I have some data like this
df = pd.DataFrame({'class':['a','a','b','b','a','a','b','c','c'],'score':[3,5,6,7,8,9,10,11,14]})
df
class score
0 a 3
1 a 5
2 b 6
3 b 7
4 a 8
5 a 9
6 b 10
7 c 11
8 c 14
I want to use groupby function extract top n% data(descending by score),i know the nlargest can make it,but the number of every group is different,so i don't know how to do it
I tried this function
top_n = 0.5
g = df.groupby(['class'])['score'].apply(lambda x:x.nlargest(int(round(top_n*len(x))),keep='all')).reset_index()
g
class level_1 score
0 a 5 9
1 a 4 8
2 b 6 10
3 b 3 7
4 c 8 14
but it can not deal with big data(more than 10 million),it is very slow,how do i speed it,thank you!
Input is a number, e.g. 9 and I want to print decimal, octal, hex and binary value from 1 to 9 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
How can I achieve this in python3 using syntax like
dm, oc, hx, bn = len(str(9)), len(bin(9)[2:]), ...
print("{:dm%d} {:oc%s}" % (i, oct(i[2:]))
I mean if number is 999 so I want decimal 10 to be printed like ' 10' and binary equivalent of 999 is 1111100111 so I want 10 like ' 1010'.
You can use str.format() and its mini-language to do the whole thing for you:
for i in range(1, 10):
print("{v} {v:>6o} {v:>6x} {v:>6b}".format(v=i))
Which will print:
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
UPDATE: To define field 'widths' in a variable you can use a format-within-format structure:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for i in range(1, 10):
print("{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}".format(v=i, w=w))
Or define a different width variable for each field type if you want them non-uniformly spaced.
Btw. since you've tagged your question with python-3.x, if you're using Python 3.6 or newer, you can use Literal String Interpolation to simplify it even more:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for v in range(1, 10):
print(f"{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}")
I want to calculate the probability of all the data in a column dataframe according to its own distribution.For example,my data like this:
data
0 1
1 1
2 2
3 3
4 2
5 2
6 7
7 8
8 3
9 4
10 1
And the output I expect like this：
data pro
0 1 0.155015
1 1 0.155015
2 2 0.181213
3 3 0.157379
4 2 0.181213
5 2 0.181213
6 7 0.048717
7 8 0.044892
8 3 0.157379
9 4 0.106164
10 1 0.155015
I also refer to another question(How to compute the probability ...) and get an example of the above.My code is as follows：
import scipy.stats
samples = [1,1,2,3,2,2,7,8,3,4,1]
samples = pd.DataFrame(samples,columns=['data'])
print(samples)
kde = scipy.stats.gaussian_kde(samples['data'].tolist())
samples['pro'] = kde.pdf(samples['data'].tolist())
print(samples)
But what I can't stand is that if my column is too long, it makes the operation slow.Is there a better way to do it in pandas?Thanks in advance.
Its own distribution does not mean kde. You can use value_counts with normalize=True
df.assign(pro=df.data.map(df.data.value_counts(normalize=True)))
data pro
0 1 0.272727
1 1 0.272727
2 2 0.272727
3 3 0.181818
4 2 0.272727
5 2 0.272727
6 7 0.090909
7 8 0.090909
8 3 0.181818
9 4 0.090909
10 1 0.272727
I'm rather new at python.
I try to have a cumulative sum for each client to see the consequential months of inactivity (flag: 1 or 0). The cumulative sum of the 1's need therefore to be reset when we have a 0. The reset need to happen as well when we have a new client. See below with example where a is the column of clients and b are the dates.
After some research, I found the question 'Cumsum reset at NaN' and 'In Python Pandas using cumsum with groupby'. I assume that I kind of need to put them together.
Adapting the code of 'Cumsum reset at NaN' to the reset towards 0, is successful:
cumsum = v.cumsum().fillna(method='pad')
reset = -cumsum[v.isnull() !=0].diff().fillna(cumsum)
result = v.where(v.notnull(), reset).cumsum()
However, I don't succeed at adding a groupby. My count just goes on...
So, a dataset would be like this:
import pandas as pd
df = pd.DataFrame({'a' : [1,1,1,1,1,1,1,2,2,2,2,2,2,2],
'b' : [1/15,2/15,3/15,4/15,5/15,6/15,1/15,2/15,3/15,4/15,5/15,6/15],
'c' : [1,0,1,0,1,1,0,1,1,0,1,1,1,1]})
this should result in a dataframe with the columns a, b, c and d with
'd' : [1,0,1,0,1,2,0,1,2,0,1,2,3,4]
Please note that I have a very large dataset, so calculation time is really important.
Thank you for helping me
Use groupby.apply and cumsum after finding contiguous values in the groups. Then groupby.cumcount to get the integer counting upto each contiguous value and add 1 later.
