I expect to describe well want I need. I have a data frame with the same columns name and another column that works as an index. The data frame looks as follows:
df = pd.DataFrame({'ID':[1,1,1,1,1,2,2,2,3,3,3,3],'X':[1,2,3,4,5,2,3,4,1,3,4,5],'Y':[1,2,3,4,5,2,3,4,5,4,3,2]})
df
Out[21]:
ID X Y
0 1 1 1
1 1 2 2
2 1 3 3
3 1 4 4
4 1 5 5
5 2 2 2
6 2 3 3
7 2 4 4
8 3 1 5
9 3 3 4
10 3 4 3
11 3 5 2
My intention is to copy X as an index or one column (it doesn't matter) and append Y columns from each 'ID' in the following way:
You can try
out = pd.concat([group.rename(columns={'Y': f'Y{name}'}) for name, group in df.groupby('ID')])
out.columns = out.columns.str.replace(r'\d+$', '', regex=True)
print(out)
ID X Y Y Y
0 1 1 1.0 NaN NaN
1 1 2 2.0 NaN NaN
2 1 3 3.0 NaN NaN
3 1 4 4.0 NaN NaN
4 1 5 5.0 NaN NaN
5 2 2 NaN 2.0 NaN
6 2 3 NaN 3.0 NaN
7 2 4 NaN 4.0 NaN
8 3 1 NaN NaN 5.0
9 3 3 NaN NaN 4.0
10 3 4 NaN NaN 3.0
11 3 5 NaN NaN 2.0
Here's another way to do it:
df_org = pd.DataFrame({'ID':[1,1,1,1,1,2,2,2,3,3,3,3],'X':[1,2,3,4,5,2,3,4,1,3,4,5]})
df = df_org.copy()
for i in set(df_org['ID']):
df1 = df_org[df_org['ID']==i]
col = 'Y'+str(i)
df1.columns = ['ID', col]
df = pd.concat([ df, df1[[col]] ], axis=1)
df.columns = df.columns.str.replace(r'\d+$', '', regex=True)
print(df)
Output:
ID X Y Y Y
0 1 1 1.0 NaN NaN
1 1 2 2.0 NaN NaN
2 1 3 3.0 NaN NaN
3 1 4 4.0 NaN NaN
4 1 5 5.0 NaN NaN
5 2 2 NaN 2.0 NaN
6 2 3 NaN 3.0 NaN
7 2 4 NaN 4.0 NaN
8 3 1 NaN NaN 1.0
9 3 3 NaN NaN 3.0
10 3 4 NaN NaN 4.0
11 3 5 NaN NaN 5.0
Another solution could be as follow.
Get unique values for column ID (stored in array s).
Use np.transpose to repeat column ID n times (n == len(s)) and evaluate the array's matches with s.
Use np.where to replace True with values from df.Y and False with NaN.
Finally, drop the orignal df.Y and rename the new columns as required.
import pandas as pd
import numpy as np
df = pd.DataFrame({'ID':[1,1,1,1,1,2,2,2,3,3,3,3],
'X':[1,2,3,4,5,2,3,4,1,3,4,5],
'Y':[1,2,3,4,5,2,3,4,5,4,3,2]})
s = df.ID.unique()
df[s] = np.where((np.transpose([df.ID]*len(s))==s),
np.transpose([df.Y]*len(s)),
np.nan)
df.drop('Y', axis=1, inplace=True)
df.rename(columns={k:'Y' for k in s}, inplace=True)
print(df)
ID X Y Y Y
0 1 1 1.0 NaN NaN
1 1 2 2.0 NaN NaN
2 1 3 3.0 NaN NaN
3 1 4 4.0 NaN NaN
4 1 5 5.0 NaN NaN
5 2 2 NaN 2.0 NaN
6 2 3 NaN 3.0 NaN
7 2 4 NaN 4.0 NaN
8 3 1 NaN NaN 5.0
9 3 3 NaN NaN 4.0
10 3 4 NaN NaN 3.0
11 3 5 NaN NaN 2.0
If performance is an issue, this method should be faster than this answer, especially when the number of unique values for ID increases.
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
Consider a simple dataframe:
import numpy as np
import pandas as pd
x = pd.DataFrame(np.arange(10).reshape(5,2))
print(x)
0 1
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9
I would like to create a hierarchically indexed dataframe of the form:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN
where the 'a' columns correspond to the original dataframe columns and the 'b' columns are blank (or nan).
