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如何创建一个具有小而简单功能的新列

如何创建一个具有小而简单功能的新列

幕布斯6054654 2023-08-22 16:53:52
我对 Pandas 很陌生,所以我不知道该怎么做。我有一个带有 Pandas 的 Python 脚本,用于从在线 .csv 中检索数据,其中包含西班牙各省的 covid 数据。我想创建一个新列来创建一个新变量。这些将是(每日新增死亡人数/100.000 人口) 有可能吗?在 csv 中,我们已经有了人口数据,因此函数为:(population/1.000.000)*daily_deaths这是我的代码,所以我不知道如何开始import requestsimport pandas as pddf1 = pd.read_csv("https://raw.githubusercontent.com/montera34/escovid19data/master/data/output/covid19-provincias-spain_consolidated.csv")df1  = pd.DataFrame(df1) #no indexAlbacete = df1.loc[df1["province"] == "Albacete"]Alicante = df1[df1['ine_code'] == 3][['date',"PCR","TestAc",'province',"new_cases","activos","hospitalized","intensive_care","deceased","cases_accumulated","recovered","cases_per_cienmil","intensive_care_per_1000000","deceassed_per_100000","hospitalized_per_100000","daily_deaths","deaths_last_week"]]Almeria  = df1.loc[df1['ine_code'] == 3]Alava    = df1.loc[df1['ine_code'] == 1]Asturias = df1.loc[df1['ine_code'] == 33]Avila    = df1.loc[df1['ine_code'] == 5]Badajoz  = df1.loc[df1['ine_code'] == 6]Baleares = df1.loc[df1['ine_code'] == 7]Barcelona= df1.loc[df1['ine_code'] == 8]Bizcaia  = df1.loc[df1['ine_code'] == 48]Burgos   = df1.loc[df1['ine_code'] == 9]Caceres  = df1.loc[df1['ine_code'] == 10]Cadiz    = df1.loc[df1['ine_code'] == 11]Cantabria= df1.loc[df1['ine_code'] == 39] Castellon= df1.loc[df1['ine_code'] == 12]Ceuta    = df1.loc[df1['ine_code'] == 51]Ciudad_R = df1.loc[df1['ine_code'] == 13]Cordoba  = df1.loc[df1['ine_code'] == 14]Cuenca   = df1.loc[df1['ine_code'] == 16]Guipuzcoa= df1.loc[df1['ine_code'] == 20]Girona   = df1.loc[df1['ine_code'] == 17] Madrid   = df1[df1['ine_code'] == 28][['date',"PCR","TestAc",'province',"new_cases","activos","hospitalized","intensive_care","deceased","cases_accumulated","recovered","cases_per_cienmil","intensive_care_per_1000000","deceassed_per_100000","hospitalized_per_100000","daily_deaths","deaths_last_week"]]
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慕斯王

TA贡献1864条经验 获得超2个赞

我无法轻松查看所有 DataFrame 列,但请尝试以下操作:

df1['percent_per_day'] = (df1['population']/1000000) * df1['daily_deaths']

在哪里:

df1['percent_per_day']是你的新专栏

df1['population']以及df1['daily_deaths']这些列在 df1 中的名称


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反对 回复 2023-08-22
?
森林海

TA贡献2011条经验 获得超2个赞

不确定这是否是您正在寻找的内容,但根据 csv 中的列,尝试在“poblacion”为人口的情况下尝试此操作

df1['death_rate'] = (df1['poblacion']/1000000) *df1['daily_deaths']


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反对 回复 2023-08-22
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