Multiply with the original row to create the AND logic cancelling all zeros and only considering positive values.
df['d'] = df.groupby('a')['c'] \
.apply(lambda x: x * (x.groupby((x != x.shift()).cumsum()).cumcount() + 1))
print(df['d'])
0 1
1 0
2 1
3 0
4 1
5 2
6 0
7 1
8 2
9 0
10 1
11 2
12 3
13 4
Name: d, dtype: int64
Another way of doing would be to apply a function after series.expanding on the groupby object which basically computes values on the series starting from the first index upto that current index.
Use reduce later to apply function of two args cumulatively to the items of iterable so as to reduce it to a single value.
from functools import reduce
df.groupby('a')['c'].expanding() \
.apply(lambda i: reduce(lambda x, y: x+1 if y==1 else 0, i, 0))
a
1 0 1.0
1 0.0
2 1.0
3 0.0
4 1.0
5 2.0
6 0.0
2 7 1.0
8 2.0
9 0.0
10 1.0
11 2.0
12 3.0
13 4.0
Name: c, dtype: float64
Timings:
%%timeit
df.groupby('a')['c'].apply(lambda x: x * (x.groupby((x != x.shift()).cumsum()).cumcount() + 1))
100 loops, best of 3: 3.35 ms per loop
%%timeit
df.groupby('a')['c'].expanding().apply(lambda s: reduce(lambda x, y: x+1 if y==1 else 0, s, 0))
1000 loops, best of 3: 1.63 ms per loop
I think you need custom function with groupby:
#change row with index 6 to 1 for better testing
df = pd.DataFrame({'a' : [1,1,1,1,1,1,1,2,2,2,2,2,2,2],
'b' : [1/15,2/15,3/15,4/15,5/15,6/15,1/15,2/15,3/15,4/15,5/15,6/15,7/15,8/15],
'c' : [1,0,1,0,1,1,1,1,1,0,1,1,1,1],
'd' : [1,0,1,0,1,2,3,1,2,0,1,2,3,4]})
print (df)
a b c d
0 1 0.066667 1 1
1 1 0.133333 0 0
2 1 0.200000 1 1
3 1 0.266667 0 0
4 1 0.333333 1 1
5 1 0.400000 1 2
6 1 0.066667 1 3
7 2 0.133333 1 1
8 2 0.200000 1 2
9 2 0.266667 0 0
10 2 0.333333 1 1
11 2 0.400000 1 2
12 2 0.466667 1 3
13 2 0.533333 1 4
def f(x):
x.ix[x.c == 1, 'e'] = 1
a = x.e.notnull()
x.e = a.cumsum()-a.cumsum().where(~a).ffill().fillna(0).astype(int)
return (x)
print (df.groupby('a').apply(f))
a b c d e
0 1 0.066667 1 1 1
1 1 0.133333 0 0 0
2 1 0.200000 1 1 1
3 1 0.266667 0 0 0
4 1 0.333333 1 1 1
5 1 0.400000 1 2 2
6 1 0.066667 1 3 3
7 2 0.133333 1 1 1
8 2 0.200000 1 2 2
9 2 0.266667 0 0 0
10 2 0.333333 1 1 1
11 2 0.400000 1 2 2
12 2 0.466667 1 3 3
13 2 0.533333 1 4 4
I have 15 datafiles with unequal row sizes, but number of columns in each file is same. e.g.
ifile1.dat ifile2.dat ifile3.dat and so on ............
0 0 0 0 1 6
1 2 5 3 2 7
2 5 6 10 4 6
5 2 8 9 5 9
10 2 10 3 8 2
In each file 1st column represents the index number.
I would like to compute average of all these files for each index number in column 1. i.e.
ofile.txt
0 0 [This is computed as (0+0)/2]
1 4 [This is computed as (2+6)/2]
2 6 [This is computed as (5+7)/2]
3 [no value]
4 6 [This is computed as (6)/1]
5 4.66 [This is computed as (2+3+9)/3]
6 10
7
8 5.5
9
10 2.5
I can't think of any simple method to do it. I was thinking of a method, but seems very lengthy. Taking the average after converting all the files with same row size, .e.g.
ifile1.dat ifile2.dat ifile3.dat and so on ............
0 0 0 0 0 0
1 2 1 1 6
2 5 2 2 7
3 3 3
4 4 4 6
5 2 5 3 5 9
6 6 10 6
7 7 7
8 8 9 8 2
9 9 9
10 2 10 3 10
$ awk '{s[$1]+=$2; c[$1]++;} END{for (i in s) print i,s[i]/c[i];}' ifile*.dat
0 0
1 4
2 6
4 6
5 4.66667
6 10
8 5.5
10 2.5
In the above code, there are two arrays, s and c. s[i] is the sum of all entries with index i and c[i] is the number of entries with index i. After we have read all the files, we print the average, s[i]/c[i], for each index i.