How to count the number of times a continuous condition is true?

In order to achieve this result, you need to iterate over the lists to determine the outputted values.

Start by assigning some base variables:

expected = []
lastTrue = None
count = 0

These will be modified throughout the loop. Next, loop through and test for each condition.

Loop through the numbers:

for i in range(len(A)):
    num = A[i]

Test for if the condition is True; if it is, save the number and reset the count:

if COND[i]:  # If True, save the number and reset count
    lastTrue = num
    count = 0

Check for if a True value has been found yet:

elif lastTrue is None:  # If haven't found a True value yet, pass
    pass

Test the number against the last true number to determine the output:

elif num < lastTrue:  # If less than the last True value, increment count
    count += 1
else:  # If larger than or equal to, reset count
    count = 0
    lastTrue = None

Add the current count to the list:

expected.append(count)

Now all you need to do is add the column and print the output.

Final code:

import pandas as pd
A=[28,30,15,25,24,13,31,19,20,11,19,21]
COND=[False,True,False,False,False,False,False,False,True,False,False,False]

expected = []
lastTrue = None
count = 0
for i in range(len(A)):
    num = A[i]
    if COND[i]:  # If True, save the number and reset count
        lastTrue = num
        count = 0
    elif lastTrue is None:  # If haven't found a True value yet, pass
        pass
    elif num < lastTrue:  # If less than the last True value, increment count
        count += 1
    else:  # If larger than or equal to, reset count
        count = 0
        lastTrue = None
    expected.append(count)

df=pd.DataFrame({'A':A,'COND':COND,'EXPECTED':expected})
print(df)

Output:

      A   COND  EXPECTED
 0   28  False         0
 1   30   True         0
 2   15  False         1
 3   25  False         2
 4   24  False         3
 5   13  False         4
 6   31  False         0
 7   19  False         0
 8   20   True         0
 9   11  False         1
 10  19  False         2
 11  21  False         0

This is an old question and you already have an answer. But I thought I add an answer from "within the Pandas framework" (i.e. avoiding explicit iteration over dataframe rows).

You could try:

df["Res"] = df.groupby(df["COND"].cumsum()).agg(
    Res=("COND", "cumcount"), A_cummax=("A", "cummax")
).assign(
    A_first=df["A"].where(df["COND"]).ffill().fillna(df["A"]),
    Res=lambda df: df["Res"].where(df["A_cummax"] <= df["A_first"], 0)
)["Res"]

• df["COND"].cumsum() identifies the groups you want to address, so grouping over it is the first step.
• Then add two columns via .agg(): (1) Res build by .cumcount() is almost the result you're looking for, but has to be adjusted a bit. (2) An auxiliary column A_cummax.
• Afterwards add a second auxiliary column A_first that identifies the resp. first A-values. The dataframe now looks like:

      Res  A_cummax  A_first
  0     0        28     28.0
  1     0        30     30.0
  2     1        30     30.0
  3     2        30     30.0
  4     3        30     30.0
  5     4        30     30.0
  6     5        31     30.0
  7     6        31     30.0
  8     0        20     20.0
  9     1        20     20.0
  10    2        20     20.0
  11    3        21     20.0
  

• Finally the adjustment of Res based on the 2 auxiliary columns: The zeros after the first A-value is exceeded.

Result for your example is as expected:

     A   COND  Res
0   28  False    0
1   30   True    0
2   15  False    1
3   25  False    2
4   24  False    3
5   13  False    4
6   31  False    0
7   19  False    0
8   20   True    0
9   11  False    1
10  19  False    2
11  21  False    0