I can certainly create a hierarchically indexed dataframe with all NaNs and loop over the columns of the original dataframe, writing them into
the new dataframe. Is there something more compact than that?
you can do with MultiIndex.from_product
extra_level = ['a', 'b']
new_cols = pd.MultiIndex.from_product([x.columns, extra_level])
x.columns = new_cols[::len(x.columns)] # select all the first element of extra_level
x = x.reindex(columns=new_cols)
print(x)
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN
Very much like #Ben.T I am using MultiIndex.from_product:
x.assign(l='a')
.set_index('l', append=True)
.unstack()
.reindex(pd.MultiIndex.from_product([x.columns.tolist(), ['a','b']]), axis=1)
Output:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN
Let's say I have a df such as this:
df = pd.DataFrame({'A': [1,2,3,4,5], 'A_z': [2,3,4,5,6], 'B': [3,4,5,6,7], 'B_z': [4,5,6,7,8],
'C': [5,6,7,8,9], 'C_z': [6,7,8,9,10]})
Which looks like this:
A A_z B B_z C C_z
0 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
What I'm looking to do is create a new df and for each letter (A,B,C) append this new df vertically with the data from the two columns per letter so that it looks like this:
Letter Letter_z
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 3 4
6 4 5
7 5 6
8 6 7
9 7 8
10 5 6
11 6 7
12 7 8
13 8 9
14 9 10
As far as I'm concerned something like this should work fine:
for col in df.columns:
if col[-1] != 'z':
new_df = new_df.append(df[[col, col + '_z']])
However this results in the following mess:
A A_z B B_z C C_z
0 1.0 2.0 NaN NaN NaN NaN
1 2.0 3.0 NaN NaN NaN NaN
2 3.0 4.0 NaN NaN NaN NaN
3 4.0 5.0 NaN NaN NaN NaN
4 5.0 6.0 NaN NaN NaN NaN
0 NaN NaN 3.0 4.0 NaN NaN
1 NaN NaN 4.0 5.0 NaN NaN
2 NaN NaN 5.0 6.0 NaN NaN
3 NaN NaN 6.0 7.0 NaN NaN
4 NaN NaN 7.0 8.0 NaN NaN
0 NaN NaN NaN NaN 5.0 6.0
1 NaN NaN NaN NaN 6.0 7.0
2 NaN NaN NaN NaN 7.0 8.0
3 NaN NaN NaN NaN 8.0 9.0
4 NaN NaN NaN NaN 9.0 10.0
What am I doing wrong? Any help would be really appreciated, cheers.
EDIT:
After the kind help from jezrael the renaming of the columns in his answer got me thinking about a possible way to do it using my original train of thought.
I can now also achieve the new df I want using the following:
for col in df:
if col[-1] != 'z':
d = df[[col, col + '_z']]
d.columns = ['Letter', 'Letter_z']
new_df = new_df.append(d)
The different columns names were clearly what was causing the problem which is something I wasn't aware of at the time. Hope this helps anyone.
One ide is use Series.str.split with expand=True for MultiIndex, then use rename for avoid NaNs and finally new columns names, reshape by DataFrame.stack, sort for correct order by DataFrame.sort_index and last remove MultiIndex:
df.columns = df.columns.str.split('_', expand=True)
df = df.rename(columns=lambda x:'Letter_z' if x == 'z' else 'Letter', level=1)
df = df.stack(0).sort_index(level=[1,0]).reset_index(drop=True)
print (df)
Letter Letter_z
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 3 4
6 4 5
7 5 6
8 6 7
9 7 8
10 5 6
11 6 7
12 7 8
13 8 9
14 9 10
Or if possible simplify problem with reshape all non z values to one column and all z values to another use numpy.ravel:
m = df.columns.str.endswith('_z')
a = df.loc[:, ~m].to_numpy().T.ravel()
b = df.loc[:, m].to_numpy().T.ravel()
df = pd.DataFrame({'Letter': a,'Letter_z': b})
print (df)
Letter Letter_z
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 3 4
6 4 5
7 5 6
8 6 7
9 7 8
10 5 6
11 6 7
12 7 8
13 8 9
14 9 10
You can use the function concat and a list comprehension:
cols = df.columns[~df.columns.str.endswith('_z')]
func = lambda x: 'letter_z' if x.endswith('_z') else 'letter'
pd.concat([df.filter(like=i).rename(func, axis=1) for i in cols])
or
cols = df.columns[~df.columns.str.endswith('_z')]
pd.concat([df.filter(like=i).set_axis(['letter', 'letter_z'], axis=1, inplace=False) for i in cols])
I have a csv that I import as a dataframe with pandas. The columns are like:
Step1:A Step1:B Step1:C Step1:D Step2:A Step2:B Step2:D Step3:B Step3:D Step3:E
0 1 2 3 4 5 6 7 8 9
Where the step and parameter are separated by ':'. I want to reshape the dataframe to look like this:
Step1 Step2 Step3
A 0 4 nan
B 1 5 7
C 2 nan nan
D 3 6 8
E nan nan 9
Now, If I want to maintain column sequential order such that I have this case:
Step2:A Step2:B Step2:C Step2:D Step1:A Step1:B Step1:D AStep3:B AStep3:D AStep3:E
0 1 2 3 4 5 6 7 8 9
Where the step and parameter are separated by ':'. I want to reshape the dataframe to look like this:
Step2 Step1 AStep3
A 0 4 nan
B 1 5 7
C 2 nan nan
D 3 6 8
E nan nan 9
Try read_csv with delim_whitespace:
df = pd.read_csv('file.csv', delim_whitespace=True)
df.columns = df.columns.str.split(':', expand=True)
df.stack().reset_index(level=0, drop=True)
output:
Step1 Step2 Step3
A 0.0 4.0 NaN
B 1.0 5.0 7.0
C 2.0 NaN NaN
D 3.0 6.0 8.0
E NaN NaN 9.0