How to check that element of a list of lists matches a condition?

You can use rlist

library(rlist)
list.filter(pairs, Name=='A')
#[[1]]
#[[1]]$Name
#[1] "A"

#[[1]]$Value
#[1] 11

Also, my original version is

 sapply(pairs, function(x) x[grep('Name',names(x))]=='A') 
 # Name  Name  Name 
 # TRUE FALSE FALSE 

You can simply use Filter if your list of lists has only one level. this will return the desired element(s):

> Filter(function(u) u$Name=='A', pairs)
#[[1]]
#[[1]]$Name
#[1] "A"

#[[1]]$Value
#[1] 11

If you just want to know whether any list component has Name=='A':

any(sapply(pairs,function(x) x$Name=='A'));
## [1] TRUE

If you want the number of list components that have Name=='A':

sum(sapply(pairs,function(x) x$Name=='A'));
## [1] 1

If you want the Value of the list component(s) that have Name=='A':

unlist(lapply(pairs,function(x) if (x$Name=='A') x$Value));
## [1] 11

If you want the sublist of components that have Name=='A':

pairs[sapply(pairs,function(x) x$Name=='A')];
## [[1]]
## [[1]]$Name
## [1] "A"
##
## [[1]]$Value
## [1] 11

If you want the first inner list that has Name=='A' (can drop the [1] if you're certain there will only be one match):

pairs[[which(sapply(pairs,function(x) x$Name=='A'))[1]]];
## $Name
## [1] "A"
##
## $Value
## [1] 11

---

Alternatively, since your data appears to be regular, you can convert to a data.frame, which will simplify all these operations:

df <- do.call(rbind,lapply(pairs,as.data.frame));
df;
##   Name Value
## 1    A    11
## 2    B    17
## 3    C    23

Here are the equivalents for df:

any(df$Name=='A');
## [1] TRUE
sum(df$Name=='A');
## [1] 1
df$Value[df$Name=='A'];
## [1] 11
subset(df,Name=='A');
##   Name Value
## 1    A    11
subset(df,Name=='A')[1,];
##   Name Value
## 1    A    11