我有一个这样的字典：
{"level": [1, 2, 3],
"conf": [-1, 1, 2],
"text": ["here", "hel", "llo"]}
我希望筛选列表以删除每个索引处的项目i，其中值"conf"中的索引不是>0。
因此，对于上述dict，输出应为：
{"level": [2, 3],
"conf": [1, 2],
"text": ["hel", "llo"]}
由于conf的第一个值不大于0。
我尝试了类似这样的方法：
new_dict = {i: [a for a in j if a >= min_conf] for i, j in my_dict.items()}
但是这只适用于一个密钥。
尝试：
from operator import itemgetter
def filter_dictionary(d):
positive_indices = [i for i, item in enumerate(d['conf']) if item > 0]
f = itemgetter(*positive_indices)
return {k: list(f(v)) for k, v in d.items()}
d = {"level": [1, 2, 3], "conf": [-1, 1, 2], "text": ["-1", "hel", "llo"]}
print(filter_dictionary(d))
输出：
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
我先尝试看哪些 'conf' 的索引是正数，然后用 itemgetter 从字典内的值中选择这些索引。
更紧凑的版本 + 不使用临时列表，而是使用生成器表达式:
def filter_dictionary(d):
f = itemgetter(*(i for i, item in enumerate(d['conf']) if item > 0))
return {k: list(f(v)) for k, v in d.items()}
dct = {k: [x for i, x in enumerate(v) if d['conf'][i] > 0] for k, v in d.items()}
输出：
>>> dct
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
使用示例数据：
d = {"level":[1,2,3], "conf":[-1,1,2], "text":["here","hel","llo"]
我会使用以下方式来保存有效元素的索引（即大于0的元素）：
kept_keys = [i for i in range(len(my_dict['conf'])) if my_dict['conf'][i] > 0]
然后你可以过滤每个列表，检查列表中某个元素的索引是否包含在kept_keys中：
{k: list(map(lambda x: x[1], filter(lambda x: x[0] in kept_keys, enumerate(my_dict[k])))) for k in my_dict}
输出：
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
import pandas as pd
data = {"level":[1,2,3], "conf":[-1,1,2], "text":["here","hel","llo"]}
df = pd.DataFrame(data)
df = df.loc[df["conf"] > 0]
result = df.to_dict(orient="list")
输出：
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
data = pd.DataFrame({
"level":[1,2,3],
"conf":[-1,1,2],
"text":["here","hel","llo"],
})
result = data.loc[data["conf"] > 0]
输出：
level conf text
1 2 1 hel
2 3 2 llo
from typing import Dict, List, Any, Set
d = {"level":[1,2,3], "conf":[-1,1,2], "text":["-1", "hel", "llo"]}
# First, we create a set that stores the indices which should be kept.
# I chose a set instead of a list because it has a O(1) lookup time.
# We only want to keep the items on indices where the value in d["conf"] is greater than 0
filtered_indexes = {i for i, value in enumerate(d.get('conf', [])) if value > 0}
def filter_dictionary(d: Dict[str, List[Any]], filtered_indexes: Set[int]) -> Dict[str, List[Any]]:
filtered_dictionary = d.copy() # We'll return a modified copy of the original dictionary
for key, list_values in d.items():
# In the next line the actual filtering for each key/value pair takes place.
# The original lists get overwritten with the filtered lists.
filtered_dictionary[key] = [value for i, value in enumerate(list_values) if i in filtered_indexes]
return filtered_dictionary
print(filter_dictionary(d, filtered_indexes))
输出：
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
试试这个，简单易懂，特别适合初学者：
a_dict = {"level": [1, 2, 3, 4, 5, 8], "conf": [-1, 1, -1, -2], "text": ["-1", "hel", "llo", "ai", 0, 9]}
# iterate backwards over the list keeping the indexes
for index, item in reversed(list(enumerate(a_dict["conf"]))):
if item <= 0:
for lists in a_dict.values():
del lists[index]
print(a_dict)
输出：
{'level': [2, 5, 8], 'conf': [1], 'text': ['hel', 0, 9]}
a = {"level":[1,2,3,4], "conf": [-1,1,2,-1],"text": ["-1","hel","llo","test"]}
# inefficient solution
# for k, v in a.items():
# if k == "conf":
# start_search = 0
# to_delete = [] #it will store the index numbers of the conf that you want to delete(conf<0)
# for element in v:
# if element < 0:
# to_delete.append(v.index(element,start_search))
# start_search = v.index(element) + 1
#more efficient and elegant solution
to_delete = [i for i, element in enumerate(a["conf"]) if element < 0]
for position in list(reversed(to_delete)):
for k, v in a.items():
v.pop(position)
结果将会是
>>> a
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
很多好的答案。这里是另一种2次遍历的方法：
mydict = {"level": [1, 2, 3], "conf": [-1, 1, 2], 'text': ["-1", "hel", "llo"]}
for i, v in enumerate(mydict['conf']):
if v <= 0:
for key in mydict.keys():
mydict[key][i] = None
for key in mydict.keys():
mydict[key] = [v for v in mydict[key] if v is not None]
print(mydict)
输出：
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
dct = {"level":[1,2,3], "conf":[-1,1,2], "text":["here","hel","llo"]}
dct = {k: np.array(v) for k, v in d.items()}
dct = {k: v[a['conf'] > 0].tolist() for k, v in a.items()}
输出：
>>> dct
{'level': [2, 3], 'conf': [1, 2], 'text': ['hel', 'llo']}
my_dict = {"level":[1,2,3], "conf":[-1,1,2],'text':["-1","hel","llo"]}
def remove_corresponding_items(d, key):
keep_indexes = [idx for idx, value in enumerate(d[key]) if value>0]
for key, lst in d.items():
d[key] = [lst[idx] for idx in keep_indexes]
remove_corresponding_items(my_dict, 'conf')
print(my_dict)
按照要求输出