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## summing vector having cells=inf values

Tag : matlab , By : cmhudson
Date : March 29 2020, 07:55 AM
I hope this helps you . I need to find out the average of the values , To find how many are Inf:
1) find the infs
``````Allvalues{1}<inf
ans =
1     1     0     1     1     1     1     1     1     1     1     1     1     1     1     1     1     1     1     1
``````
``````sum(~(Allvalues{1}<inf))
ans =
1
``````

## Total sum of a numeric vector conditional on values from a character vector

Tag : r , By : rhyhann
Date : March 29 2020, 07:55 AM
Hope this helps x is a numeric vector and y is a character vector, both of the same length. I have to obtain the total sum of x for each character in y. So for example if y has the letters of the alphabet, the algorithm must sum all the values of x that indexed by a "A" and then all the values indexed by a "B", etc. It seems it can be done with the aggregate function but I haven't been able to figure out how. , Test data, in a dataframe for convenience rather than two vectors:
``````> d = data.frame(x=runif(10),y=sample(LETTERS[1:3],10,TRUE))
> d
x y
1  0.25927547 B
2  0.95012667 C
3  0.85133149 C
4  0.64658480 B
etc
``````
``````> tapply(d\$x,d\$y,sum)
A        B        C
1.547225 1.891884 2.666552
``````
``````> sum(d\$x[d\$y=="A"])
 1.547225
> sum(d\$x[d\$y=="B"])
 1.891884
> sum(d\$x[d\$y=="C"])
 2.666552
``````
``````> d=data.frame(x=runif(10),y=sample(LETTERS[1:3],10,TRUE),z=sample(LETTERS[1:3],10,TRUE))
> d
x y z
1  0.4166217 A C
2  0.5816940 B C
3  0.9915231 A C
4  0.7177323 B C
etc
``````
``````> aggregate(x~y+z,d,sum)
y z         x
1 C A 1.6392171
2 B B 0.9389463
3 C B 0.3330299
4 A C 2.3748477
5 B C 1.2994263
``````
``````> sum(d\$x[d\$y=="C" & d\$z=="A"])
 1.639217
> sum(d\$x[d\$y=="B" & d\$z=="B"])
 0.9389463
``````

## Summing values in a vector based on duplicate values in another R

Tag : r , By : user181445
Date : March 29 2020, 07:55 AM
This might help you Let's say I have two vectors below: , Using data.table package:
``````DT = data.table(r,s)
DT[, `:=`(r=c(rep(NA, .N-1L), r[1L]),
s=c(rep(NA, .N-1L), sum(s))
),
by=r]

> DT
#      r   s
#  1: NA  NA
#  2: NA  NA
#  3:  4  43
#  4: NA  NA
#  5: NA  NA
#  6:  6  11
#  7: NA  NA
#  8:  8  97
#  9:  9   9
# 10: NA  NA
# 11:  2 170
# 12:  3  78
``````

## Generate a vector: when values in one vector match another vector, input the value of another vector at the same positio

Tag : r , By : Keonne Rodriguez
Date : March 29 2020, 07:55 AM
will help you In your example data df1\$outlet is already equal to concordance\$outlet at all points. I assume you want something that works even when that's not true, so let's scramble the data first to check that what we do works in all cases:
``````set.seed(123)
df1 = df1[sample(1:nrow(df1)), , drop = FALSE]
``````
``````df1 = merge(df1, concordance, by="outlet", sort = FALSE)
``````

## Split a vector and summing values

Tag : r , By : Cadu
Date : March 29 2020, 07:55 AM
hope this fix your issue I'm a R newbie. I've got a vector , Starting with this vector...
``````> vec
 105  29  41  70  77   0  56  49  63   0 105
``````
``````> vec == 0
 FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE FALSE
``````
``````> cumsum(vec==0)
 0 0 0 0 0 1 1 1 1 2 2
``````
``````> split(vec, cumsum(vec==0))
\$`0`
 105  29  41  70  77

\$`1`
  0 56 49 63

\$`2`
   0 105
``````
``````> sapply(split(vec, cumsum(vec==0)),sum)
0   1   2
322 168 105
`````` 