I am tryning to sum all values in a range of columns from the third to last of several thousand columns using:
day3prep['D3counts'] = day3prep.sum(day3prep.iloc[:, 2:].sum(axis=1))
dataframe is formated as:
ID G1 Z1 Z2 ...ZN
0 50 13 12 ...62
1 51 62 23 ...19
dataframe with summed column:
ID G1 Z1 Z2 ...ZN D3counts
0 50 13 12 ...62 sum(Z1:ZN in row 0)
1 51 62 23 ...19 sum(Z1:ZN in row 1)
I've changed the NaNs to 0's. The datatype is float but I am getting the error:
'Series' objects are mutable, thus they cannot be hashed
You only need this part:
day3prep['D3counts'] = day3prep.iloc[:, 2:].sum(axis=1)
With some random numbers:
import pandas as pd
import random
random.seed(42)
day3prep = pd.DataFrame({'ID': random.sample(range(10), 5), 'G1': random.sample(range(10), 5),
'Z1': random.sample(range(10), 5), 'Z2': random.sample(range(10), 5), 'Z3': random.sample(range(10), 5)})
day3prep['D3counts'] = day3prep.iloc[:, 2:].sum(axis=1)
Output:
> day3prep
ID G1 Z1 Z2 Z3 D3counts
0 1 2 0 8 8 16
1 0 1 9 0 6 15
2 4 8 1 3 3 7
3 9 4 7 5 7 19
4 6 3 6 6 4 16
Related
I have a df of 15 x 4 and I'm trying to compute the maximum gradient in a North (N) minus South (S) direction for each row using a "S" and "N" value for each min or max in the rows below. I'm not sure that this is the best pythonic way to do this. My df "ms" looks like this:
minSlats minNlats maxSlats maxNlats
0 57839.4 54917.0 57962.6 56979.9
0 57763.2 55656.7 58120.0 57766.0
0 57905.2 54968.6 58014.3 57031.6
0 57796.0 54810.2 57969.0 56848.2
0 57820.5 55156.4 58019.5 57273.2
0 57542.7 54330.6 58057.6 56145.1
0 57829.8 54755.4 57978.8 56777.5
0 57796.0 54810.2 57969.0 56848.2
0 57639.4 54286.6 58087.6 56140.1
0 57653.3 56182.7 57996.5 57975.8
0 57665.1 56048.3 58069.7 58031.4
0 57559.9 57121.3 57890.8 58043.0
0 57689.7 55155.5 57959.4 56440.8
0 57649.4 56076.5 58043.0 58037.4
0 57603.9 56290.0 57959.8 57993.9
My loop structure looks like this:
J = len(ms)
grad = pd.DataFrame()
for i in range(J):
if ms.maxSlats.iloc[i] > ms.maxNlats.iloc[i]:
gr = ( ms.maxSlats.iloc[i] - ms.minNlats.iloc[i] ) * -1
grad[gr] = [i+1, i]
elif ms.maxNlats.iloc[i] > ms.maxSlats.iloc[i]:
gr = ms.maxNlats.iloc[i] - ms.minSlats.iloc[i]
grad[gr] = [i+1, i]
grad = grad.T # need to transpose
print(grad)
I obtain the correct answer but I'm wondering if there is a cleaner way to do this to obtain the same answer below:
grad.T
Out[317]:
0 1
-3045.6 1 0
-2463.3 2 1
-3045.7 3 2
-3158.8 8 7
-2863.1 5 4
-3727.0 6 5
-3223.4 7 6
-3801.0 9 8
-1813.8 10 9
-2021.4 11 10
483.1 12 11
-2803.9 13 12
-1966.5 14 13
390.0 15 14
thank you,
Use np.where to compute gradient and keep only last duplicated index.
grad = np.where(ms.maxSlats > ms.maxNlats, (ms.maxSlats - ms.minNlats) * -1,
ms.maxNlats - ms.minSlats)
df = pd.DataFrame({'A': pd.RangeIndex(1, len(ms)+1),
'B': pd.RangeIndex(0, len(ms))},
index=grad)
df = df[~df.index.duplicated(keep='last')]
>>> df
A B
-3045.6 1 0
-2463.3 2 1
-3045.7 3 2
-2863.1 5 4
-3727.0 6 5
-3223.4 7 6
-3158.8 8 7
-3801.0 9 8
-1813.8 10 9
-2021.4 11 10
483.1 12 11
-2803.9 13 12
-1966.5 14 13
390.0 15 14
I have a dataframe with 1600 columns.
The dataframe df looks like where the column names are 1, 3 , 2:
Row Labels 1 3 2
41730Type1 9 6 5
41730Type2 14 12 20
41731Type1 2 15 5
41731Type2 3 20 12
41732Type1 8 10 5
41732Type2 8 18 16
I need to create the following dataframe df2 pythonically:
Row Labels (1, 2) (1, 3) (2, 3)
41730Type1 -4 -3 1
41730Type2 6 -2 -8
41731Type1 3 13 10
41731Type2 9 17 8
41732Type1 -3 2 5
41732Type2 8 10 2
where e.g. column (1, 2) is created by df[2] - df[1]
The column names for df2 are created by pairing the column headers of df1 such that second element of each name is greater than first e.g. (1, 2), (1, 3), (2, 3)
The second challenge is can pandas dataframe support 1.3 million columns?
