Is there a shorthand function for obtaining the number of valid entries in each column of a data set?

Question

I actually wrote a code for this already.

Name_Count <- 
  data.frame(sum(A = !is.na(Name_Database[ ,1]))) %>%
  data.frame(sum(B = !is.na(Name_Database[ ,2]))) %>%
  data.frame(sum(C = !is.na(Name_Database[ ,3]))) %>%
  data.frame(sum(D = !is.na(Name_Database[ ,4]))) %>%
  data.frame(sum(E = !is.na(Name_Database[ ,5]))) %>%
  data.frame(sum(F = !is.na(Name_Database[ ,6]))) %>%
  data.frame(sum(G = !is.na(Name_Database[ ,7]))) %>%
  data.frame(sum(H = !is.na(Name_Database[ ,8]))) %>%
  data.frame(sum(I = !is.na(Name_Database[ ,9]))) %>%
  data.frame(sum(J = !is.na(Name_Database[ ,10]))) %>%
  data.frame(sum(K = !is.na(Name_Database[ ,11]))) %>%
  data.frame(sum(L = !is.na(Name_Database[ ,12]))) %>%
  data.frame(sum(M = !is.na(Name_Database[ ,13]))) %>%
  data.frame(sum(N = !is.na(Name_Database[ ,14]))) %>%
  data.frame(sum(O = !is.na(Name_Database[ ,15]))) %>%
  data.frame(sum(P = !is.na(Name_Database[ ,16]))) %>%
  data.frame(sum(Q = !is.na(Name_Database[ ,17]))) %>%
  data.frame(sum(R = !is.na(Name_Database[ ,18]))) %>%
  data.frame(sum(S = !is.na(Name_Database[ ,19]))) %>%
  data.frame(sum(T = !is.na(Name_Database[ ,20]))) %>%
  data.frame(sum(U = !is.na(Name_Database[ ,21]))) %>%
  data.frame(sum(V = !is.na(Name_Database[ ,22]))) %>%
  data.frame(sum(W = !is.na(Name_Database[ ,23]))) %>%
  data.frame(sum(X = !is.na(Name_Database[ ,24]))) %>%
  data.frame(sum(Y = !is.na(Name_Database[ ,25]))) %>%
  data.frame(sum(Z = !is.na(Name_Database[ ,26])))

#> Name_Count
#> A  B  C   D  E  F  G  H  I  J  K L  M  N O P  Q  R  S  T U  V  W X Y  Z
#> 1 23 28 45 141 18 47 42 12 27 21 49 8 50 28 3 9 15 40 94 25 5 35 13 3 5 18

This code counts the number of valid entries for each column of names. Each column is organized by each letter.

Even though the code works, it's extremely bloated and I want to find a way to optimize it. Thank you in advance for your assistance.