I am trying to get the max start for each id, this is the table that I have:
id descrip start
0 0000 x 4
1 0000 y 60
2 1111 x 7
3 1111 x 0
4 2222 z 452
5 3333 x 36622
6 3333 t 32
And this is what I want:
id descrip start
0 0000 y 60
1 1111 x 7
2 2222 z 452
3 3333 x 36622
I tried doing this
df.loc[df.reset_index().groupby(['id'])['start'].idxmax()]
But i have been getting this error:
KeyError: 'Passing list-likes to .loc or [] with any missing labels is no longer supported, see https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#deprecate-loc-reindex-listlike'
Related
If I have a data frame like this:
id descrip
0 0000 x
1 0000 y
2 0000 z
3 1111 x
4 1111 z
5 2222 z
6 3333 x
7 3333 y
And I want to basically keep rows based on a priority of the descrip column, where if there is a z, then that is preferred over a y, which is preferred over an x.
So I basically want this:
id descrip
0 0000 z
1 1111 z
2 2222 z
3 3333 y
Not sure how I would approach this
df.groupby('id')['descrip'].max().reset_index()
id descrip
0 0 z
1 1111 z
2 2222 z
3 3333 y
Its always good to keep a track of what is exactly preferred over what.
Lets say the ordering was different ie: y<z<x where x is the most prefered. Then we could do:
df['descrip'] = df.descrip.astype('category').cat.reorder_categories(['y', 'z', 'x']).\
cat.as_ordered()
df.groupby('id')['descrip'].max().reset_index()
id descrip
0 0 x
1 1111 x
2 2222 z
3 3333 x
I'm trying to do this program where given a number N, one has to print out the decimal, octal, hexadecimal and binary for all the numbers in range 1 to N. The trouble is that the platform requires the solution in a particular format.
Suppose the number is 17, so the output should be like :
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 A 1010
11 13 B 1011
12 14 C 1100
13 15 D 1101
14 16 E 1110
15 17 F 1111
16 20 10 10000
17 21 11 10001
For 7 it would be like :
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
If you notice, the above is required to be printed in a way that the decimal, octal and hexadecimal numbers need a minimum of 2 spaces at their left whereas the binary numbers need at least one space at their left. Now, as the length of the numbers increase the space needs to be given accordingly such that the minimum space is there even for the max length number. So, how do I print them using a variable space? So far I have tried this :
Code
def print_formatted(number):
space=len(str(bin(number))[2:])
for i in range(1,number+1):
print('{:2d}'.format(i), end='')
print('{:>3s}'.format(str(oct(i))[2:]), end='')
print('{:>3s}'.format(str(hex(i))[2:]), end='')
print('{:>'+str(space)+'s}'.format(str(bin(i))[2:]))
print_formatted(17)
Here, I just tried doing the required with just the binary numbers but it's giving me an error
print('{:>'+str(space)+'s}'.format(str(bin(i))[2:]))
ValueError: Single '}' encountered in format string
Is there any fix/alternative for this?
Your problem is operator order - the + for string concattenation is weaker then the method call in
'{:>' + str(space) + 's}'.format(str(bin(i))[2:])
. Thats why you call the .format(...) only on "s}" - not the whole string. And thats where the
ValueError: Single '}' encountered in format string
comes from.
Putting the complete formatstring into parenthesis before applying .format to it fixes that.
You also need 1 more space for binary and can skip some str() that are not needed:
def print_formatted(number):
space=len(str(bin(number))[2:])+1 # fix here
for i in range(1,number+1):
print('{:2d}'.format(i), end='')
print('{:>3s}'.format(oct(i)[2:]), end='')
print('{:>3s}'.format(hex(i)[2:]), end='')
print(('{:>'+str(space)+'s}').format(bin(i)[2:])) # fix here
print_formatted(17)
Output:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 a 1010
11 13 b 1011
12 14 c 1100
13 15 d 1101
14 16 e 1110
15 17 f 1111
16 20 10 10000
17 21 11 10001
From your given output above you might need to prepend this by 2 spaces - not sure if its a formatting error in your output above or part of the restrictions.
You could also shorten this by using f-strings (and removing superflous str() around bin, oct, hex: they all return a strings already).
Then you need to calculate the the numbers you use to your space out your input values:
def print_formatted(number):
de,bi,oc,he = len(str(number)), len(bin(number)), len(oct(number)), len(hex(number))
for i in range(1,number+1):
print(f' {i:{de}d}{oct(i)[2:]:>{oc}s}{hex(i)[2:]:>{he}s}{bin(i)[2:]:>{bi}s}')
print_formatted(26)
to accomodate other values then 17, f.e. 128:
1 1 1 1
2 2 2 10
3 3 3 11
...
