在pandas数据帧中将字符串2.90K转换为2900或5.2M到5200000

Question

在pandas数据帧中处理数据需要一些帮助.我们非常欢迎任何帮助.

我有CSV格式的OHCLV数据.我已将文件加载到pandas dataframe中.

如何将音量列从2.90K转换为2900或将5.2M转换为5200000.该列可以包含数千K的K和数百万的M.

import pandas as pd

file_path = '/home/fatjoe/UCHM.csv'
df = pd.read_csv(file_path, parse_dates=[0], index_col=0)
df.columns = [
"closing_price", 
"opening_price", 
"high_price", 
"low_price",
"volume",
"change"]

df['opening_price'] = df['closing_price']
df['opening_price'] = df['opening_price'].shift(-1)
df = df.replace('-', 0)
df = df[:-1]
print(df.head())

Console:
 Date
 2016-09-23          0
 2016-09-22      9.60K
 2016-09-21     54.20K
 2016-09-20    115.30K
 2016-09-19     18.90K
 2016-09-16    176.10K
 2016-09-15     31.60K
 2016-09-14     10.00K
 2016-09-13      3.20K

Answer 1

DataFrame.replace 和 pd.eval

我喜欢 MaxU 的回答。您可以使用pd.eval以下方法大大缩短此时间：

df['Val'].replace({'K': '*1e3', 'M': '*1e6'}, regex=True).map(pd.eval).astype(int)

0        100
1    9600000
2      54200
3     115300
4      18900
5     176100
6      31600
7      10000
8    3200000
Name: Val, dtype: int64

稍作修改也会使这种情况不区分大小写：

repl_dict = {'[kK]': '*1e3', '[mM]': '*1e6', '[bB]': '*1e9', }
df['Val'].replace(repl_dict, regex=True).map(pd.eval)

0        100.0
1    9600000.0
2      54200.0
3     115300.0
4      18900.0
5     176100.0
6      31600.0
7      10000.0
8    3200000.0
Name: Val, dtype: float64

解释

假设“Val”是一列字符串，replace操作产生，

df['Val'].replace({'K': '*1e3', 'M': '*1e6'}, regex=True)

0           100
1      9.60*1e6
2     54.20*1e3
3    115.30*1e3
4     18.90*1e3
5    176.10*1e3
6     31.60*1e3
7     10.00*1e3
8      3.20*1e6
Name: Val, dtype: object

这是一个pd.eval可以求值的算术表达式！

_ .map(pd.eval)

0        100.0
1    9600000.0
2      54200.0
3     115300.0
4      18900.0
5     176100.0
6      31600.0
7      10000.0
8    3200000.0
Name: Val, dtype: float64

Answer 2

假设您有以下DF:

In [30]: df
Out[30]:
         Date      Val
0  2016-09-23      100
1  2016-09-22    9.60M
2  2016-09-21   54.20K
3  2016-09-20  115.30K
4  2016-09-19   18.90K
5  2016-09-16  176.10K
6  2016-09-15   31.60K
7  2016-09-14   10.00K
8  2016-09-13    3.20M

你可以这样做:

In [31]: df.Val = (df.Val.replace(r'[KM]+$', '', regex=True).astype(float) * \
   ....:           df.Val.str.extract(r'[\d\.]+([KM]+)', expand=False)
   ....:             .fillna(1)
   ....:             .replace(['K','M'], [10**3, 10**6]).astype(int))

In [32]: df
Out[32]:
         Date        Val
0  2016-09-23      100.0
1  2016-09-22  9600000.0
2  2016-09-21    54200.0
3  2016-09-20   115300.0
4  2016-09-19    18900.0
5  2016-09-16   176100.0
6  2016-09-15    31600.0
7  2016-09-14    10000.0
8  2016-09-13  3200000.0

说明:

In [36]: df.Val.replace(r'[KM]+$', '', regex=True).astype(float)
Out[36]:
0    100.0
1      9.6
2     54.2
3    115.3
4     18.9
5    176.1
6     31.6
7     10.0
8      3.2
Name: Val, dtype: float64

In [37]: df.Val.str.extract(r'[\d\.]+([KM]+)', expand=False)
Out[37]:
0    NaN
1      M
2      K
3      K
4      K
5      K
6      K
7      K
8      M
Name: Val, dtype: object

In [38]: df.Val.str.extract(r'[\d\.]+([KM]+)', expand=False).fillna(1)
Out[38]:
0    1
1    M
2    K
3    K
4    K
5    K
6    K
7    K
8    M
Name: Val, dtype: object

In [39]: df.Val.str.extract(r'[\d\.]+([KM]+)', expand=False).fillna(1).replace(['K','M'], [10**3, 10**6]).astype(int)
Out[39]:
0          1
1    1000000
2       1000
3       1000
4       1000
5       1000
6       1000
7       1000
8    1000000
Name: Val, dtype: int32

Answer 3

def value_to_float(x):
    if type(x) == float or type(x) == int:
        return x
    if 'K' in x:
        if len(x) > 1:
            return float(x.replace('K', '')) * 1000
        return 1000.0
    if 'M' in x:
        if len(x) > 1:
            return float(x.replace('M', '')) * 1000000
        return 1000000.0
    if 'B' in x:
        return float(x.replace('B', '')) * 1000000000
    return 0.0

    df."Insert data series column" = df."Insert data series column" .apply(value_to_float)

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