I want to count how many column data (pd.Dataframe) before Nan data. My data:
df
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Id
A 1 1 2 3 3 NaN NaN NaN NaN NaN NaN NaN NaN NaN
B 6 6 7 7 8 9 10 NaN NaN NaN NaN NaN NaN NaN
C 1 2 3 3 4 5 6 6 7 7 8 9 10 NaN
my desire output:
df_result
count
Id
A 5
B 7
C 13
thank you in advance for the answer.
Use:
print (df)
0 1 2 3 4 5 6 7 8 9 10 11 12 13
A 1 1 2 3 3 NaN NaN NaN NaN NaN NaN NaN NaN 54.0
B 6 6 7 7 8 9.0 10.0 NaN NaN NaN NaN NaN 5.0 NaN
C 1 2 3 3 4 5.0 6.0 6.0 7.0 7.0 8.0 9.0 10.0 NaN
df = df.isnull().cumsum(axis=1).eq(0).sum(axis=1)
print (df)
A 5
B 7
C 13
dtype: int64
Detail:
First check NaNs:
print (df.isnull())
0 1 2 3 4 5 6 7 8 9 \
A False False False False False True True True True True
B False False False False False False False True True True
C False False False False False False False False False False
10 11 12 13
A True True True False
B True True False True
C False False False True
Get cumsum - Trues are processes like 1, False like 0
print (df.isnull().cumsum(axis=1))
0 1 2 3 4 5 6 7 8 9 10 11 12 13
A 0 0 0 0 0 1 2 3 4 5 6 7 8 8
B 0 0 0 0 0 0 0 1 2 3 4 5 5 6
C 0 0 0 0 0 0 0 0 0 0 0 0 0 1
Compare by 0:
print (df.isnull().cumsum(axis=1).eq(0))
0 1 2 3 4 5 6 7 8 9 10 \
A True True True True True False False False False False False
B True True True True True True True False False False False
C True True True True True True True True True True True
11 12 13
A False False False
B False False False
C True True False
Sum boolean mask - Trues like 1s:
print (df.isnull().cumsum(axis=1).eq(0).sum(axis=1))
A 5
B 7
C 13
dtype: int64
Related
a code from Kaggle, which is said to remove outliners:
outliers_mask = (ft.abs() > ft.abs().quantile(outl_thresh)).any(axis=1)
Would not Any return a boolean item? either a an item being in a list or not?
So what the code says is, save in the mask all absolute values in Ft which are above the quantile (introduced by another variable)? What does the Any stand for? what for? thank you.
I think first part return DataFrame filled by boolean True or/and False:
(ft.abs() > ft.abs().quantile(outl_thresh))
so is added DataFrame.any for test if at least one True per rows to boolean Series.
df = pd.DataFrame({'a':[False, False, True],
'b':[False, True, True],
'c':[False, False, True]})
print (df)
a b c
0 False False False
1 False True False
2 True True True
print (df.any(axis=1))
0 False <- no True per rows
1 True <- one True per rows
2 True <- three Trues per rows
dtype: bool
Similar method for test if all values are Trues is DataFrame.all:
print (df.all(axis=1))
0 False
1 False
2 True
dtype: bool
Reason is for filtering by boolean indexing is necessary boolean Series, not boolean DataFrame.
Another sample data:
np.random.seed(2021)
ft = pd.DataFrame(np.random.randint(100, size=(10, 5))).sub(20)
print (ft)
0 1 2 3 4
0 65 37 -20 74 66
1 24 42 71 9 1
2 73 4 -8 50 50
3 13 -13 -19 77 6
4 46 28 79 43 29
5 -4 30 34 32 73
6 -15 29 18 -6 51
7 65 50 21 1 5
8 -10 16 -1 37 62
9 70 -5 20 56 33
outl_thresh = 0.95
print (ft.abs().quantile(outl_thresh))
0 71.65
1 46.40
2 75.40
3 75.65
4 69.85
Name: 0.95, dtype: float64
print((ft.abs() > ft.abs().quantile(outl_thresh)))
0 1 2 3 4
0 False False False False False
1 False False False False False
2 True False False False False
3 False False False True False
4 False False True False False
5 False False False False True
6 False False False False False
7 False True False False False
8 False False False False False
9 False False False False False
outliers_mask = (ft.abs() > ft.abs().quantile(outl_thresh)).any(axis=1)
print (outliers_mask)
0 False
1 False
2 True
3 True
4 True
5 True
6 False
7 True
8 False
9 False
dtype: bool
df1 = ft[outliers_mask]
print (df1)
0 1 2 3 4
2 73 4 -8 50 50
3 13 -13 -19 77 6
4 46 28 79 43 29
5 -4 30 34 32 73
7 65 50 21 1 5
0 1 2 3 4
df2 = ft[~outliers_mask]
print (df2)
0 1 2 3 4
0 65 37 -20 74 66
1 24 42 71 9 1
6 -15 29 18 -6 51
8 -10 16 -1 37 62
9 70 -5 20 56 33
I had search stackoverflow about this, all are so complex.
I want to output the row and column info about all cells that is inf or NaN.
