Question

我的代码在下面

将pd.to_numeric应用于应该为int或float但作为对象出现的列。我们可以像应用np.where

一样将更多内容转换为熊猫吗？

if df.dtypes.all() == 'object':
    df=df.apply(pd.to_numeric,errors='coerce').fillna(df)
else:
    df = df

Answer 1

一个简单的衬里是{ "message": "candidate job listing", "status": true, "data": [ { "id": 1, "company_name_shown_job": "xyz company", "job_title": "Android Developer", "job_description": null, "experience_tags": [ { "id": 2, "employer_job_post_id": 1, "experience_in": "Android" }, { "id": 3, "employer_job_post_id": 1, "experience_in": "Php" } ] }, { "id": 2, "company_name_shown_job": "Abc company", "job_title": "Web Developer", "job_description": "<p>Lorem ipsum dolor sit amet,laborum.</p>", "experience_tags": [ { "id": 4, "employer_job_post_id": 2, "experience_in": "Swift" }, { "id": 5, "employer_job_post_id": 2, "experience_in": "Java" } ] } ]和assign，它将重新分配现有的列

selest_dtypes

df.assign(**df.select_dtypes('O').apply(pd.to_numeric,errors='coerce').fillna(df))：

np.where

示例（选中df[:] = (np.where(df.dtypes=='object', df.apply(pd.to_numeric,errors='coerce').fillna(df),df)列）：

Price

d = {'CusID': {0: 1, 1: 2, 2: 3},
     'Name': {0: 'Paul', 1: 'Mark', 2: 'Bill'},
     'Shop': {0: 'Pascal', 1: 'Casio', 2: 'Nike'},
     'Price': {0: '24000', 1: 'a', 2: '900'}}
df = pd.DataFrame(d)

print(df)
   CusID  Name    Shop  Price
0      1  Paul  Pascal  24000
1      2  Mark   Casio      a
2      3  Bill    Nike    900

df.to_dict()
{'CusID': {0: 1, 1: 2, 2: 3},
 'Name': {0: 'Paul', 1: 'Mark', 2: 'Bill'},
 'Shop': {0: 'Pascal', 1: 'Casio', 2: 'Nike'},
 'Price': {0: '24000', 1: 'a', 2: '900'}}

Answer 2

您的if / else等效值为df.mask

df_out = df.mask(df.dtypes =='O', df.apply(pd.to_numeric, errors='coerce')
                                    .fillna(df))

如何在熊猫中将if / else转换为np.where

2 个答案: