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How to add column for every month and generate number i.e. 1,2,3..etc

I have a huge csv file of dataframe. However, I don't have the date column. I only have the sales for every month from Jan-2022 until Dec-2034. Below is the example of my dataframe:
import pandas as pd
data = [[6661, 'Mobile Phone', 43578, 5000, 78564, 52353, 67456, 86965, 43634, 32546, 56332, 58944, 98878, 68588, 43634, 3463, 74533, 73733, 64436, 45426, 57333, 89762, 4373, 75457, 74845, 86843, 59957, 74563, 745335, 46342, 463473, 52352, 23622],
[6672, 'Play Station', 4475, 2546, 5757, 2352, 57896, 98574, 53536, 56533, 88645, 44884, 76585, 43575, 74573, 75347, 57573, 5736, 53737, 35235, 5322, 54757, 74573, 75473, 77362, 21554, 73462, 74736, 1435, 4367, 63462, 32362, 56332],
[6631, 'Laptop', 35347, 36376, 164577, 94584, 78675, 76758, 75464, 56373, 56343, 54787, 7658, 76584, 47347, 5748, 8684, 75373, 57573, 26626, 25632, 73774, 847373, 736646, 847457, 57346, 43732, 347346, 75373, 6473, 85674, 35743, 45734],
[6600, 'Camera', 14365, 60785, 25436, 46747, 75456, 97644, 63573, 56433, 25646, 32548, 14325, 64748, 68458, 46537, 7537, 46266, 7457, 78235, 46223, 8747, 67453, 4636, 3425, 4636, 352236, 6622, 64625, 36346, 46346, 35225, 6436],
[6643, 'Lamp', 324355, 143255, 696954, 97823, 43657, 66686, 56346, 57563, 65734, 64484, 87685, 54748, 9868, 573, 73472, 5735, 73422, 86352, 5325, 84333, 7473, 35252, 7547, 73733, 7374, 32266, 654747, 85743, 57333, 46346, 46266]]
ds = pd.DataFrame(data, columns = ['ID', 'Product', 'SalesJan-22', 'SalesFeb-22', 'SalesMar-22', 'SalesApr-22', 'SalesMay-22', 'SalesJun-22', 'SalesJul-22', 'SalesAug-22', 'SalesSep-22', 'SalesOct-22', 'SalesNov-22', 'SalesDec-22', 'SalesJan-23', 'SalesFeb-23', 'SalesMar-23', 'SalesApr-23', 'SalesMay-23', 'SalesJun-23', 'SalesJul-23', 'SalesAug-23', 'SalesSep-23', 'SalesOct-23', 'SalesNov-23', 'SalesDec-23', 'SalesJan-24', 'SalesFeb-24', 'SalesMar-24', 'SalesApr-24', 'SalesMay-24', 'SalesJun-24', 'SalesJul-24']
Since I have more than 10 monthly sales column, I want to loop the date after each of the month sales column. Then, the first 6 months will generate number 1, while the next 12 months will generate number 2, then another 12 months will generate number 3, another subsequent 12 months will generate number 4 and so on.
Below shows the sample of result that I want:
Is there any way to perform the loop and adding the date column beside each of the sales month?

Here is the simplest approach I can think of:
for i, col in enumerate(ds.columns[2:]):
ds.insert(2 * i + 2, col.removeprefix("Sales"), (i - 6) // 12 + 2)

