pandas 在熊猫数据框中将字符串 2.90K 转换为 2900 或 5.2M 为 5200000
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Convert the string 2.90K to 2900 or 5.2M to 5200000 in pandas dataframe
提问by Joseph M Njuguna
Need some help on processing data inside a pandas dataframe. Any help is most welcome.
在处理Pandas数据帧内的数据时需要一些帮助。任何帮助都是最受欢迎的。
I have OHCLV data in CSV format. I have loaded the file in to pandas dataframe.
我有 CSV 格式的 OHCLV 数据。我已将文件加载到Pandas数据框中。
How do I convert the volume column from 2.90K to 2900 or 5.2M to 5200000. The column can contain both K in form of thousands and M in millions.
如何将体积列从 2.90K 转换为 2900 或 5.2M 至 5200000。该列可以包含以千为单位的 K 和以百万为单位的 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
回答by MaxU
assuming you have the following DF:
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
you can do it this way:
你可以这样做:
In [31]: df.Val = (df.Val.replace(r'[KM]+$', '', regex=True).astype(float) * \
....: df.Val.str.extract(r'[\d\.]+([KM]+)', expand=False)
....: .fillna(1)
....: .replace(['K','M'], [10**3, 10**6]).astype(int))
In [32]: df
Out[32]:
Date Val
0 2016-09-23 100.0
1 2016-09-22 9600000.0
2 2016-09-21 54200.0
3 2016-09-20 115300.0
4 2016-09-19 18900.0
5 2016-09-16 176100.0
6 2016-09-15 31600.0
7 2016-09-14 10000.0
8 2016-09-13 3200000.0
Explanation:
解释:
In [36]: df.Val.replace(r'[KM]+$', '', 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
回答by ChiTownDataDude
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)
`
`
回答by cs95
I like MaxU's answer. You can considerably shorten this using 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
Slight modification will make this case insensitive as well:
稍微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
Assuming "Val" is a column of strings, the replace
operation yields,
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
This is an arithmetic expression which pd.eval
can evaluate!
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