I Have one for condition, and can't do if statement to count cell condition

Question

I’m trying to do this condition to count how many times (cells) have values less than 300 seconds (5 minutes) but it returns me an error, the condition is about a DF below, that I pull from my csv:

df =
    0
0   NaN
1   79.0
2   140.0
3   131.0
4   72.0
... ...
16341   349.0
16342   795.0
16343   787.0
16344   410.0
16345   1221.0

I try this way:

for key, value in data_df.items() :
  df = data_df[key].fillna(0, inplace=False).astype(int)   
  if key <= 300 :
   df_count = value.count( value.values() <= 300)
   display(df_count)

but the error message says:

TypeError: ‘numpy.ndarray’ object is not callable

How can I solve my IF statement to execute the Display() code?

EDITED FOR BETTER RESULTS

I Should asked this before but i thought that the sollution would be alike IF statement.

I need to make it in range, the condition is to count every cell in a range about 5 to 5 minutes, untill I get 1 hour complete, So the conditions is about:

0 to 300 (5minutes)
300 to 600 (5 to 10 minutes)
600 to 900 (10 minutes to 15 minutes)

.
.
.

until i get one hour complete, I tryed:
df_count300to600 = data_df[key].fillna(0).le( key > 300 && key< 600 )

How can I solve at this point?

Asked By: Felipe Abrão Mariotti

||

Source

Answer 1

IIUC, you just need:

count = df[0].fillna(0).le(300).sum()

Example output: 5

How it works

df[0].fillna(0).le(300) creates a boolean Series, sum counts the True:

            0  df['0'].fillna(0).le(300)
0         NaN                       True
1        79.0                       True
2       140.0                       True
3       131.0                       True
4        72.0                       True
16341   349.0                      False
16342   795.0                      False
16343   787.0                      False
16344   410.0                      False
16345  1221.0                      False

Answered By: mozway

I Have one for condition, and can't do if statement to count cell condition

Question:

Answers:

How it works