Here is a vectorial approach (using insert repeatedly is inefficient):
# convert (valid) columns to datetime
cols = pd.to_datetime(ds.columns, format='Sales%b-%y', errors='coerce')
# identify valid dates
m = cols.notna()
# get year
y = cols[m].year
# calculate number (1 for first 6 months, then +1 per 12 months)
num = ((cols[m].month+12*(y-y.min()))+5)//12+1
# slice dates columns, assign the number, rename
df2 = (ds.loc[:, m].assign(**dict(zip(ds.columns[m], num)))
.rename(columns=lambda x: x[5:])
)
# get new order of columns
idx = np.r_[np.zeros((~m).sum()), np.tile(np.arange(m.sum()), 2)+1]
# concat and reorder
out = pd.concat([ds, df2], axis=1).iloc[:, np.argsort(idx)]
print(out)
output:
ID Product SalesJan-22 Jan-22 SalesFeb-22 Feb-22 SalesMar-22 Mar-22 SalesApr-22 Apr-22 SalesMay-22 May-22 SalesJun-22 Jun-22 SalesJul-22 Jul-22 SalesAug-22 Aug-22 Sep-22 SalesSep-22 Oct-22 SalesOct-22 SalesNov-22 Nov-22 Dec-22 SalesDec-22 Jan-23 SalesJan-23 Feb-23 SalesFeb-23 SalesMar-23 Mar-23 Apr-23 SalesApr-23 SalesMay-23 May-23 SalesJun-23 Jun-23 Jul-23 SalesJul-23 SalesAug-23 Aug-23 Sep-23 SalesSep-23 SalesOct-23 Oct-23 Nov-23 SalesNov-23 Dec-23 SalesDec-23 Jan-24 SalesJan-24 Feb-24 SalesFeb-24 Mar-24 SalesMar-24 Apr-24 SalesApr-24 May-24 SalesMay-24 SalesJun-24 Jun-24 SalesJul-24 Jul-24
0 6661 Mobile Phone 43578 1 5000 1 78564 1 52353 1 67456 1 86965 1 43634 2 32546 2 2 56332 2 58944 98878 2 2 68588 2 43634 2 3463 74533 2 2 73733 64436 2 45426 2 3 57333 89762 3 3 4373 75457 3 3 74845 3 86843 3 59957 3 74563 3 745335 3 46342 3 463473 52352 3 23622 4
1 6672 Play Station 4475 1 2546 1 5757 1 2352 1 57896 1 98574 1 53536 2 56533 2 2 88645 2 44884 76585 2 2 43575 2 74573 2 75347 57573 2 2 5736 53737 2 35235 2 3 5322 54757 3 3 74573 75473 3 3 77362 3 21554 3 73462 3 74736 3 1435 3 4367 3 63462 32362 3 56332 4
2 6631 Laptop 35347 1 36376 1 164577 1 94584 1 78675 1 76758 1 75464 2 56373 2 2 56343 2 54787 7658 2 2 76584 2 47347 2 5748 8684 2 2 75373 57573 2 26626 2 3 25632 73774 3 3 847373 736646 3 3 847457 3 57346 3 43732 3 347346 3 75373 3 6473 3 85674 35743 3 45734 4
3 6600 Camera 14365 1 60785 1 25436 1 46747 1 75456 1 97644 1 63573 2 56433 2 2 25646 2 32548 14325 2 2 64748 2 68458 2 46537 7537 2 2 46266 7457 2 78235 2 3 46223 8747 3 3 67453 4636 3 3 3425 3 4636 3 352236 3 6622 3 64625 3 36346 3 46346 35225 3 6436 4
4 6643 Lamp 324355 1 143255 1 696954 1 97823 1 43657 1 66686 1 56346 2 57563 2 2 65734 2 64484 87685 2 2 54748 2 9868 2 573 73472 2 2 5735 73422 2 86352 2 3 5325 84333 3 3 7473 35252 3 3 7547 3 73733 3 7374 3 32266 3 654747 3 85743 3 57333 46346 3 46266 4

Here's a little solution : (I put the year unstead of your 1, 2, ... incrementation since i thought it is more representative, but you can change it easily)
idx_counter = 0
for idx, col in enumerate(ds.columns):
if col.startswith('Sales'):
date = col.replace('Sales', '')
year = col.split('-')[1]
ds.insert(loc=idx + 1 + idx_counter, column=date, value=[year] * ds.shape[0])
idx_counter += 1
output:
ID Product SalesJan-22 Jan-22 SalesFeb-22 Feb-22 SalesMar-22 Mar-22 SalesApr-22 Apr-22 ... SalesMar-24 Mar-24 SalesApr-24 Apr-24 SalesMay-24 May-24 SalesJun-24 Jun-24 SalesJul-24 Jul-24
0 6661 Mobile Phone 43578 22 5000 22 78564 22 52353 22 ... 745335 24 46342 24 463473 24 52352 24 23622 24
1 6672 Play Station 4475 22 2546 22 5757 22 2352 22 ... 1435 24 4367 24 63462 24 32362 24 56332 24
2 6631 Laptop 35347 22 36376 22 164577 22 94584 22 ... 75373 24 6473 24 85674 24 35743 24 45734 24
3 6600 Camera 14365 22 60785 22 25436 22 46747 22 ... 64625 24 36346 24 46346 24 35225 24 6436 24
4 6643 Lamp 324355 22 143255 22 696954 22 97823 22 ... 654747 24 85743 24 57333 24 46346 24 46266 24

