Question

我正在尝试在具有其他值的嵌套字典中舍入浮点值。我看着这个线程-Rounding decimals in nested data structures in Python

尝试实现以下功能：

import collections
import numbers

def formatfloat(x):
    return "%.3g" % float(x)

def pformat(dictionary, function):
    if isinstance(dictionary, dict):
        return type(dictionary)((key, pformat(value)) for key, value in dictionary.items())
    if isinstance(dictionary, collections.Container):
        return type(dictionary)(pformat(value) for value in dictionary)
    if isinstance(dictionary, numbers.Number):
        return formatfunc(dictionary)
    return dictionary

x={'a':[1.05600000001,2.34581736481,[8.1111111112,9.999990111111]], 
   'b':[3.05600000001,4.34581736481,[5.1111111112,6.999990111111]]}

pformat(x, formatfloat)

我收到此错误：

TypeError: pformat() missing 1 required positional argument: 'function'

答案被标记为正确，所以我想知道是否需要更改任何内容。这正是我要为字典实现的内容。任何帮助表示赞赏！

Answer 1

这里只是语法错误。

您在内部调用函数pformat时不需要任何参数。您仅传递 value 而没有传递 function

return type(dictionary)((key, pformat(value)) for key, value in dictionary.items())

在最后一个if语句中，您调用了未实现的功能formatfunc。也许是拼写错误，您需要调用formatfloat函数。

如果解决此问题，结果：

{'a': ['1.06', '2.35', ['8.11', '10']], 'b': ['3.06', '4.35', ['5.11', '7']]}

Answer 2

您需要在pformat之后添加一个参数：

之前

pformat(value))

之后：

pformat(value, function))

然后更改输入错误：将formatfunc更改为功能

这是工作代码：

import collections
import numbers

def formatfloat(x):
    return "%.3g" % float(x)

def pformat(dictionary, function):
    if isinstance(dictionary, dict):
        return type(dictionary)((key, pformat(value, function)) for key, value in dictionary.items())
    if isinstance(dictionary, collections.Container):
        return type(dictionary)(pformat(value, function) for value in dictionary)
    if isinstance(dictionary, numbers.Number):
        return function(dictionary)
    return dictionary

x={'a':[1.05600000001,2.34581736481,[8.1111111112,9.999990111111]], 
   'b':[3.05600000001,4.34581736481,[5.1111111112,6.999990111111]]}

pformat(x, formatfloat)

结果：

{'a': ['1.06', '2.35', ['8.11', '10']], 'b': ['3.06', '4.35', ['5.11', '7']]}

Answer 3

我写的有点不同：

x={'a':[1.05600000001,2.34581736481,[8.1111111112,9.999990111111]], 
   'b':[3.05600000001,4.34581736481,[5.1111111112,6.999990111111]]}


def recursive_rounding(keys, values):
    to_return = {}
    for key, value in zip(keys, values):
        print key, value
        if isinstance(value, dict):
            rounded_value = recursive_rounding(value.keys(), value.values())
        elif isinstance(value, (tuple, list)):
            rounded_value = [round_by_type(x) for x in value]
        else:
            rounded_value = round_by_type(value)
        print key, value
        to_return[round_by_type(key)] = rounded_value
    return to_return

def round_by_type(to_round):
    if isinstance(to_round, (int, float)):
        return round(to_round, 2)
    elif isinstance(to_round, (list, tuple)):
        return [round_by_type(x) for x in to_round]
    return to_round

recursive_rounding(x.keys(), x.values())

我超级快地编写了它，可以对其进行一些清理和改进，但是只是为了向您展示如何从不同的角度来处理它

输出是：

# Result: {'a': [1.06, 2.35, [8.11, 10.0]], 'b': [3.06, 4.35, [5.11, 7.0]]} #

圆形递归地嵌套在嵌套字典中

3 个答案: