大约有 47,000 项符合查询结果（耗时：0.0536秒） [XML]
How to get a value from a cell of a dataframe?
...ue using the column name:
In [3]: sub_df
Out[3]:
A B
2 -0.133653 -0.030854
In [4]: sub_df.iloc[0]
Out[4]:
A -0.133653
B -0.030854
Name: 2, dtype: float64
In [5]: sub_df.iloc[0]['A']
Out[5]: -0.13365288513107493
...
~x + ~y == ~(x + y) is always false?
...
|
edited Jun 20 '12 at 13:47
answered Jun 20 '12 at 2:35
...
How do you round a number to two decimal places in C#?
... 2); //returns 1.99
decimal b = 1.995555M;
Math.Round(b, 2); //returns 2.00
You might also want to look at bankers rounding / round-to-even with the following overload:
Math.Round(a, 2, MidpointRounding.ToEven);
There's more information on it here.
...
How to convert 1 to true or 0 to false upon model fetch
... or false into a boolean/tinyint field in the database, which uses 1 or 0 .
4 Answers
...
.NET String.Format() to add commas in thousands place for a number
...
String.Format("{0:n}", 1234); // Output: 1,234.00
String.Format("{0:n0}", 9876); // No digits after the decimal point. Output: 9,876
share
|
...
Regular expression to match URLs in Java
...
106
Try the following regex string instead. Your test was probably done in a case-sensitive manner....
Check time difference in Javascript
...cript Date objects to get their difference:
// use a constant date (e.g. 2000-01-01) and the desired time to initialize two dates
var date1 = new Date(2000, 0, 1, 9, 0); // 9:00 AM
var date2 = new Date(2000, 0, 1, 17, 0); // 5:00 PM
// the following is to handle cases where the times are on the ...
Getting key with maximum value in dictionary?
...
You can use operator.itemgetter for that:
import operator
stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iteritems(), key=operator.itemgetter(1))[0]
And instead of building a new list in memory use stats.iteritems(). The key parameter to the max() function is a function that computes a key th...
Javascript roundoff number to nearest 0.5
Can someone give me an idea how can i round off a number to the nearest 0.5.
I have to scale elements in a web page according to screen resolution and for that i can only assign font size in pts to 1, 1.5 or 2 and onwards etc.
...
How can I obtain the element-wise logical NOT of a pandas Series?
...~s:
In [7]: s = pd.Series([True, True, False, True])
In [8]: ~s
Out[8]:
0 False
1 False
2 True
3 False
dtype: bool
Using Python2.7, NumPy 1.8.0, Pandas 0.13.1:
In [119]: s = pd.Series([True, True, False, True]*10000)
In [10]: %timeit np.invert(s)
10000 loops, best of 3: 91.8 µs...