I have the following dataframe, which the value should be increasing. Originally the dataframe has some unknown values.
index
value
0
1
1
2
3
2
4
5
6
7
4
8
9
10
3
11
3
12
13
14
15
5
Based on the assumsion that the value should be increasing, I would like to remove the value at index 10 and 11. This would be the desired dataframe:
index
value
0
1
1
2
3
2
4
5
6
7
4
8
9
12
13
14
15
5
Thank you very much
Assuming NaN in the empty cells (if not, temporarily replace them with NaN), use boolean indexing:
# if not NaNs uncomment below
# and use s in place of df['value'] afterwards
# s = pd.to_numeric(df['value'], errors='coerce')
# is the cell empty?
m1 = df['value'].isna()
# are the values strictly increasing?
m2 = df['value'].ge(df['value'].cummax())
out = df[m1|m2]
Output:
index value
1 1 NaN
2 2 NaN
3 3 2.0
4 4 NaN
5 5 NaN
6 6 NaN
7 7 4.0
8 8 NaN
9 9 NaN
12 12 NaN
13 13 NaN
14 14 NaN
15 15 5.0
Try this:
def del_df(df):
df_no_na = df.dropna().reset_index(drop = True)
num_tmp = df_no_na['value'][0] # First value which is not NaN.
del_index_list = [] # indicies to delete
for row_index in range(1, len(df_no_na)):
if df_no_na['value'][row_index] > num_tmp : #Increasing
num_tmp = df_no_na['value'][row_index] # to compare following two values.
else : # Not increasing(same or decreasing)
del_index_list.append(df_no_na['index'][row_index]) # index to delete
df_goal = df.drop([df.index[i] for i in del_index_list])
return df_goal
output:
index value
0 0 1.0
1 1 NaN
2 2 NaN
3 3 2.0
4 4 NaN
5 5 NaN
6 6 NaN
7 7 4.0
8 8 NaN
9 9 NaN
12 12 NaN
13 13 NaN
14 14 NaN
15 15 5.0
Related
I have a one column dataframe which looks like this:
Neive Bayes
0 8.322087e-07
1 3.213342e-24
2 4.474122e-28
3 2.230054e-16
4 3.957606e-29
5 9.999992e-01
6 3.254807e-13
7 8.836033e-18
8 1.222642e-09
9 6.825381e-03
10 5.275194e-07
11 2.224289e-06
12 2.259303e-09
13 2.014053e-09
14 1.755933e-05
15 1.889681e-04
16 9.929193e-01
17 4.599619e-05
18 6.944654e-01
19 5.377576e-05
I want to pivot it to wide format but with specific intervals. The first 9 rows should make up 9 columns of the first row, and continue this pattern until the final table has 9 columns and has 9 times fewer rows than now. How would I achieve this?
Using pivot_table:
df.pivot_table(columns=df.index % 9, index=df.index // 9, values='Neive Bayes')
0 1 2 3 4 \
0 8.322087e-07 3.213342e-24 4.474122e-28 2.230054e-16 3.957606e-29
1 6.825381e-03 5.275194e-07 2.224289e-06 2.259303e-09 2.014053e-09
2 6.944654e-01 5.377576e-05 NaN NaN NaN
5 6 7 8
0 0.999999 3.254807e-13 8.836033e-18 1.222642e-09
1 0.000018 1.889681e-04 9.929193e-01 4.599619e-05
2 NaN NaN NaN NaN
Construct multiindex, set_index and unstack
iix = pd.MultiIndex.from_arrays([np.arange(df.shape[0]) // 9,
np.arange(df.shape[0]) % 9])
df_wide = df.set_index(iix)['Neive Bayes'].unstack()
Out[204]:
0 1 2 3 4 \
0 8.322087e-07 3.213342e-24 4.474122e-28 2.230054e-16 3.957606e-29
1 6.825381e-03 5.275194e-07 2.224289e-06 2.259303e-09 2.014053e-09
2 6.944654e-01 5.377576e-05 NaN NaN NaN
5 6 7 8
0 0.999999 3.254807e-13 8.836033e-18 1.222642e-09
1 0.000018 1.889681e-04 9.929193e-01 4.599619e-05
2 NaN NaN NaN NaN
I want to add a list as a column to the df dataframe. The list has a different size than the column length.
df =
A B C
1 2 3
5 6 9
4
6 6
8 4
2 3
4
6 6
8 4
D = [11,17,18]
I want the following output
df =
A B C D
1 2 3 11
5 6 9 17
4 18
6 6
8 4
2 3
4
6 6
8 4
I am doing the following to extend the list to the size of the dataframe by adding "nan"
# number of nan value require for the list to match the size of the column
extend_length = df.shape[0]-len(D)
# extend the list
D.extend(extend_length * ['nan'])
# add to the dataframe
df["D"] = D
A B C D
1 2 3 11
5 6 9 17
4 18
6 6 nan
8 4 nan
2 3 nan
4 nan
6 6 nan
8 4 nan
Where "nan" is treated like string but I want it to be empty ot "nan", thus, if I search for number of valid cell in D column it will provide output of 3.
