无法将字符串转换为浮点Python破折号

Question

我试图在短跑回调中将字符串转换为浮点值，但当我运行代码时，我在短跑应用程序中遇到错误：lati=float（lati[-1]） ValueError:无法将字符串转换为float:'float64）但我在终端中没有收到此错误

首先，我需要做的是提取给定的纬度（和经度）数字。因此，我需要它将其转换为字符串并拆分它，因为我无法找到更好的方法使用pandas从csv文件中获取此数字

输出：

# converting to string:
12    41.6796
Name: latitude, dtype: float64

# splitting:
['12', '', '', '', '41.6796']

# converting to float:
41.6796

这是实际代码：

@app.callback(Output('text-output', 'children'),
    [Input('submit-val', 'n_clicks')],
    [State('search-input', 'value')])
def updateText(n_clicks, searchVar):
    df = pd.read_csv("powerplant.csv")
    df = df[df.name == searchVar]

    # converting to string
    lati = str(df['latitude'])
    longi = str(df['longitude'])

    # splitting it
    lati = lati.split('\n', 1)
    lati = lati[0].split(' ', 4)
    longi = longi.split('\n', 1)
    longi = longi[0].split(' ', 4)

    #converting to float
    lati = float(lati[-1])
    longi = float(longi[-1]) 

实际上，我在其他脚本中测试了这段代码，效果很好。有没有更好的方法来提取纬度和经度数字

这些数据可以从https://datasets.wri.org/dataset/globalpowerplantdatabase;下载，这里是一个摘录

country,country_long,name,gppd_idnr,capacity_mw,latitude,longitude,primary_fuel,other_fuel1,other_fuel2,other_fuel3,commissioning_year,owner,source,url,geolocation_source,wepp_id,year_of_capacity_data,generation_gwh_2013,generation_gwh_2014,generation_gwh_2015,generation_gwh_2016,generation_gwh_2017,estimated_generation_gwh
AFG,Afghanistan,Kajaki Hydroelectric Power Plant Afghanistan,GEODB0040538,33.0,32.3220,65.1190,Hydro,,,,,,GEODB,http://globalenergyobservatory.org,GEODB,1009793,2017,,,,,,
AFG,Afghanistan,Mahipar Hydroelectric Power Plant Afghanistan,GEODB0040541,66.0,34.5560,69.4787,Hydro,,,,,,GEODB,http://globalenergyobservatory.org,GEODB,1009795,2017,,,,,,
ALB,Albania,Shkopet,WRI1002173,24.0,41.6796,19.8305,Hydro,,,,1963.0,,Energy Charter Secretariat,http://www.energycharter.org/fileadmin/DocumentsMedia/IDEER/IDEER-Albania_2013_en.pdf,GEODB,1021238,,,,,,,79.22851153039832
ALB,Albania,Ulez,WRI1002174,25.0,41.6796,19.8936,Hydro,,,,1958.0,,Energy Charter Secretariat,http://www.energycharter.org/fileadmin/DocumentsMedia/IDEER/IDEER-Albania_2013_en.pdf,GEODB,1021241,,,,,,,82.52969951083159

Answer 1

您看到的是一个pandas.Series对象，它包含一行数据，您试图分割它的__repr__以获取值。没有这个必要。我不熟悉plotly的Python版本，但我看到您有一个回调，因此我将其包装到一个函数中（我不确定是否存在找不到名称的情况）：

import pandas as pd

def get_by_name(name):
    df = pd.read_csv('powerplants.csv')
    df = df[df['name'] == name]
    if not df.empty:
        return df[['latitude', 'longitude']].values.tolist()[0]
    return None, None


lat, lon = get_by_name('Kajaki Hydroelectric Power Plant Afghanistan')

Answer 2

问题在于访问数据帧中的值的方式。Pandas允许您访问数据，而无需解析字符串表示

您可以在对.loc的一次调用中访问行和列 如果您知道将有一个值，那么可以调用squeeze方法

>>> import pandas as pd
>>> from io import StringIO
>>> # data shortened for brievity
>>> df = pd.read_csv(StringIO("""country,country_long,name,gppd_idnr,capacity_mw,latitude,longitude
... AFG,Afghanistan,Kajaki Hydroelectric Power Plant Afghanistan,GEODB0040538,33.0,32.3220,65.1190
... AFG,Afghanistan,Mahipar Hydroelectric Power Plant Afghanistan,GEODB0040541,66.0,34.5560,69.4787
... ALB,Albania,Shkopet,WRI1002173,24.0,41.6796,19.8305
... ALB,Albania,Ulez,WRI1002174,25.0,41.6796,19.8936"""))
>>> searchVar = "Ulez"
>>> df.loc[df["name"] == searchVar, "latitude"] # here you have a pd.Series
3    41.6796
Name: latitude, dtype: float64
>>> df.loc[df["name"] == searchVar, "latitude"].squeeze() # here you have a scalar
41.6796
>>> df.loc[df["name"] == searchVar, "longitude"].squeeze()
19.8936

如果出于某种原因，您有多行具有相同的名称，那么您将返回一个系列，而不是标量。但在这种情况下，失败可能是您想要的，而不是传递模棱两可的数据