8 10 8 1000
...
16 20 10 10000
...
32 40 20 100000
...
64 100 40 1000000
...
128 200 80 10000000
I want to create a new Id column to a data frame which should start from 0000 and increments
Expecting output:
You can use this
df['id'] = pd.Series(np.arange(len(df))).astype(str).str.zfill(4)
input
Place Number Code
0 X A 1
1 Y B 2
2 X C 3
3 Y D 0
4 X F 1
5 Y G 2
6 X H 5
7 Y I 4
output
Place Number Code id
0 X A 1 0000
1 Y B 2 0001
2 X C 3 0002
3 Y D 0 0003
4 X F 1 0004
5 Y G 2 0005
6 X H 5 0006
7 Y I 4 0007
I have a dataframe which is grouped at product store day_id level Say it looks like the below and I need to create a column with rolling sum
prod store day_id visits
111 123 1 2
111 123 2 3
111 123 3 1
111 123 4 0
111 123 5 1
111 123 6 0
111 123 7 1
111 123 8 1
111 123 9 2
need to create a dataframe as below
prod store day_id visits rolling_4_sum cond
111 123 1 2 6 1
111 123 2 3 5 1
111 123 3 1 2 1
111 123 4 0 2 1
111 123 5 1 4 0
111 123 6 0 4 0
111 123 7 1 NA 0
111 123 8 1 NA 0
111 123 9 2 NA 0
i am looking for create a
cond column: that recursively checks a condition , say if rolling_4_sum is greater than 5 then make the next 4 rows as 1 else do nothing ,i.e. even if the condition is not met retain what was already filled before , do this check for each row until 7 th row.
How can i achieve this using python ? i am trying
d1['rolling_4_sum'] = d1.groupby(['prod', 'store']).visits.rolling(4).sum()
but getting an error.
The formation of rolling sums can be done with rolling method, using boxcar window:
df['rolling_4_sum'] = df.visits.rolling(4, win_type='boxcar', center=True).sum().shift(-2)
The shift by -2 is because you apparently want the sums to be placed at the left edge of the window.
Next, the condition about rolling sums being less than 4:
df['cond'] = 0
for k in range(1, 4):
df.loc[df.rolling_4_sum.shift(k) < 7, 'cond'] = 1
A new column is inserted and filled with 0; then for each k=1,2,3,4, look k steps back; if the sum then less than 7, then set the condition to 1.
This maybe real simple solution but I am new to python 3 and I have a dataframe with multiple columns. I would like to add a new column to the existing dataframe - which does the following calculation i.e.
New Column = Max((Column A/Column B), (Column C/Column D), (Column E/Column F))
I can do a max based on the following code but wanted to check how can I do div alongwith it.
df['Max'] = df[['Column A','Column B','Column C', 'Column D', 'Column E', 'Column F']].max(axis=1)
Column A Column B Column C Column D Column E Column F Max
3600 36000 22 11 3200 3200 36000
2300 2300 13 26 1100 1200 2300
1300 13000 15 33 1000 1000 13000
Thanks
You can div the df by itself by slicing the columns in steps and then take the max:
In [105]:
df['Max'] = df.ix[:,df.columns[::2]].div(df.ix[:,df.columns[1::2]].values, axis=1).max(axis=1)
df
Out[105]:
Column A Column B Column C Column D Column E Column F Max
0 3600 36000 22 11 3200 3200 2
1 2300 2300 13 26 1100 1200 1
2 1300 13000 15 33 1000 1000 1
Here are the intermediate values:
In [108]:
df.ix[:,df.columns[::2]].div(df.ix[:,df.columns[1::2]].values, axis=1)
Out[108]:
Column A Column C Column E
0 0.1 2.000000 1.000000
1 1.0 0.500000 0.916667
2 0.1 0.454545 1.000000
You can try something like as follows
df['Max'] = df.apply(lambda v: max(v['A'] / v['B'].astype(float), v['C'] / V['D'].astype(float), v['E'] / v['F'].astype(float)), axis=1)
Example
In [14]: df
Out[14]:
A B C D E F
0 1 11 1 11 12 98
1 2 22 2 22 67 1
2 3 33 3 33 23 4
3 4 44 4 44 11 10
In [15]: df['Max'] = df.apply(lambda v: max(v['A'] / v['B'].astype(float), v['C'] /
v['D'].astype(float), v['E'] / v['F'].astype(float)), axis=1)
In [16]: df
Out[16]:
A B C D E F Max
0 1 11 1 11 12 98 0.122449
1 2 22 2 22 67 1 67.000000
2 3 33 3 33 23 4 5.750000
3 4 44 4 44 11 10 1.100000