Python：当满足条件时，如何使用for循环删除和转换字典中某些项的形式

字典在上面，我想删除值为0的键，并将数组形式转换为float。想要的输出是：

d = {0: {0: {0: 0, 1: 0},
         1: {0: 0,
             2: np.array([ 0.02981036])}},
     9: {0: {0: 0, 1: 0},
         7: {0: 0,
             4: np.array([ 0.01502996]),
             9: 0},
         8: {0: 0,
             1: 0},
         9: {0: 0, 1: 0}}}

我尝试使用for循环删除0项，但始终得到keyrerror:0。我是python的新手，请帮我上船。谢谢

这里的是dict或dict of dict。使用json可能是个好主意。但这里有一个解决方案的草稿版本

另外，由于您试图在不同级别上迭代键时删除键，因此可能会遇到RuntimeError问题。为此，我们将需要深度复制dict，重复原始内容并从副本中删除

源代码

递归解决方案：

import numpy as np
import copy

d = {0: {0: {0: 0, 1: 0},
         1: {0: 0,
             2: np.array([ 0.02981036])}},
     9: {0: {0: 0, 1: 0},
         7: {0: 0,
             4: np.array([ 0.01502996]),
             9: 0},
         8: {0: 0,
             1: 0},
         9: {0: 0, 1: 0}}}

d2 = copy.deepcopy(d)
for k1,v1 in d.iteritems():    
    for k2,v2 in v1.iteritems():
        for k3,v3 in v2.iteritems():
            #print k1,k2,k3,v3
            if v3 == 0:
                d2[k1][k2].pop(k3)
            if type(v3) is np.ndarray:
                v3 = v3.tolist()
                d2[k1][k2][k3] = float(''.join(str(i) for i in v3))

        if len(d2[k1][k2]) == 0:
            d2[k1].pop(k2)

 >>> d2
{0: {1: {2: 0.02981036}}, 9: {7: {4: 0.01502996}}}

---

您可以尝试删除值为0的所有零：

import numpy

def clean(d):
    new = {}
    for k, v in d.items():
        if v == 0:
            pass
        elif isinstance(v, numpy.ndarray):
            new_element = float(v[0])
            new[k] = new_element
        else:
            new_element = clean(v)

            if new_element != {}:
                new[k] = new_element

    return new

输出：

import numpy as np
from copy import deepcopy

d = {0: {0: {0: 0, 1: 0},
         1: {0: 0,
             2: np.array([ 0.02981036])}},
     9: {0: {0: 0, 1: 0},
         7: {0: 0,
             4: np.array([ 0.01502996]),
             9: 0},
         8: {0: 0,
             1: 0},
         9: {0: 0, 1: 0}}}

deep_copy=deepcopy(d)
for key,value in d.items():
    for key1,value1 in value.items():
        for key2,value2 in value1.items():
            if value2==0:
                del deep_copy[key][key1][key2]






new_deep_copy=deepcopy(deep_copy)
for key,value in deep_copy.items():
    for key1,value1 in value.items():
        for key2,value2 in value1.items():
            if isinstance(value2,np.ndarray):
                new_deep_copy[key][key1][key2]=float(value2)


        if value1=={}:
            del new_deep_copy[key][key1]


print(new_deep_copy)

---

请向我们展示您得到的keyrorror的代码这是可行的，但它非常昂贵，您可以用递归替换这些循环注意，在Python3中dict.iteritems已被dict.items替换。我理解。OP.对时间/空间复杂性没有限制，我使用的是2.7，因此dict.iteritems。正如我提到的，这是一次抓挠/偷懒的尝试。

import numpy

def clean(d):
    new = {}
    for k, v in d.items():
        if v == 0:
            pass
        elif isinstance(v, numpy.ndarray):
            new_element = float(v[0])
            new[k] = new_element
        else:
            new_element = clean(v)

            if new_element != {}:
                new[k] = new_element

    return new

import numpy as np
from copy import deepcopy

d = {0: {0: {0: 0, 1: 0},
         1: {0: 0,
             2: np.array([ 0.02981036])}},
     9: {0: {0: 0, 1: 0},
         7: {0: 0,
             4: np.array([ 0.01502996]),
             9: 0},
         8: {0: 0,
             1: 0},
         9: {0: 0, 1: 0}}}

deep_copy=deepcopy(d)
for key,value in d.items():
    for key1,value1 in value.items():
        for key2,value2 in value1.items():
            if value2==0:
                del deep_copy[key][key1][key2]






new_deep_copy=deepcopy(deep_copy)
for key,value in deep_copy.items():
    for key1,value1 in value.items():
        for key2,value2 in value1.items():
            if isinstance(value2,np.ndarray):
                new_deep_copy[key][key1][key2]=float(value2)


        if value1=={}:
            del new_deep_copy[key][key1]


print(new_deep_copy)

{0: {1: {2: 36.0}}, 9: {7: {4: 0.01502996}}}