This should do the trick.
import math
new_cols = []
old_cols = [x for x in df.columns if x.startswith('Sales')]
for i, col in enumerate(old_cols):
new_cols.append(col[5:])
if i < 6:
val = 1
else:
val = ((i+6)/12)+1
df[col[5:]] = math.floor(val)
df[['ID', 'Product'] + [x for y in zip(old_cols, new_cols) for x in y]]

Find the second largest date in df python [duplicate]

I am using pandas to analyse some election results. I have a DF, Results, which has a row for each constituency and columns representing the votes for the various parties (over 100 of them):
In[60]: Results.columns
Out[60]:
Index(['Constituency', 'Region', 'Country', 'ID', 'Type', 'Electorate',
'Total', 'Unnamed: 9', '30-50', 'Above',
...
'WP', 'WRP', 'WVPTFP', 'Yorks', 'Young', 'Zeb', 'Party', 'Votes',
'Share', 'Turnout'],
dtype='object', length=147)
So...
In[63]: Results.head()
Out[63]:
Constituency Region Country ID Type \
PAID
1 Aberavon Wales Wales W07000049 County
2 Aberconwy Wales Wales W07000058 County
3 Aberdeen North Scotland Scotland S14000001 Burgh
4 Aberdeen South Scotland Scotland S14000002 Burgh
5 Aberdeenshire West & Kincardine Scotland Scotland S14000058 County
Electorate Total Unnamed: 9 30-50 Above ... WP WRP WVPTFP \
PAID ...
1 49821 31523 NaN NaN NaN ... NaN NaN NaN
2 45525 30148 NaN NaN NaN ... NaN NaN NaN
3 67745 43936 NaN NaN NaN ... NaN NaN NaN
4 68056 48551 NaN NaN NaN ... NaN NaN NaN
5 73445 55196 NaN NaN NaN ... NaN NaN NaN
Yorks Young Zeb Party Votes Share Turnout
PAID
1 NaN NaN NaN Lab 15416 0.489040 0.632725
2 NaN NaN NaN Con 12513 0.415052 0.662230
3 NaN NaN NaN SNP 24793 0.564298 0.648550
4 NaN NaN NaN SNP 20221 0.416490 0.713398
5 NaN NaN NaN SNP 22949 0.415773 0.751528
[5 rows x 147 columns]
The per-constituency results for each party are given in the columns Results.ix[:, 'Unnamed: 9': 'Zeb']
I can find the winning party (i.e. the party which polled highest number of votes) and the number of votes it polled using:
RawResults = Results.ix[:, 'Unnamed: 9': 'Zeb']
Results['Party'] = RawResults.idxmax(axis=1)
Results['Votes'] = RawResults.max(axis=1).astype(int)
But, I also need to know how many votes the second-place party got (and ideally its index/name). So is there any way in pandas to return the second highest value/index in a set of columns for each row?

To get the highest values of a column, you can use nlargest() :
df['High'].nlargest(2)
The above will give you the 2 highest values of column High.
You can also use nsmallest() to get the lowest values.