Adding the list as a Series will handle this directly.
D = [11,17,18]
df.loc[:, 'D'] = pd.Series(D)
A simple pd.concat on df and series of D as follows:
pd.concat([df, pd.Series(D, name='D')], axis=1)
or
df.assign(D=pd.Series(D))
Out[654]:
A B C D
0 1 2.0 3.0 11.0
1 5 6.0 9.0 17.0
2 4 NaN NaN 18.0
3 6 NaN 6.0 NaN
4 8 NaN 4.0 NaN
5 2 NaN 3.0 NaN
6 4 NaN NaN NaN
7 6 NaN 6.0 NaN
8 8 NaN 4.0 NaN
I just want to know how to get the sum of the last 5th values based on id from every rows.
df:
id values
-----------------
a 5
a 10
a 10
b 2
c 2
d 2
a 5
a 10
a 20
a 10
a 15
a 20
expected df:
id values sum(x.tail(5))
-------------------------------------
a 5 NaN
a 10 NaN
a 10 NaN
b 2 NaN
c 2 NaN
d 2 NaN
a 5 NaN
a 10 NaN
a 20 40
a 10 55
a 15 55
a 20 60
For simplicity, I'm trying to find the sum of values from the last 5th rows from every rows with id a only.
I tried to use code df.apply(lambda x: x.tail(5)), but that only showed me last 5 rows from the very last row of the entire df. I want to get the sum of last nth rows from every and each rows. Basically it's like rolling_sum for time series data.
you can calculate the sum of the last 5 as like this:
df["rolling As"] = df[df['id'] == 'a'].rolling(window=5).sum()["values"]
(this includes the current row as one of the 5. not sure if that is what you want)
id values rolling As
0 a 5 NaN
1 a 10 NaN
2 a 10 NaN
3 b 2 NaN
4 c 2 NaN
5 d 5 NaN
6 a 10 NaN
7 a 20 55.0
8 a 10 60.0
9 a 10 60.0
10 a 15 65.0
11 a 20 75.0
If you don't want it included. you can shift
df["rolling"] = df[df['id'] == 'a'].rolling(window=5).sum()["values"].shift()
to give:
id values rolling
0 a 5 NaN
1 a 10 NaN
2 a 10 NaN
3 b 2 NaN
4 c 2 NaN
5 d 5 NaN
6 a 10 NaN
7 a 20 NaN
8 a 10 55.0
9 a 10 60.0
10 a 15 60.0
11 a 20 65.0
Try using groupby, transform, and rolling:
df['sum(x.tail(5))'] = df.groupby('id')['values']\
.transform(lambda x: x.rolling(5, min_periods=5).sum().shift())
Output:
id values sum(x.tail(5))
1 a 5 NaN
2 a 10 NaN
3 a 10 NaN
4 b 2 NaN
5 c 2 NaN
6 d 2 NaN
7 a 5 NaN
8 a 10 NaN
9 a 20 40.0
10 a 10 55.0
11 a 15 55.0
12 a 20 60.0
Here is a test dataframe. I want to use the relationship between EmpID and MgrID to further map the manager of MgrID in a new column.
Test_df = pd.DataFrame({'EmpID':['1','2','3','4','5','6','7','8','9','10'],
'MgrID':['4','4','4','6','8','8','10','10','10','12']})
Test_df
If I create a dictionary for the initial relationship, I will be able to create the first link of the chain, but I affraid I need to loop through each of the new columns to create a new one.
ID_Dict = {'1':'4',
'2':'4',
'3':'4',
'4':'6',
'5':'8',
'6':'8',
'7':'10',
'8':'10',
'9':'10',
'10':'12'}
Test_df['MgrID_L2'] = Test_df['MgrID'].map(ID_Dict)
Test_df
What is the most efficient way to do this?
Thank you!
