此值不是真实的,只是df外观的一个示例
import pandas as pd
g = {"id": "1515", "cost": "100"}
b = {"id": "1515", "cost": "100"}
f = {"id": "1515", "cost": "100"}
c = {"id": "1515", "cost": "100"}
o = {"id": "1515", "cost": "100"}
all_vendors = pd.concat([g, b, f, c, o])
数据类型
all_vendors.dtypes
Campaign object
campaignid float64
Campaign_name object
Cost object
Month object
Year & month object
dtype: object
尝试
# 1 attempt
all_vendors.Cost.sum()
TypeError: can only concatenate str (not "float") to str
# 2 attempt
all_vendors.Cost.astype(str)
all_vendors.Cost.sum()
TypeError: can only concatenate str (not "float") to str
# 3 attempt
all_vendors.Cost.astype(float)
all_vendors.Cost.sum()
ValueError: could not convert string to float: '100'
您的问题是您没有将astype
呼叫重新分配给DataFrame
:
import pandas as pd
data = {
"id": ['1,515','1,515','1,515','1,515','1,515'],
"cost": ['1,000','1,000','1,000','1,000','1,000']
}
all_vendors = pd.DataFrame.from_dict(data)
all_vendors['cost'] = all_vendors.cost.str.replace(',','').astype(float)
print(all_vendors.cost.sum())
# Output: 500
如注释中所述,使用str.replace
删除字符串中的所有逗号
我将此值设为500时就可以使用了:
df_list = [pd.DataFrame(data={"id": ["1515"], "cost": ["100"]}) for i in range(5)]
pd.concat(df_list).cost.astype(float).sum()
只要它们是数据帧,并且您将字符串转换为浮点数,它看起来就很好。
检查是否有帮助。这将提供ID的总数。
import pandas as pd
g = pd.DataFrame({"id": ["1515"], "cost": ["100"]})
b = pd.DataFrame({"id": ["1515"], "cost": ["100"]})
f = pd.DataFrame({"id": ["1515"], "cost": ["100"]})
c = pd.DataFrame({"id": ["1515"], "cost": ["100"]})
o = pd.DataFrame({"id": ["1515"], "cost": ["100"]})
all_vendors = pd.concat([g, b, f, c, o])
a=pd.DataFrame.from_records(all_vendors).astype(float).groupby('id').sum().T.to_dict()
print(a)
您首先需要将数据帧转换为浮点数,以便能够使用小数点添加数字,因为您使用DataFrame.astype
,所以>
DataFrame.astype
如果字符串中有',',则需要:
import pandas as pd g = pd.DataFrame({"id": ["1515"], "cost": ["100"]}) b = pd.DataFrame({"id": ["1515"], "cost": ["100"]}) f = pd.DataFrame({"id": ["1515"], "cost": ["100"]}) c = pd.DataFrame({"id": ["1515"], "cost": ["100"]}) o = pd.DataFrame({"id": ["1515"], "cost": ["100"]}) all_vendors = pd.concat([g, b, f, c, o])
然后您计算总和:
all_vendors['cost']=all_vendors['cost'].str.replace(',','')
输出:
all_vendors.astype(float).cost.sum()
如果要使用浮点型数据框,则需要为其分配:
500.0
输出:
all_vendors2=all_vendors.astype(float)
all_vendros2.cost.sum()