You can replace np.inf to missing values and test them by DataFrame.isna and last test at least one True by DataFrame.any passed to DataFrame.loc for SubDataFrame:
df = pd.DataFrame({
'A':list('abcdef'),
'B':[4,5,4,5,5,np.inf],
'C':[7,np.nan,9,4,2,3],
'D':[1,3,5,7,1,0],
'E':[np.inf,3,6,9,2,np.nan],
'F':list('aaabbb')
})
print (df)
A B C D E F
0 a 4.0 7.0 1 inf a
1 b 5.0 NaN 3 3.0 a
2 c 4.0 9.0 5 6.0 a
3 d 5.0 4.0 7 9.0 b
4 e 5.0 2.0 1 2.0 b
5 f inf 3.0 0 NaN b
m = df.replace(np.inf, np.nan).isna()
print (m)
A B C D E F
0 False False False False True False
1 False False True False False False
2 False False False False False False
3 False False False False False False
4 False False False False False False
5 False True False False True False
df = df.loc[m.any(axis=1), m.any()]
print (df)
B C E
0 4.0 7.0 inf
1 5.0 NaN 3.0
5 inf 3.0 NaN
Or if need index and columns names in DataFrame use DataFrame.stack with Index.to_frame:
s = df.replace(np.inf, np.nan).stack(dropna=False)
df1 = s[s.isna()].index.to_frame(index=False)
print (df1)
0 1
0 0 E
1 1 C
2 5 B
3 5 E
My code to get the duplicates, how to negate the below meaning
df.duplicated(subset='col', keep='last').sum()
len(df['col'])-len(df['col'].drop_duplicates())
I think you need DataFrame.duplicated with keep=False for all duplicates, invert mask and sum for count Trues:
df = pd.DataFrame({'col':[1,2,2,3,3,3,4,5,5]})
print (df.duplicated(subset='col', keep=False))
0 False
1 True
2 True
3 True
4 True
5 True
6 False
7 True
8 True
dtype: bool
print (~df.duplicated(subset='col', keep=False))
0 True
1 False
2 False
3 False
4 False
5 False
6 True
7 False
8 False
dtype: bool
print ((~df.duplicated(subset='col', keep=False)).sum())
2
Another solution with Series.drop_duplicates and keep=False with length of Series:
print (df['col'].drop_duplicates(keep=False))
0 1
6 4
Name: col, dtype: int64
print (len(df['col'].drop_duplicates(keep=False)))
2
I have the following MultiIndex dataframe:
Close ATR condition
Date Symbol
1990-01-01 A 24 1 True
B 72 1 False
C 40 3 False
D 21 5 True
1990-01-02 A 65 4 True
B 19 2 True
C 43 3 True
D 72 1 False
1990-01-03 A 92 5 False
B 32 3 True
C 52 2 False
D 33 1 False
I perform the following calculation on this dataframe:
data.loc[data.index.levels[0][0], 'Shares'] = 0
data.loc[data.index.levels[0][0], 'Closed_P/L'] = 0
data = data.reset_index()
Equity = 10000
def calcs(x):
global Equity
# Skip first date
if x.index[0]==0: return x
# calculate Shares where condition is True
x.loc[x['condition'] == True, 'Shares'] = np.floor((Equity * 0.02 / x['ATR']).astype(float))
# other calulations
x['Closed_P/L'] = x['Shares'] * x['Close']
Equity += x['Closed_P/L'].sum()
return x
data = data.groupby('Date').apply(calcs)
data['Equity'] = data.groupby('Date')['Closed_P/L'].transform('sum')
data['Equity'] = data.groupby('Symbol')['Equity'].cumsum() + Equity
data = data.set_index(['Date','Symbol'])
The output is:
Close ATR condition Shares Closed_P/L Equity
Date Symbol
1990-01-01 A 24 1.2 True 0.0 0.0 10000.0
B 72 1.4 False 0.0 0.0 10000.0
C 40 3 False 0.0 0.0 10000.0
D 21 5 True 0.0 0.0 10000.0
1990-01-02 A 65 4 True 50.0 3250.0 17988.0
B 19 2 True 100.0 1900.0 17988.0
C 43 3 True 66.0 2838.0 17988.0
D 72 1 False NaN NaN 17988.0
1990-01-03 A 92 5 False NaN NaN 21796.0
B 32 3 True 119.0 3808.0 21796.0
C 52 2 False NaN NaN 21796.0
D 33 1 False NaN NaN 21796.0
I want to forward fill Shares values - grouped by Symbol - in case condition evaluates to False (except for first date). So the Shares value on 1990-01-02 for D should be 0 (because on 1990-01-01 the Shares value for D was 0 and the condition on 1990-01-02 is False). Also values for Shares on 1990-01-03 for A, C and D should be 50, 66 and 0 respectively based on the logic described above. How can I do that?
I have the following datasets of boolean columns
date hr energy
0 5-Feb-18 False False
1 29-Jan-18 False False
2 6-Dec-17 True False
3 16-Nov-17 False False
4 14-Nov-17 True True
5 25-Oct-17 False False
6 24-Oct-17 False False
7 5-Oct-17 False False
8 3-Oct-17 False False
9 26-Sep-17 False False
10 13-Sep-17 True False
11 7-Sep-17 False False
12 31-Aug-17 False False
I want to multiply each boolean column by 1 to turn it into a dummy
I tried:
df = df.iloc[:, 1:]
for col in df:
col = col*1
but the columns remain boolean, why?
Just using
df.iloc[:,1:]=df.iloc[:,1:].astype(int)
df
Out[477]:
date hr energy
0 5-Feb-18 0 0
1 29-Jan-18 0 0
2 6-Dec-17 1 0
3 16-Nov-17 0 0
4 14-Nov-17 1 1
5 25-Oct-17 0 0
6 24-Oct-17 0 0
7 5-Oct-17 0 0
8 3-Oct-17 0 0
9 26-Sep-17 0 0
10 13-Sep-17 1 0
11 7-Sep-17 0 0
12 31-Aug-17 0 0
For future cases other than True or False, If you want to convert categorical into numerical you could always use the replace function.
df.iloc[:,1:]=df.iloc[:,1:].replace({True:1,False:0})