Here's a way with a simple while loop. Note I changed the name of MgrID to MgrID_1
Test_df = pd.DataFrame({'EmpID':['1','2','3','4','5','6','7','8','9','10'],
'MgrID_1':['4','4','4','6','8','8','10','10','10','12']})
d = Test_df.set_index('EmpID').MgrID_1.to_dict()
s = 2
while s:
Test_df['MgrID_'+str(s)] = Test_df['MgrID_'+str(s-1)].map(d)
if Test_df['MgrID_'+str(s)].isnull().all():
Test_df = Test_df.drop(columns='MgrID_'+str(s))
s = 0
else:
s+=1
Ouptut: Test_df
EmpID MgrID_1 MgrID_2 MgrID_3 MgrID_4 MgrID_5
0 1 4 6 8 10 12
1 2 4 6 8 10 12
2 3 4 6 8 10 12
3 4 6 8 10 12 NaN
4 5 8 10 12 NaN NaN
5 6 8 10 12 NaN NaN
6 7 10 12 NaN NaN NaN
7 8 10 12 NaN NaN NaN
8 9 10 12 NaN NaN NaN
9 10 12 NaN NaN NaN NaN
I am trying to join a fragment of a dataframe with another one. The structure of the dataframe to join is simplified below:
left:
ID f1 TIME
1 10 1
3 10 1
7 10 1
9 10 2
2 10 2
1 10 2
3 10 2
right:
ID f2 f3
1 0 11
7 9 11
I need to select the left dataset by time, and I need to attached the right one, the result I would like to have is the following:
left:
ID f1 TIME f2 f3
1 10 1 0 11
3 10 1 nan nan
7 10 1 9 11
9 10 2 nan nan
2 10 2 nan nan
1 10 2 nan nan
3 10 2 nan nan
Currently I am usually joining dataframes in this way:
left = left.join(right.set_index('ID'), on='ID')
In this case I am using:
left[left.TIME == 1] = left[left.TIME == 1].join(right.set_index('ID'), on='ID')
I have also tried with merge, but the result is the left dataframe without any of the other columns.
Finally the structure of my script need to do this for every unique TIME in the dataframe, thus:
for t in numpy.unique(left.TIME):
#do join on the fragment left.TIME == t
If I save the returned value from the join function in a new dataframe everything works fine, but trying to add the value at the left dataframe does not work.
EDIT: The IDs of the left dataset can be present multiple times, but not inside the same TIME value.
You can filter first by boolean indexing, merge and concat last:
df1 = left[left['TIME']==1]
#alternative
#df1 = left.query('TIME == 1')
df2 = left[left['TIME']!=1]
#alternative
#df2 = left.query('TIME != 1')
df = pd.concat([df1.merge(right, how='left'), df2])
print (df)
ID TIME f1 f2 f3
0 1 1 10 0.0 11.0
1 3 1 10 NaN NaN
2 7 1 10 9.0 11.0
3 9 2 10 NaN NaN
4 2 2 10 NaN NaN
5 1 2 10 NaN NaN
6 3 2 10 NaN NaN
EDIT: merge create default indices, so possible solution is create column first and then set to index:
print (left)
ID f1 TIME
10 1 10 1
11 3 10 1
12 7 10 1
13 9 10 2
14 2 10 2
15 1 10 2
16 3 10 2
#df = left.merge(right, how='left')
df1 = left[left['TIME']==1]
df2 = left[left['TIME']!=1]
df = pd.concat([df1.reset_index().merge(right, how='left').set_index('index'), df2])
print (df)
ID TIME f1 f2 f3
10 1 1 10 0.0 11.0
11 3 1 10 NaN NaN
12 7 1 10 9.0 11.0
13 9 2 10 NaN NaN
14 2 2 10 NaN NaN
15 1 2 10 NaN NaN
16 3 2 10 NaN NaN
EDIT:
After discussion after modify input data is possible use:
df = left.merge(right, how='left', on=['ID','TIME'])
This is one way:
res = left.drop_duplicates('ID')\
.merge(right, how='left')\
.append(left[left.duplicated(subset=['ID'])])
# ID TIME f1 f2 f3
# 0 1 1 10 0.0 11.0
# 1 3 1 10 NaN NaN
# 2 7 1 10 9.0 11.0
# 3 9 2 10 NaN NaN
# 4 2 2 10 NaN NaN
# 5 1 2 10 NaN NaN
# 6 3 2 10 NaN NaN
Note that columns f2 and f3 become float since NaN is considered a float.