问题
I would like to know if there is a faster and more "pythonic" way of doing the following, e.g. using some built in methods. Given a pandas DataFrame or numpy array of floats, if the value is equal or smaller than 0.5 I need to calculate the reciprocal value and multiply with -1 and replace the old value with the newly calculated one. "Transform" is probably a bad choice of words, please tell me if you have a better/more accurate description.
Thank you for your help and support!!
Data:
import numpy as np
import pandas as pd
dicti = {"A" : np.arange(0.0, 3, 0.1),
"B" : np.arange(0, 30, 1),
"C" : list("ELVISLIVES")*3}
df = pd.DataFrame(dicti)
my function:
def transform_colname(df, colname):
series = df[colname]
newval_list = []
for val in series:
if val <= 0.5:
newval = (1/val)*-1
newval_list.append(newval)
else:
newval_list.append(val)
df[colname] = newval_list
return df
function call:
transform_colname(df, colname="A")
**--> I'm summing up the results here, since comments wouldn't allow to post code (or I don't know how to do it).**
Thank you all for your fast and great answers!!
using ipython "%timeit" with "real" data:
my function: 10 loops, best of 3: 24.1 ms per loop
from jojo:
def transform_colname_v2(df, colname):
series = df[colname]
df[colname] = np.where(series <= 0.5, 1/series*-1, series)
return df
100 loops, best of 3: 2.76 ms per loop
from FooBar:
def transform_colname_v3(df, colname):
df.loc[df[colname] <= 0.5, colname] = - 1 / df[colname][df[colname] <= 0.5]
return df
100 loops, best of 3: 3.32 ms per loop
from dmvianna:
def transform_colname_v4(df, colname):
df[colname] = df[colname].where(df[colname] <= 0.5, (1/df[colname])*-1)
return df
100 loops, best of 3: 3.7 ms per loop
Please tell/show me if you would implement your code in a different way!
One final QUESTION: (answered) How could "FooBar" and "dmvianna" 's versions be made "generic"? I mean, I had to write the name of the column into the function (since using it as a variable didn't work). Please explain this last point! --> thanks jojo, ".loc" isn't the right way, but very simple df[colname] is sufficient. changed the functions above to be more "generic". (also changed ">" to be "<=", and updated timing)
Thank you very much!!
回答1:
If we are talking about arrays:
import numpy as np
a = np.array([0.1, 0.2, 0.3, 0.4, 0.5, 0.6], dtype=np.float)
print 1 / a[a <= 0.5] * (-1)
This will, however only return the values smaller than 0.5.
Alternatively use np.where:
import numpy as np
a = np.array([0.1, 0.2, 0.3, 0.4, 0.5, 0.6], dtype=np.float)
print np.where(a < 0.5, 1 / a * (-1), a)
Talking about pandas DataFrame:
As in @dmvianna's answer (so give some credit to him ;) ), adapting it to pd.DataFrame:
df.a = df.a.where(df.a > 0.5, (1 / df.a) * (-1))
回答2:
The typical trick is to write a general mathematical operation to apply to the whole column, but then use indicators to select rows for which we actually apply it:
df.loc[df.A < 0.5, 'A'] = - 1 / df.A[df.A < 0.5]
In[13]: df
Out[13]:
A B C
0 -inf 0 E
1 -10.000000 1 L
2 -5.000000 2 V
3 -3.333333 3 I
4 -2.500000 4 S
5 0.500000 5 L
6 0.600000 6 I
7 0.700000 7 V
8 0.800000 8 E
9 0.900000 9 S
10 1.000000 10 E
11 1.100000 11 L
12 1.200000 12 V
13 1.300000 13 I
14 1.400000 14 S
15 1.500000 15 L
16 1.600000 16 I
17 1.700000 17 V
18 1.800000 18 E
19 1.900000 19 S
20 2.000000 20 E
21 2.100000 21 L
22 2.200000 22 V
23 2.300000 23 I
24 2.400000 24 S
25 2.500000 25 L
26 2.600000 26 I
27 2.700000 27 V
28 2.800000 28 E
29 2.900000 29 S
回答3:
As in @jojo's answer, but using pandas:
df.A = df.A.where(df.A > 0.5, (1/df.A)*-1)
or
df.A.where(df.A > 0.5, (1/df.A)*-1, inplace=True) # this should be faster
.where docstring:
Definition: df.A.where(self, cond, other=nan, inplace=False, axis=None, level=None, try_cast=False, raise_on_error=True)
Docstring: Return an object of same shape as self and whose corresponding entries are from self where cond is True and otherwise are from other.
来源：https://stackoverflow.com/questions/25326649/python-numpy-pandas-change-values-on-condition