使用字典方法搜索字符串格式中包含的值 - Python答案

【问题标题】：Search for value contained in a string format using dictionary method - Python使用字典方法搜索字符串格式中包含的值 - Python
【发布时间】：2021-11-01 15:52:13
【问题描述】：

我已经为我编写了一个函数来从包含在我的数据框中的列中的“值”（在我的字典中定义）中提取“键”。这个问题最好在下面的代码中解释。

我的问题是我的输出列只返回搜索到的最后一个“键”。我想归还所有的钥匙。

不胜感激！

df = pd.DataFrame({'chat':['swimming is great for you! volleyball is fun too.',
                       'i like to cycle and run during my free time',
                      'apart from netball, i am good at basketball and diving too']})

sport_dic = {'team': ['basketball', 'soccer','volley','netball','hockey'],
 'individual': ['run','swim','cycl','div']}

我的功能：

def type_search(string,dic):
    for key,value in dic.items():
        return_x = []
        for v in value:
            if v in string:
                return_x.append(key)
                return(return_x)

df['chat_type'] = df['chat'].apply(lambda x: type_search(x,sport_dic))

我从我的函数中得到了什么：

df = {'chat': {0: 'swimming is great for you! volleyball is fun too.',
  1: 'i like to cycle and run during my free time',
  2: 'apart from netball, i am good at basketball and diving too'},

 'chat_type': {0: ['team'], 1: ['individual'], 2: ['team']}}

我的期望：

df = {'chat': {0: 'swimming is great for you! volleyball is fun too.',
  1: 'i like to cycle and run during my free time',
  2: 'apart from netball, i am good at basketball and diving too'},

 'chat_type': {0: ['individual','team'], 1: ['individual','individual'], 2: ['team','team','diving']}}

【问题讨论】：

标签： python pandas string dataframe function

【解决方案1】：

试试这个：

df = pd.DataFrame({'chat':['swimming is great for you! volleyball is fun too.',
                       'i like to cycle and run during my free time',
                      'apart from netball, i am good at basketball and diving too']})

sport_dic = {'team': ['basketball', 'soccer','volley','netball','hockey'],'individual': ['run','swim','cycl','div']}


def type_search(string,dic):
    return_x = []
    for key,value in dic.items():
        for v in value:
            if v in string:
                return_x.append(key)
    return(return_x)

df['chat_type'] = df['chat'].apply(lambda x: type_search(x,sport_dic))

df['chat_type']

输出：

0          [team, individual]
1    [individual, individual]
2    [team, team, individual]
Name: chat_type, dtype: object

如果您想获得预期的输出，请使用此行并获得您想要的输出：

{i: df['chat_type'].tolist()[i] for i in range(len(df['chat_type'].tolist()))}

希望输出：

{0: ['team', 'individual'],
 1: ['individual', 'individual'],
 2: ['team', 'team', 'individual']}

【讨论】：