We can do combinations for the column , then create the dict and concat it back
import itertools
l=itertools.combinations(df.columns,2)
d={'{0[0]}|{0[1]}'.format(x) : df[x[0]]-df[x[1]] for x in [*l] }
newdf=pd.concat(d,axis=1)
1|3 1|2 3|2
RowLabels
41730Type1 3 4 1
41730Type2 2 -6 -8
41731Type1 -13 -3 10
41731Type2 -17 -9 8
41732Type1 -2 3 5
41732Type2 -10 -8 2
itertools combinations seems the obvious choice, same as #YOBEN_S, a different route to the solution, using numpy arrays and dictionary
from itertools import combinations
new_data = combinations(df.to_numpy().T,2)
new_cols = combinations(df.columns, 2)
result = {key : np.subtract(arr1,arr2)
if key[0] > key[1]
else np.subtract(arr2,arr1)
for (arr1, arr2), key
in zip(new_data,new_cols)}
outcome = pd.DataFrame.from_dict(result,orient='index').sort_index().T
outcome
(1, 2) (1, 3) (3, 2)
0 -4 -3 1
1 6 -2 -8
2 3 13 10
3 9 17 8
4 -3 2 5
5 8 10 2
I have a 11729 rows × 8 columns DataFrame, I'd like to convert it to a 11729 × 30 × 8 matrix with MultiIndex, which 30 means every 30 lines of 11729 rows from 0 to 11728 - 30
for a shorter example:
the origin 2d DataFrame looks like:
col0 col1
0 1 2
1 3 4
2 5 6
3 7 8
4 9 10
the 3d MultiIndex DataFrame which I want to get would looks like:
col0 col1
0 c0 1 2
c1 3 4
c2 5 6
1 c0 3 4
c1 5 6
c2 7 8
2 c0 5 6
c1 7 8
c2 9 10
which means (0,c0)~(0,c2) from 0~2 rows in origin DataFrame, (1,c0)~(1,c2) from 1~3 rows in origin DataFrame, (2,c0)~(2,c2) from 2~4 rows in origin DataFrame.
I'm using the following code to convert the origin 2d DataFrame to MultiIndex 3d DataFrame:
multi_index = pd.MultiIndex(levels=[[],[]],
labels=[[],[]],
names=['', ''])
df = pd.DataFrame(index=multi_index, columns=origin_df.columns)
for i in range(n):
for j in range(i, len(origin_df) - (n - i)):
print("i{}/n{},j{}".format(i, n, j)) # print progress
df.loc[(j, 'c%d' % i), :] = origin_df.loc[origin_df.index[j]].tolist()
for i in range(n, len(origin_df)):
df.loc[(i, 'y'), :] = origin_df.loc[origin_df.index[i]].tolist()
return df
My problem is the insertion speed is getting slow while running.
At first the progress output is fast, but getting slower and slower.
How could I optimize this operation?
You should not be adding one by one. Here's what I would do:
# toy data:
df = pd.DataFrame(np.arange(11792*8).reshape(-1,8));
window = 30
new_len = len(df) - window + 1
# create new dataframe, ignoring the index
new_df = pd.concat(df.iloc[i:i+window] for i in range(new_len))
# modify the index
new_df.index = pd.MultiIndex.from_product([np.arange(new_len), [f'c{i}' for i in range(window)]])
That took about 1 second on a 6600k. With your sample data, the output is:
col0 col1
0 c0 1 2
c1 3 4
c2 5 6
1 c0 3 4
c1 5 6
c2 7 8
2 c0 5 6
c1 7 8
c2 9 10
Let's have a dataframe df and a series s1 in pandas
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randn(10000,1000))
s1 = pd.Series(range(0,10000))
How can I modify df so that the column 42 become equal to s1?
How can I modify df so that the columns between 42 and 442 become equal to s1?
I would like to know the simplest way to do that but also a way to do that in place.
