如何将这些包含字符串的列转换为float，同时保持其含义？

Question

我想将列Actual、Forecast、Previous转换为float，以便对它们进行计算。csv还包含一些应该保留的nan

csv文件如下所示：

2018-01-04 04:30:00,GBP,Low Impact Expected,Mortgage Approvals,65K,64K,65K
2018-01-04 04:51:00,EUR,Low Impact Expected,Spanish 10-y Bond Auction,1.53|1.8,,1.49|2.0
2018-01-04 05:01:00,EUR,Low Impact Expected,French 10-y Bond Auction,0.79|1.4,,0.36|1.9
2018-01-04 07:30:00,USD,Low Impact Expected,Challenger Job Cuts y/y,-3.6%,,30.1%

到目前为止，我已经尝试过：

df.columns = ['Date','Currency','Impact','Event','Actual','Forecast','Previous']

df = df[~(df['Actual'].isin('|','<']))]
#df = df[~df.Actual.str.contains("|")]

df['Actual'] = df['Actual'].str.replace('%', '')
df['Forecast'] = df['Forecast'].str.replace('%', '')
df['Previous'] = df['Previous'].str.replace('%', '')

df['Actual'] = df['Actual'].str.replace('K', '000')
df['Forecast'] = df['Forecast'].str.replace('K', '000')
df['Previous'] = df['Previous'].str.replace('K', '000')

for i in df['Actual']: float(i)
for i in df['Forecast']: float(i)
for i in df['Previous']: float(i)

去除和<；不要工作。互联网上的许多建议似乎与文件中的值有关

我也不知道如何替换%，同时移动小数点，这样数字表示就正确了

希望有人能帮忙。谢谢

Answer 1

这不是最漂亮的方法，但我相信这就是你想要的：

from io import StringIO
import pandas as pd
df = pd.read_table(StringIO("""2018-01-04 04:30:00,GBP,Low Impact Expected,Mortgage Approvals,65K,64K,65K
2018-01-04 04:51:00,EUR,Low Impact Expected,Spanish 10-y Bond Auction,1.53|1.8,,1.49|2.0
2018-01-04 05:01:00,EUR,Low Impact Expected,French 10-y Bond Auction,0.79|1.4,,0.36|1.9
2018-01-04 07:30:00,USD,Low Impact Expected,Challenger Job Cuts y/y,-3.6%,,30.1%"""), names=['Date','Currency','Impact','Event','Actual','Forecast','Previous'], sep=',')
df = df.loc[~df['Actual'].str.contains('[\|\<]')]
for col in ['Actual', 'Forecast', 'Previous']:
    df.loc[pd.notnull(df[col]) & df[col].str.contains('\%'), col] = pd.to_numeric(df.loc[pd.notnull(df[col]) & df[col].str.contains('\%'), col].str.replace('%', '')) / 100
    df.loc[pd.notnull(df[col]) & df[col].str.endswith('K'), col] = pd.to_numeric(df.loc[pd.notnull(df[col]) & df[col].str.endswith('K'), col].str.replace('K', '')) * 1000