[英]Extracting values of a dictionary within a list
據我了解,我在列表中有以下字典:
[{'week': 3, 'timing': '07:30'}, {'week': 4, 'timing': '20:30'},{},....]
我想提取時間和星期並將它們放在單獨的列表中。 但是,由於外面有一個列表,因此無法識別鑰匙。 我試圖做的是以下內容:
for item in list:
new_list =list( item.values() )[0]
但它顯示了一個錯誤,因為我認為其中一些是空白的。 目前,我收到錯誤:IndexError: list index out of range。 我怎樣才能將它們提取到兩個單獨的列表中,並在它們空白的地方有一個 NaN?
以下代碼將有助於分別提取周和時間值
check = [{'week': 3, 'timing': '07:30'}, {'week': 4, 'timing': '20:30'}]
week_list = []
timing_list=[]
for i in check:
for k,v in i.items():
if k == 'week':
week_list.append(v)
else:
timing_list.append(v)
Output
```[3, 4]
['07:30', '20:30']
您首先創建兩個列表並附加項目。 迭代應該在列表中的每個對象上完成,每個對象都是字典本身。 假設包含字典的列表稱為full_list
,因為將其命名為list
對 python 來說不太方便。
week_list = []
timing_list = []
for i in full_list:
week_list.append(i['week'])
timing_list.append(i['timing'])
例如,如果有空值的字典、空字典或只有week
但沒有timing
字典,我喜歡使用:
import numpy as np
for i in full_list:
try:
week_list.append(i['week'])
except KeyError:
week_list.append(np.nan)
try:
timing_list.append(i['timing'])
except KeyError:
timing_list.append(np.nan)
這樣,只要鍵不存在,您就會將NaN
值附加到您的列表中,如果您以后對列表執行操作,這將很有幫助。
完整示例:
full_list = [{'week': 3, 'timing': '07:30'}, {'week': 4, 'timing': '20:30'},{},{'week':4},{'timing':'09:21'},{}]
week_list = []
timing_list = []
import numpy as np
for i in full_list:
try:
week_list.append(i['week'])
except KeyError:
week_list.append(np.nan)
try:
timing_list.append(i['timing'])
except KeyError:
timing_list.append(np.nan)
print(week_list)
print(timing_list)
輸出:
[3, 4, nan, 4, nan, nan]
['07:30', '20:30', nan, nan, '09:21', nan]
如果字典是這樣的:
full_dict = {'person_1':{'week': 3, 'timing': '07:30'},'person_2':{'week': 4, 'timing': '20:30'},'person_3':{}}
每個key
都應該進行迭代。 因此循環將是:
for i in full_dict.keys():
and exactly the same code as before
我使用此代碼分別提取它們。
weeks = [ a['week'] for a in new_list ]
timings = [ a['timing'] for a in new_list ]
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