Here is a NumPy solution:
In [120]: df
Out[120]:
a b c d e f g h
0 1.334444 0.322029 0.302296 -0.841236 -0.360488 -0.860188 -0.157942 1.522082
1 2.056572 0.991643 0.160067 -0.066473 0.235132 0.533202 1.282371 -2.050731
2 0.955586 -0.966734 0.055210 -0.993924 -0.553841 0.173793 -0.534548 -1.796006
3 1.201001 1.067291 -0.562357 -0.794284 -0.554820 -0.011836 0.519928 0.514669
4 -0.243972 -0.048144 0.498007 0.862016 1.284717 -0.886455 -0.757603 0.541992
5 0.739435 -0.767399 1.574173 1.197063 -1.147961 -0.903858 0.011073 -1.404868
6 -1.258282 -0.049719 0.400063 0.611456 0.443289 -1.110945 1.352029 0.215460
7 0.029121 -0.771431 -0.285119 -0.018216 0.408425 -1.458476 -1.363583 0.155134
8 1.427226 -1.005345 0.208665 -0.674917 0.287929 -1.259707 0.220420 -1.087245
9 0.452589 0.214592 -1.875423 0.487496 2.411265 0.062324 -0.327891 0.256577
In [121]: np.sort(df.values)[:,-2:]
Out[121]:
array([[ 1.33444404, 1.52208164],
[ 1.28237078, 2.05657214],
[ 0.17379254, 0.95558613],
[ 1.06729107, 1.20100071],
[ 0.86201603, 1.28471676],
[ 1.19706331, 1.57417327],
[ 0.61145573, 1.35202868],
[ 0.15513379, 0.40842477],
[ 0.28792928, 1.42722604],
[ 0.48749578, 2.41126532]])
or as a pandas Data Frame:
In [122]: pd.DataFrame(np.sort(df.values)[:,-2:], columns=['2nd-largest','largest'])
Out[122]:
2nd-largest largest
0 1.334444 1.522082
1 1.282371 2.056572
2 0.173793 0.955586
3 1.067291 1.201001
4 0.862016 1.284717
5 1.197063 1.574173
6 0.611456 1.352029
7 0.155134 0.408425
8 0.287929 1.427226
9 0.487496 2.411265
or a faster solution from #Divakar:
In [6]: df
Out[6]:
a b c d e f g h
0 0.649517 -0.223116 0.264734 -1.121666 0.151591 -1.335756 -0.155459 -2.500680
1 0.172981 1.233523 0.220378 1.188080 -0.289469 -0.039150 1.476852 0.736908
2 -1.904024 0.109314 0.045741 -0.341214 -0.332267 -1.363889 0.177705 -0.892018
3 -2.606532 -0.483314 0.054624 0.979734 0.205173 0.350247 -1.088776 1.501327
4 1.627655 -1.261631 0.589899 -0.660119 0.742390 -1.088103 0.228557 0.714746
5 0.423972 -0.506975 -0.783718 -2.044002 -0.692734 0.980399 1.007460 0.161516
6 -0.777123 -0.838311 -1.116104 -0.433797 0.599724 -0.884832 -0.086431 -0.738298
7 1.131621 1.218199 0.645709 0.066216 -0.265023 0.606963 -0.194694 0.463576
8 0.421164 0.626731 -0.547738 0.989820 -1.383061 -0.060413 -1.342769 -0.777907
9 -1.152690 0.696714 -0.155727 -0.991975 -0.806530 1.454522 0.788688 0.409516
In [7]: a = df.values
In [8]: a[np.arange(len(df))[:,None],np.argpartition(-a,np.arange(2),axis=1)[:,:2]]
Out[8]:
array([[ 0.64951665, 0.26473378],
[ 1.47685226, 1.23352348],
[ 0.17770473, 0.10931398],
[ 1.50132666, 0.97973383],
[ 1.62765464, 0.74238959],
[ 1.00745981, 0.98039898],
[ 0.5997243 , -0.0864306 ],
[ 1.21819904, 1.13162068],
[ 0.98982033, 0.62673128],
[ 1.45452173, 0.78868785]])

Here is an interesting approach. What if we replace the maximum value with the minimum value and calculate. Although it is a quick hack and, not recommended!
first_highest_value_index = df.idxmax()
second_highest_value_index = df.replace(df.max(),df(min)).idxmax()
first_highest_value = df[first_highest_value_index]
second_highest_value = df[second_highest_value_index]

You could just sort your results, such that the first rows will contain the max. Then you can simply use indexing to get the first n places.
RawResults = Results.ix[:, 'Unnamed: 9': 'Zeb'].sort_values(by='votes', ascending=False)
RawResults.iloc[0, :] # First place
RawResults.iloc[1, :] # Second place
RawResults.iloc[n, :] # nth place

