Question

输入：

dict1 = {'Category': ['item1','item2','item3','item4'],
 'Freq': [71984.0, 8129.0, 3140.0, 0.0]}

输出：

dict2 = {'Category': ['item1','item2','item3'],
 'Freq': [71984.0, 8129.0, 3140.0]}

要执行的任务：
制作一个新字典，其中不包含所有0.0频率值和类别中的各个项目。

Answer 1

这是使用* Try: Run with --stacktrace option to get the stack trace. Run with --info or --debug option to get more log output. Run with --scan to get full insights. * Get more help at https://help.gradle.org BUILD FAILED in 1m 59s Command: C:\Users\IB\Desktop\Android\myapp\berry_networks\Dart test\Scaffold\intro_to_scaffold\android\gradlew.bat app:properties Please review your Gradle project setup in the android/ folder. Exited (sigterm)`

的一种方法

例如：

zip

输出：

dict1 = {'Category': ['item1','item2','item3','item4'],'Freq': [71984.0, 8129.0, 3140.0, 0.0]}
Category, Freq = zip(*[(m,n) for m,n in zip(*dict1.values()) if n])
result = {"Category":Category, "Freq":Freq}

print(result)

Answer 2

如果可以使用pandas，则可以执行以下操作：

首先通过将dict1传递到DataFrame构造函数中来创建一个熊猫DataFrame：

import pandas as pd
df = pd.DataFrame(dict1)
print(df)
#  Category     Freq
#0    item1  71984.0
#1    item2   8129.0
#2    item3   3140.0
#3    item4      0.0

现在仅保留Freq!=0中的行：

print(df[df['Freq']!=0])
#  Category     Freq
#0    item1  71984.0
#1    item2   8129.0
#2    item3   3140.0

要将其重新放入字典中：

dict2 = df[df['Freq']!=0].to_dict(orient='list')
print(dict2)
#{'Category': ['item1', 'item2', 'item3'], 'Freq': [71984.0, 8129.0, 3140.0]}

使用此方法的优点是可以轻松应用于两个以上的列表。

Answer 3

下面的代码可以解决问题。

dict1 = {'Category': ['item1','item2','item3','item4'],
 'Freq': [71984.0, 8129.0, 3140.0, 0.0]}

zipped = zip(*dict1.values())
filtered = filter(lambda x : x[1] != 0.0, zipped)
result = map(list, zip(*filtered))
print {
    'Category': result[0],
    'Freq': result[1]
}

Answer 4

Rakesh答案的版本稍干净

frequency_key = 'Freq'
category_key = 'Category'
freq,category = zip(*[[(i),(j)] for i,j in zip(dict1.get(frequency_key),dict1.get(category_key)) if i!=0])
{category_key:list(category),frequency_key:list(freq)}

输出

{'Category': ['item1', 'item2', 'item3'], 'Freq': [71984.0, 8129.0, 3140.0]}

Answer 5

dict1 = {
    'Category': ['item1','item2','item3','item4'],
    'Freq': [71984.0, 8129.0, 3140.0, 0.0]
}

indices_to_delete = []
for i, e in enumerate(dict1['Freq']):
    if e == 0:
        indices_to_delete.append(i)

dict2 = {
    key: [
        v
        for i, v in enumerate(dict1[key])
        if i not in indices_to_delete
    ]
    for key in dict1.keys()
}

dict2

返回：

{'Category': ['item1', 'item2', 'item3'], 'Freq': [71984.0, 8129.0, 3140.0]}

Answer 6

也许有一种更优雅的解决方法，但这确实可以解决问题：

<key>GADApplicationIdentifier</key>
<string>APP_KEY</string>

输出：

dict1 = {'Category': ['item1','item2','item3','item4'],
 'Freq': [71984.0, 8129.0, 3140.0, 0.0]}

# create an empty list to save indices that you want to delete from both lists
del_ind = []

# iterate over the list of Frequencies
for x in dict1['Freq']:
  # if the value euquals 0.0, append the index of this value to your list
  if x == 0.0:
    ind = dict1['Freq'].index(x)
    del_ind.append(ind)

# then iterate over the list of indices you want to delete and use the pop method to delete those entries from both of your lists
for x in del_ind:
  dict1['Category'].pop(x)
  dict1['Freq'].pop(x)

print(dict1)

需要根据python中的条件修改Dictionary及其元素

6 个答案: