subtraction

Subtracting two pointers giving unexpected result

随声附和 提交于 2019-12-31 05:17:25
问题 #include <stdio.h> int main() { int *p = 100; int *q = 92; printf("%d\n", p - q); //prints 2 } Shouldn't the output of above program be 8? Instead I get 2. 回答1: Undefined behavior aside, this is the behavior that you get with pointer arithmetic: when it is legal to subtract pointers, their difference represents the number of data items between the pointers. In case of int which on your system uses four bytes per int , the difference between pointers that are eight-bytes apart is (8 / 4) ,

Subtract a column from one pandas dataframe from another

冷暖自知 提交于 2019-12-30 09:36:46
问题 I'm sorry for a dumb question, but I cannot find any way to do this easily. I have two pandas data frames in Python 2.7, which are indexed by tenor: In [136]: rates Out[136]: A A- BBB+ BBB BBB- BB 3M 0.3150 0.3530 0.4960 0.6460 0.7910 1.9070 6M 0.3070 0.3560 0.5330 0.6740 0.8740 1.9170 1Y 0.3810 0.4470 0.6380 0.8970 1.1220 1.9900 2Y 0.7040 0.8690 1.0080 1.3510 1.6150 2.3230 3Y 1.0650 1.2870 1.4350 1.7950 2.0960 2.7590 4Y 1.5980 1.7920 1.9540 2.2660 2.6600 3.5890 5Y 2.0890 2.2660 2.4390 2.7890

How to subtract two strings?

喜夏-厌秋 提交于 2019-12-30 01:40:08
问题 I have a long string, which is basically a list like str="lamp, bag, mirror," (and other items) I was wondering if I can add or subtract some items, in other programming languages I can easily do: str=str-"bag," and get str="lamp, mirror," this doesnt work in python (I'm using 2.7 on a W8 pc) Is there a way to split the string across say "bag," and somehow use that as a subtraction? Then I still need to figure out how to add. 回答1: you could also just do print "lamp, bag, mirror".replace("bag,

how to convert the minutes into hours and minutes with subtracted time(subtracted time values)

£可爱£侵袭症+ 提交于 2019-12-29 09:22:21
问题 I want to subtract the two different 24 hours time format. I had tried with following : var startingTimeValue = 04:40; var endTimeValue = 00:55; var hour = startingTimeValue.split(":"); var hour1 = endTimeValue.split(":"); var th = 1 * hour[0] - 1 * hour1[0]; var tm = 1 * hour[1] - 1 * hour1[1]; var time = th+":"+tm; This code is working fine if second minutes is not greater than the first.but other case it will return minus values. The above code sample values result : time1 : 04:40 time2 :

Subtracting dates with Ruby

不想你离开。 提交于 2019-12-29 06:37:07
问题 I'm just having a look at ruby and was playing with the date/time thing. irb(main):001:0> jamis_DOB = Time.mktime(2003, 10, 22, 06, 59) => Wed Oct 22 06:59:00 +0300 2003 irb(main):002:0> age = Time.now - jamis_DOB => 222934108.172989 irb(main):005:0> age_in_years = (((age / 3600) / 24) / 365).to_i => 7 So my example is not so good as age_in_years won't know if there are leap years as those years add up. I've been through some googled time/date tutorials and haven't found an easy way to just

PHP Subtract First Character of String

我怕爱的太早我们不能终老 提交于 2019-12-29 01:43:10
问题 I know this is simple, and I've done it before, but I've forgotten how to do it. How might I, in php, subtract the first letter of a string? For example: FOOBAR would become OOBAR I think its something like $string[-1]; 回答1: substr($string, 1) http://php.net/manual/en/function.substr.php 回答2: substr ( $string , 1 ,strlen($string) ) 回答3: If you are echoing the string out, then use: $string[0] = ''; It's over an order of magnitude faster than: substr($string, 1); However, for anything else (eg

subtracting two dataframes

喜欢而已 提交于 2019-12-28 23:42:22
问题 df1: City, 2015-12-31, 2016-01-31, ... YYZ 562.14, -701.18, ... DFW 562.14, -701.18, ... YYC 562.14, -701.18, ... df2: City, 2015-12-31, 2016-01-31, ... SFO 562.14, -701.18, ... PDX 562.14, -701.18, ... LAX 562.14, -701.18, ... I want to subtract df1 from df2. i.e. subtract values in respective date columns. I tried the following: df2.subtract(df1, fill_value=0) But I receive the following error: TypeError: unsupported operand type(s) for -: 'str' and 'float' I think the error is because the

How do I subtract 11 from a list and mod 121 the same list all at once?

给你一囗甜甜゛ 提交于 2019-12-25 21:24:10
问题 This is for prime number theory. This theory which actually shows up as an axiom factors numbers like 2^n-1 where n=11 or n=23 or n=29. Of course it would not factor n=7 or n=31 when those are Mersenne prime numbers. Hopefully I can get help here I wish to do this all at once. I have this list below. I want to -11 from the entire list and then in the next instance mod 121 the list right after the subtraction while maintaining the enumerated list at the same time. Can this be done? I'm getting

How do I subtract 11 from a list and mod 121 the same list all at once?

无人久伴 提交于 2019-12-25 21:24:03
问题 This is for prime number theory. This theory which actually shows up as an axiom factors numbers like 2^n-1 where n=11 or n=23 or n=29. Of course it would not factor n=7 or n=31 when those are Mersenne prime numbers. Hopefully I can get help here I wish to do this all at once. I have this list below. I want to -11 from the entire list and then in the next instance mod 121 the list right after the subtraction while maintaining the enumerated list at the same time. Can this be done? I'm getting

Oracle sql - date subtraction within a function

不打扰是莪最后的温柔 提交于 2019-12-25 08:08:45
问题 (moved from previous post) - sorry if this is seen as a repeat! Hi everyone, Just having some issues with date calculations and subtracting from a date within a function. I am confused with the data types I should be using as I am having to convert from date to_char for example. I'm not sure whether to have the return type as varchar2 or date? Here is the function that receives a car_id, looks in the table for that car, pulls out the cars arrive date, stores it in a date variable does the