而不是按键排序。

所以，当你致电for k in sorted(d, key=d.get, reverse=True)时会发生什么是字典中的每个键都被传递给dictionairy的“get”方法 - 而不是检索值。

如果您想按键排序，则根本不需要将任何值传递给“key”。做就是了： for k in sorted(d, reverse=True):

然而，像“10.1”这样的版本将首先出现在“2.0”之上，因为它是一个字符串比较 - 将版本号拆分为“。”并将每个部分转换为数字将产生正确的比较：

for k in sorted(d, key=lambda x: tuple(int(n) for n in x.split(".")) , reverse=True):

当key是版本时，按键对dict进行排序

3 个答案: