how use np.append with np.where method?-CodePudding

I'm trying to convert this code :

df = pd.DataFrame({'values': [np.nan,-1.2,3,6,7,2]})

def f_evol (df) : 
list1 = []  
for i in range (len(df)) : 
    if df['values'].isnull()[i] : 
        list1.append(0)
    else : 
        if df['values'][i] > 0 : 
            k = list1[i-1]   1
            list1.append(k)
        if df['values'][i] < 0: 
            k = list1[i-1] - 1 
            list1.append(k)
return list1

df['count'] = f_evol(df)

on this one :

def v_evol (df) :
k = [1]
a = np.where(df['values'].isnull().shift(0).values, 
             k,
             np.where(df['values'].shift(0).values > 0,
                      k.append(k[-1]   1),
                      np.where(df['values'].shift(0).values < 0, 
                              k.append(k[-1] - 1),
                              np.nan
                              )
                     )
            )
return a

i failed to append the last value of k like in the first code ... i tried, to replace ‘k[0] 1’ by ‘np.concatenate()’, by ‘np.append()’ without succes... b

CodePudding user response：

With numpy

This will give you want you want without any explicit iteration (generally, you want to avoid for loops when working with arrays and dataframes):

np.sign(df["values"].fillna(0)).cumsum()

Demo:

In [3]: df["count"] = np.sign(df["values"].fillna(0)).cumsum()

In [4]: df
Out[4]:
   values  count
0     NaN    0.0
1    -1.2   -1.0
2     3.0    0.0
3     6.0    1.0
4     7.0    2.0
5     2.0    3.0

You can certainly cast the "count" column to an integer dtype if you want.

Without numpy

This will produce the same result without the numpy dependency, if that matters:

(df["values"] // df["values"].abs()).fillna(0).cumsum()