I think you need first same length Series with DataFrame, here 20:
np.random.seed(456)
df = pd.DataFrame(np.random.randn(20,10))
#print (df)
s1 = pd.Series(range(0,20))
#print (s1)
#set column by Series
df[8] = s1
#set Series to range of columns
cols = df.loc[:, 3:5].columns
df[cols] = pd.concat([s1] * len(cols), axis=1)
print (df)
0 1 2 3 4 5 6 7 8 9
0 -0.668129 -0.498210 0.618576 0 0 0 0.301966 0.449483 0 -0.315231
1 -2.015971 -1.130231 -1.111846 1 1 1 1.915676 0.920348 1 1.157552
2 -0.106208 -0.088752 -0.971485 2 2 2 -0.366948 -0.301085 2 1.141635
3 -1.309529 -0.274381 0.864837 3 3 3 0.670294 0.086347 3 -1.212503
4 0.120359 -0.358880 1.199936 4 4 4 0.389167 1.201631 4 0.445432
5 -1.031109 0.067133 -1.213451 5 5 5 -0.636896 0.013802 5 1.726135
6 -0.491877 0.254206 -0.268168 6 6 6 0.671070 -0.633645 6 1.813671
7 0.080433 -0.882443 1.152671 7 7 7 0.249225 1.385407 7 1.010374
8 0.307274 0.806150 0.071719 8 8 8 1.133853 -0.789922 8 -0.286098
9 -0.767206 1.094445 1.603907 9 9 9 0.083149 2.322640 9 0.396845
10 -0.740018 -0.853377 -2.039522 10 10 10 0.764962 -0.472048 10 -0.071255
11 -0.238565 1.077573 2.143252 11 11 11 1.542892 2.572560 11 -0.803516
12 -0.139521 -0.992107 -0.892619 12 12 12 0.259612 -0.661760 12 -1.508976
13 -1.077001 0.381962 0.205388 13 13 13 -0.023986 -1.293080 13 1.846402
14 -0.714792 -0.728496 -0.127079 14 14 14 0.606065 -2.320500 14 -0.992798
15 -0.127113 -0.563313 -0.101387 15 15 15 0.647325 -0.816023 15 -0.309938
16 -1.151304 -1.673719 0.074930 16 16 16 -0.392157 0.736714 16 1.142983
17 -1.247396 -0.471524 1.173713 17 17 17 -0.005391 0.426134 17 0.781832
18 -0.325111 0.579248 0.040363 18 18 18 0.361926 0.036871 18 0.581314
19 -1.057501 -1.814500 0.109628 19 19 19 -1.738658 -0.061883 19 0.989456
Timings
Another solutions, but it seems concat solution is fastest:
np.random.seed(456)
df = pd.DataFrame(np.random.randn(1000,1000))
#print (df)
s1 = pd.Series(range(0,1000))
#print (s1)
#set column by Series
df[8] = s1
#set Series to range of columns
cols = df.loc[:, 42:442].columns
print (df)
In [310]: %timeit df[cols] = np.broadcast_to(s1.values[:, np.newaxis], (len(df),len(cols)))
1 loop, best of 3: 202 ms per loop
In [311]: %timeit df[cols] = np.repeat(s1.values[:, np.newaxis], len(cols), axis=1)
1 loop, best of 3: 208 ms per loop
In [312]: %timeit df[cols] = np.array([s1.values]*len(cols)).transpose()
10 loops, best of 3: 175 ms per loop
In [313]: %timeit df[cols] = pd.concat([s1] * len(cols), axis=1)
10 loops, best of 3: 53.8 ms per loop
First, let me set the stage.
I start with a pandas dataframe klmn, that looks like this:
In [15]: klmn
Out[15]:
K L M N
0 0 a -1.374201 35
1 0 b 1.415697 29
2 0 a 0.233841 18
3 0 b 1.550599 30
4 0 a -0.178370 63
5 0 b -1.235956 42
6 0 a 0.088046 2
7 0 b 0.074238 84
8 1 a 0.469924 44
9 1 b 1.231064 68
10 2 a -0.979462 73
11 2 b 0.322454 97
Next I split klmn into two dataframes, klmn0 and klmn1, according to the value in the 'K' column:
In [16]: k0 = klmn.groupby(klmn['K'] == 0)
In [17]: klmn0, klmn1 = [klmn.ix[k0.indices[tf]] for tf in (True, False)]
In [18]: klmn0, klmn1
Out[18]:
( K L M N
0 0 a -1.374201 35
1 0 b 1.415697 29
2 0 a 0.233841 18
3 0 b 1.550599 30
4 0 a -0.178370 63
5 0 b -1.235956 42
6 0 a 0.088046 2
7 0 b 0.074238 84,
K L M N
8 1 a 0.469924 44
9 1 b 1.231064 68
10 2 a -0.979462 73
11 2 b 0.322454 97)
Finally, I compute the mean of the M column in klmn0, grouped by the value in the L column:
In [19]: m0 = klmn0.groupby('L')['M'].mean(); m0
Out[19]:
L
a -0.307671
b 0.451144
Name: M
Now, my question is, how can I subtract m0 from the M column of the klmn1 sub-dataframe, respecting the value in the L column? (By this I mean that m0['a'] gets subtracted from the M column of each row in klmn1 that has 'a' in the L column, and likewise for m0['b'].)
One could imagine doing this in a way that replaces the the values in the M column of klmn1 with the new values (after subtracting the value from m0). Alternatively, one could imagine doing this in a way that leaves klmn1 unchanged, and instead produces a new dataframe klmn11 with an updated M column. I'm interested in both approaches.
If you reset the index of your klmn1 dataframe to be that of the column L, then your dataframe will automatically align the indices with any series you subtract from it:
In [1]: klmn1.set_index('L')['M'] - m0
Out[1]:
L
a 0.777595
a -0.671791
b 0.779920
b -0.128690
Name: M
Option #1:
df1.subtract(df2, fill_value=0)
Option #2:
df1.subtract(df2, fill_value=None)