Here is a solution using nlargest function:
>>> df
a b c
0 4 20 2
1 5 10 2
2 3 40 5
3 1 50 10
4 2 30 15
>>> def give_largest(col,n):
... largest = col.nlargest(n).reset_index(drop = True)
... data = [x for x in largest]
... index = [f'{i}_largest' for i in range(1,len(largest)+1)]
... return pd.Series(data,index=index)
...
...
>>> def n_largest(df, axis, n):
... '''
... Function to return the n-largest value of each
... column/row of the input DataFrame.
... '''
... return df.apply(give_largest, axis = axis, n = n)
...
>>> n_largest(df,axis = 1, n = 2)
1_largest 2_largest
0 20 4
1 10 5
2 40 5
3 50 10
4 30 15
>>> n_largest(df,axis = 0, n = 2)
a b c
1_largest 5 50 15
2_largest 4 40 10

import numpy as np
import pandas as pd
df = pd.DataFrame({
'a': [4, 5, 3, 1, 2],
'b': [20, 10, 40, 50, 30],
'c': [25, 20, 5, 15, 10]
})
def second_largest(df):
return (df.nlargest(2).min())
print(df.apply(second_largest))
a 4
b 40
c 20
dtype: int64

df
a b c d e f g h
0 1.334444 0.322029 0.302296 -0.841236 -0.360488 -0.860188 -0.157942 1.522082
1 2.056572 0.991643 0.160067 -0.066473 0.235132 0.533202 1.282371 -2.050731
2 0.955586 -0.966734 0.055210 -0.993924 -0.553841 0.173793 -0.534548 -1.796006
3 1.201001 1.067291 -0.562357 -0.794284 -0.554820 -0.011836 0.519928 0.514669
4 -0.243972 -0.048144 0.498007 0.862016 1.284717 -0.886455 -0.757603 0.541992
5 0.739435 -0.767399 1.574173 1.197063 -1.147961 -0.903858 0.011073 -1.404868
6 -1.258282 -0.049719 0.400063 0.611456 0.443289 -1.110945 1.352029 0.215460
7 0.029121 -0.771431 -0.285119 -0.018216 0.408425 -1.458476 -1.363583 0.155134
8 1.427226 -1.005345 0.208665 -0.674917 0.287929 -1.259707 0.220420 -1.087245
9 0.452589 0.214592 -1.875423 0.487496 2.411265 0.062324 -0.327891 0.256577
tranpose and use nlargest in a for loop to get the results order by each line:
df1=df.T
results=list()
for col in df1.columns: results.append(df1[col].nlargest(len(df.columns))
the results var is a list of pandas objects, where the first item on the list will be the df's first row sorted in descending order and so on. Since each item on the list is a pandas object, it carries df's column as index (it was transposed), so you will get the values and the df's columns name of each row sorted
results
[h 1.522082
a 1.334444
b 0.322029
c 0.302296
g -0.157942
e -0.360488
d -0.841236
f -0.860188
Name: 0, dtype: float64,
a 2.056572
g 1.282371
b 0.991643
f 0.533202
e 0.235132
c 0.160067
d -0.066473
h -2.050731
Name: 1, dtype: float64,
....

