How to get Count[] to count positive integers?

Questions
1

Try as I may, I can not get Count[] to count positive integers. My data

change = FinancialData["UPST", 
  "Change", {DatePlus[Today, {-70, "BusinessDay"}], {2021, 10}}]

Try to get a pure List

cc = change[[2, 1]] // Normal // Flatten

Any thing I do, the result is 0, can not get a true response

Count[#, # == Positive] & /@ cc // QuantityMagnitude
2

Other combinations I have tried without success

Count[cc, Positive]
Count[cc, "-"]
Count[#, Positive] & /@ cc // QuantityMagnitude
Count[#, "-" ] & /@ cc // QuantityMagnitude
Count[#, u_ /; u_ == Positive] & /@ cc // QuantityMagnitude
Count[#, Positive[#]] & /@ cc
Count[Positive[#]] & /@ cc   (* as least the function Positive[] \
works*)
Count[Positive[#], True] & /@ cc   
Length@Select[Positive[#]] & /@ cc // QuantityMagnitude

I used QuantityMagnitude[] to strip off the units ($)
cc looks like this

In[416]:= cc // QuantityMagnitude

Out[416]= {-20.915, -0.814995, -1.66, -4.02, 7.3, -2.45, -2.51, -1.7, \
5.49, -1.72, -1.22, 1.69, -0.959999, 1.97, -0.980003, -2.71, -1.21, \
-2.6, 2.35, 0.25, 9.89999, -10.52, -6.25, -0.0400009, 0.700005, 1.49, \
2.35, 1.52, -2.24, -0.340004, 5.25, -6.07, 6.04, 2.08, -3.11, 12.65, \
-4.73001, -0.699989, 5.18999, -1.03999, 5.06, -1.51001, 35.52, 6.92, \
25.17, -2.39999, -0.710007, 12.09, -18.4, 1.77, 7.32001, 16.49, \
-5.31, 2.61, 6.42999, -7.53999, 13.48, -2.31, 12.99, 7.48999, 15.41, \
11.63, 15.27, -19.14, 7.54001, -8.95999, 10.05, 14.17, 10.06, \
-9.85001}

cc is just a list of positive and negative numbers

3

Count[cc // QuantityMagnitude, u_ /; u < 0]

So cc//QuantityMagnitude is the first argument to Count and that first argument should get evaluated resulting in a list of numbers. And u_/;u<0 is a pattern that is going to get used on each of those items in that list. And Count will total up the number where it matched.

If we keep at this then one or the other of us will trip over something and figure this out.

4

Thanks for helping Bill

I tried the () on all the combinations I tried before and the one you suggested but Count[] or Length[] does not give a count. The inner code I get all zeros, or all ones, or all twos or just plain 0.

Count[cc, Positive]
Count[cc, "-"]
Count[#, Positive] & /@ (cc // QuantityMagnitude)
Count[#, "-" ] & /@ (cc // QuantityMagnitude)
Count[#, u_ /; u_ == Positive] & /@ (cc // QuantityMagnitude)
Count[#, Positive[#]] & /@ cc
Count[Positive[#]] & /@ cc   (* as least the function Positive[] \
works*)
Count[Positive[#], True] & /@ cc   
Length@Select[Positive[#]] & /@ (cc // QuantityMagnitude)
5

I did, or I thought I did, but I have learned that if you go through the motions, what you did, it is not always what you thought you did and this is another one of those cases. I thought I copied your code

In[558]:= Count[#, u /; u < 0] & /@ (cc // QuantityMagnitude)

Out[558]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0}

so I checked again, and what I thought I did, I did not do, So yes, copying your code correctly gives me a correct result. Thank you very much, I am sorry for the mistake.

In[559]:= Count[cc // QuantityMagnitude, u_ /; u < 0]

Out[559]= 34
6

Another option

CountsBy[cc, Positive]
(* <|True -> 33, False -> 36|> *)

The result is different because of the {-70, "BusinessDay"}