Convert the string 2.90K to 2900 or 5.2M to 5200000 in pandas dataframe
DataFrame.replace
with pd.eval
I like MaxU's answer. You can considerably shorten this using pd.eval
:
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:
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
Explanation
Assuming "Val" is a column of strings, the replace
operation yields,
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!
_ .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
To further generalize cs95's answer I would do this:
df['Val'].replace({'K': '*1e3', 'M': '*1e6', '-':'-1'}, regex=True).map(pd.eval).astype(int)
since on some numeric values pd.eval has to multiply '-' by some other number which will result an error. (could not convert string to float '-')
assuming you have the following 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:
In [31]: df.Val = (df.Val.replace(r'[KM]+$', '', regex=True).astype(float) * \
....: df.Val.str.extract(r'[\d\.]+([KM]+)', expand=False)
....: .fillna(1)
....: .replace(['K','M'], [10**3, 10**6]).astype(int))
In [32]: df
Out[32]:
Date Val
0 2016-09-23 100.0
1 2016-09-22 9600000.0
2 2016-09-21 54200.0
3 2016-09-20 115300.0
4 2016-09-19 18900.0
5 2016-09-16 176100.0
6 2016-09-15 31600.0
7 2016-09-14 10000.0
8 2016-09-13 3200000.0
Explanation:
In [36]: df.Val.replace(r'[KM]+$', '', 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
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['col'] = df['col'].apply(value_to_float)