Permute the last column only using python

I have tried
import numpy as np
import paandas as pd
df1 = pd.read_csv('Trans_ZS_Control_64')
df = df1.apply(np.random.permutation)
This permuted the entire data but i want to permute the value of last column only upto 100 times.
How do I proceed for this.
Input Data
0.051424 0.535067 0.453645 0.161857 -0.017189 -0.001850 0.481861 0.711553 0.083747 0.583215 ... 0.541249 0.048360 0.370659 0.890987 0.723995 -0.014502 1.295998 0.150719 0.885673 1
-0.067129 0.673519 0.212407 0.195590 -0.034868 -0.231418 0.480255 0.643735 -0.054970 0.511684 ... 0.524751 0.206757 0.578314 0.614924 0.230632 -0.074980 0.747007 0.047382 1.413796 1
-0.994564 -0.881392 -1.150127 -0.589125 -0.663275 -0.955622 -1.088923 -1.210452 -0.922861 -0.689851 ... -0.442188 -1.294110 -0.934985 -1.085506 -0.808874 -0.779111 -1.032484 -1.026208 -0.248476 1
-0.856323 -0.619472 -1.113073 -0.691285 -0.515566 -1.080643 -0.513487 -0.912825 -1.010245 -0.870335 ... -0.941149 -1.012917 -1.647812 -0.654150 -0.735166 -0.984510 -0.949168 -1.052115 -0.052492 1
.. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
-0.145871 0.832727 -0.003379 0.327546 1.409891 0.840316 0.700613 0.184477 0.962488 0.200397 ... -0.337530 0.988197 0.751663 0.480126 0.663302 -0.522189 0.512744 -0.063515 1.125415 0
0.972923 0.857971 -0.195672 0.190443 1.652155 0.763571 0.604728 0.115846 0.942269 0.453387 ... -0.522834 0.985770 0.570573 0.438632 0.737030 -0.445704 0.387023 0.031686 1.266407 0
0.281427 1.060266 0.172624 0.258344 1.544505 0.859777 0.689876 0.439106 0.955198 0.335523 ... -0.442724 0.929343 0.707809 0.290670 0.688595 -0.438848 0.762695 -0.105879 0.944989 0
0.096601 1.112720 0.105861 -0.133927 1.526764 0.773759 0.661673 -0.007070 0.884725 0.478899 ... -0.404426 0.966646 0.994733 0.418965 0.862612 -0.174580 0.407309 -0.010520 1.044876 0
-0.298780 1.036580 0.131270 0.019826 1.381928 0.879310 0.619529 -0.022691 0.982060 -0.039355 ... -0.702316 0.985320 0.457767 0.215949 0.752685 -0.405060 0.166226 -0.216972 1.021018 0
Expected output: randomly permute the last column
0.051424 0.535067 0.453645 0.161857 -0.017189 -0.001850 0.481861 0.711553 0.083747 0.583215 ... 0.541249 0.048360 0.370659 0.890987 0.723995 -0.014502 1.295998 0.150719 0.885673 0
-0.067129 0.673519 0.212407 0.195590 -0.034868 -0.231418 0.480255 0.643735 -0.054970 0.511684 ... 0.524751 0.206757 0.578314 0.614924 0.230632 -0.074980 0.747007 0.047382 1.413796 0
-0.994564 -0.881392 -1.150127 -0.589125 -0.663275 -0.955622 -1.088923 -1.210452 -0.922861 -0.689851 ... -0.442188 -1.294110 -0.934985 -1.085506 -0.808874 -0.779111 -1.032484 -1.026208 -0.248476 1
-0.856323 -0.619472 -1.113073 -0.691285 -0.515566 -1.080643 -0.513487 -0.912825 -1.010245 -0.870335 ... -0.941149 -1.012917 -1.647812 -0.654150 -0.735166 -0.984510 -0.949168 -1.052115 -0.052492 1
.. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
-0.145871 0.832727 -0.003379 0.327546 1.409891 0.840316 0.700613 0.184477 0.962488 0.200397 ... -0.337530 0.988197 0.751663 0.480126 0.663302 -0.522189 0.512744 -0.063515 1.125415 0
0.972923 0.857971 -0.195672 0.190443 1.652155 0.763571 0.604728 0.115846 0.942269 0.453387 ... -0.522834 0.985770 0.570573 0.438632 0.737030 -0.445704 0.387023 0.031686 1.266407 1
0.281427 1.060266 0.172624 0.258344 1.544505 0.859777 0.689876 0.439106 0.955198 0.335523 ... -0.442724 0.929343 0.707809 0.290670 0.688595 -0.438848 0.762695 -0.105879 0.944989 0
0.096601 1.112720 0.105861 -0.133927 1.526764 0.773759 0.661673 -0.007070 0.884725 0.478899 ... -0.404426 0.966646 0.994733 0.418965 0.862612 -0.174580 0.407309 -0.010520 1.044876 0
-0.298780 1.036580 0.131270 0.019826 1.381928 0.879310 0.619529 -0.022691 0.982060 -0.039355 ... -0.702316 0.985320 0.457767 0.215949 0.752685 -0.405060 0.166226 -0.216972 1.021018 1

No sure if this is what you meant, but you could do it like this
import pandas as pd
import numpy as np
v = np.arange(0,10)
df = pd.DataFrame({'c1': v, 'c2': v, 'c3': v})
df
this would create the following df:
c1 c2 c3
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
to permute the last column you could run this:
df1 = df.copy()
df1.c3 = np.random.permutation(df1.c3)
df1
resulting in:
c1 c2 c3
0 0 0 5
1 1 1 9
2 2 2 2
3 3 3 6
4 4 4 0
5 5 5 4
6 6 6 8
7 7 7 7
8 8 8 1
9 9 9 3
I hope it helps

