python list append方法

標籤：

keyValueResult = {‘a‘: 1, ‘b‘: 2}sendData = []def set_push_format(ip):    data_format = {        "endpoint": "test-endpoint",        "metric": "test-metric",        "timestamp": 54,        "step": 60,        "value": 1,        "counterType": "GAUGE",        "tags": "",    }    # print keyValueResult    for key_value in keyValueResult:        data_format = data_format.copy()          # print key_value        data_format[‘endpoint‘] = ip        data_format[‘metric‘] = key_value        data_format[‘value‘] = keyValueResult.get(key_value)        print ‘data_format:‘ + key_value        print data_format        # 字典賦值給列表，構建JSON檔案格式        sendData.append(data_format)        print ‘sendData:‘ + key_value        print sendDataif __name__ == "__main__":    set_push_format(‘192.168.137.10‘)    print ‘final‘    print sendData

　　該句必須加上，不然append的全是同一個字典！

 

別人遇到的類似問題

問題：將資料庫中查出的資料（列表中包含元組）轉換為列表中字典。原資料結構，從資料庫查出：

cur = [("t1", "d1"), ("t2", "d2")]

轉換後資料結構：

[{‘description‘: ‘d1‘, ‘title‘: ‘t1‘}, {‘description‘: ‘d2‘, ‘title‘: ‘t2‘}]

方法一，使用append, 出現錯誤結果

cur = [("t1", "d1"), ("t2", "d2")] post_dict = {}posts = []for row in cur:    post_dict[‘title‘] = row[0]    post_dict[‘description‘] = row[1]    print "post_dict:",post_dict    posts.append(post_dict)    print "posts:",posts

方法一運行結果：

post_dict: {‘description‘: ‘d1‘, ‘title‘: ‘t1‘}posts: [{‘description‘: ‘d1‘, ‘title‘: ‘t1‘}]post_dict: {‘description‘: ‘d2‘, ‘title‘: ‘t2‘}posts: [{‘description‘: ‘d2‘, ‘title‘: ‘t2‘}, {‘description‘: ‘d2‘, ‘title‘: ‘t2‘}]

方法二，使用列表解析,結果正常

cur = [("a", "a1"), ("b", "b1")] posts = []posts = [dict(title=row[0], description=row[1]) for row in cur]print "posts:",posts

方法二運行結果，正常

posts: [{‘description‘: ‘d1‘, ‘title‘: ‘t1‘}, {‘description‘: ‘d2‘, ‘title‘: ‘t2‘}]

採納

方法一中，你的post_dict是一個字典對象，for迴圈的操作都是在更新這個對象的key和value，自始至終就這一個對象，append多少次都一樣。

把字典對象放在迴圈內建立即可：

cur = [("t1", "d1"), ("t2", "d2")] posts = []for row in cur:    post_dict = {}    post_dict[‘title‘] = row[0]    post_dict[‘description‘] = row[1]    print "post_dict:",post_dict    posts.append(post_dict)    print "posts:",posts

優先考慮列表解析，另，本例的tupel列表可以用迴圈解包，大致如下：

In [1]: cur = [("t1", "d1"), ("t2", "d2")]In [2]: r = [{‘description‘: description, ‘title‘: title} for description, title in cur]In [3]: rOut[3]: [{‘description‘: ‘t1‘, ‘title‘: ‘d1‘}, {‘description‘: ‘t2‘, ‘title‘: ‘d2‘}

方法一的迴圈中，post_dict始終指向的是同一個對象。 在for迴圈中，使用匿名對象就可以了：

 

for row in cur:    posts.append({‘title‘:row[0],‘description‘:row[1]})

python list append方法