Python：将函数应用于tem列表

Question

我有以下代码片段可以帮助我获取googletrends数据（请参见https://github.com/GeneralMills/pytrends）：

trend_payload = {'q': 'Dogs, Cats, Catfood, Dogfood','date': '01/2015 12m'}
trend = pytrend.trend(trend_payload)
df = pytrend.trend(trend_payload, return_type='dataframe')
df

由于这个查询的缺点是googletrends基于查询的数据对所有数据进行规范化，所以我更喜欢将每个查询都作为单个调用，并将df链接到彼此旁边。我想到了这样一个函数：

queries = ['Cats', 'Dogs', 'Catfood','Dogfood']

function(queries)
    trend_payload = {'q': queries, 'date': '01/2015 12m'}
    trend = pytrend.trend(trend_payload)
    df = pytrend.trend(trend_payload, return_type='dataframe')

# then put every df of each query next to each other

我该怎么做？你知道吗

Answer 1

我只需将DFs连接为jimifiki has already proposed：

df = pd.concat([pytrend.trend({'q': x, 'date': '01/2015 12m'},
                              return_type='dataframe')
                for x in queries], axis=1)

或在功能上：

def get_trends(queries, dt):
    return pd.concat([pytrend.trend({'q': x, 'date': dt},
                                    return_type='dataframe')
                      for x in queries], axis=1)

df = get_trends(queries, '01/2015 12m')

演示：

In [24]: df = get_trends(queries, '01/2015 12m')

In [25]: df
Out[25]:
            cats   dogs  catfood  dogfood
Date
2015-01-04  74.0   85.0     65.0     47.0
2015-01-11  74.0   84.0     60.0     52.0
2015-01-18  72.0   82.0     49.0     57.0
2015-01-25  69.0   78.0     45.0     37.0
2015-02-01  73.0   77.0     51.0     52.0
...          ...    ...      ...      ...
2015-11-29  83.0   80.0     47.0     49.0
2015-12-06  80.0   79.0     70.0     50.0
2015-12-13  83.0   84.0     67.0     49.0
2015-12-20  89.0   91.0     61.0     58.0
2015-12-27  90.0  100.0     58.0     45.0

[52 rows x 4 columns]

Answer 2

你可以这样做：

queries = ['Cats', 'Dogs', 'Catfood','Dogfood']

def function(queries):
    trend_payload = {'q': queries, 'date': '01/2015 12m'}
    trend = pytrend.trend(trend_payload)
    df = pytrend.trend(trend_payload, return_type='dataframe')
    return df 

list_of_df = [function([query,]) for query in queries] 

然后必须concat列表中的数据帧。你知道吗

您可以更优雅地调用：

list_of_df = map(function, queries)

在这种情况下，您应该重写function，以便它接受单个项。 如果不想修改function，可以这样写：

list_of_df = map(lambda x: function([x,]), queries)