Just create a dataframe from your last column and permutate that. It seems like permutating individual columns with apply doesn't work the way you expect it to.
import numpy as np
import pandas as pd
df = pd.read_csv('Trans_ZS_Control_64')
column_to_change = pd.DataFrame(df['last_column_name'])
for i in range(100):
column_to_change = column_to_change.apply(np.random.permutation)
df['last_column_name'] = column_to_change

How to create Traingular moving average in python using for loop

I use python pandas to caculate the following formula
(https://i.stack.imgur.com/XIKBz.png)
I do it in python like this :
EURUSD['SMA2']= EURUSD['Close']. rolling (2).mean()
EURUSD['TMA2']= ( EURUSD['Close'] + EURUSD[SMA2']) / 2
The proplem is long coding when i calculated TMA 100 , so i need to use " for loop " to easy change TMA period .
Thanks in advance
Edited :
I had found the code but there is an error :
values = []
for i in range(1,201): values.append(eurusd['Close']).rolling(window=i).mean() values.mean()

TMA is average of averages.
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.rand(10, 5))
print(df)
# df['mean0']=df.mean(0)
df['mean1']=df.mean(1)
print(df)
df['TMA'] = df['mean1'].rolling(window=10,center=False).mean()
print(df)
Or you can easily print it.
print(df["mean1"].mean())
Here is how it looks:
0 1 2 3 4
0 0.643560 0.412046 0.072525 0.618968 0.080146
1 0.018226 0.222212 0.077592 0.125714 0.595707
2 0.652139 0.907341 0.581802 0.021503 0.849562
3 0.129509 0.315618 0.711265 0.812318 0.757575
4 0.881567 0.455848 0.470282 0.367477 0.326812
5 0.102455 0.156075 0.272582 0.719158 0.266293
6 0.412049 0.527936 0.054381 0.587994 0.442144
7 0.063904 0.635857 0.244050 0.002459 0.423960
8 0.446264 0.116646 0.990394 0.678823 0.027085
9 0.951547 0.947705 0.080846 0.848772 0.699036
0 1 2 3 4 mean1
0 0.643560 0.412046 0.072525 0.618968 0.080146 0.365449
1 0.018226 0.222212 0.077592 0.125714 0.595707 0.207890
2 0.652139 0.907341 0.581802 0.021503 0.849562 0.602470
3 0.129509 0.315618 0.711265 0.812318 0.757575 0.545257
4 0.881567 0.455848 0.470282 0.367477 0.326812 0.500397
5 0.102455 0.156075 0.272582 0.719158 0.266293 0.303313
6 0.412049 0.527936 0.054381 0.587994 0.442144 0.404901
7 0.063904 0.635857 0.244050 0.002459 0.423960 0.274046
8 0.446264 0.116646 0.990394 0.678823 0.027085 0.451842
9 0.951547 0.947705 0.080846 0.848772 0.699036 0.705581
0 1 2 3 4 mean1 TMA
0 0.643560 0.412046 0.072525 0.618968 0.080146 0.365449 NaN
1 0.018226 0.222212 0.077592 0.125714 0.595707 0.207890 NaN
2 0.652139 0.907341 0.581802 0.021503 0.849562 0.602470 NaN
3 0.129509 0.315618 0.711265 0.812318 0.757575 0.545257 NaN
4 0.881567 0.455848 0.470282 0.367477 0.326812 0.500397 NaN
5 0.102455 0.156075 0.272582 0.719158 0.266293 0.303313 NaN
6 0.412049 0.527936 0.054381 0.587994 0.442144 0.404901 NaN
7 0.063904 0.635857 0.244050 0.002459 0.423960 0.274046 NaN
8 0.446264 0.116646 0.990394 0.678823 0.027085 0.451842 NaN
9 0.951547 0.947705 0.080846 0.848772 0.699036 0.705581 0.436115

Problem to implement count, groupby, np.repeat and agg with pandas

I have similar dataframe pandas:
df = pd.DataFrame({'x': np.random.rand(61800), 'y':np.random.rand(61800), 'z':np.random.rand(61800)})
I need to work out my dataset for the following result:
extract = df.assign(count=np.repeat(range(10),10)).groupby('count',as_index=False).agg(['mean','min', 'max'])
But if i use np.repeat(range(150),150)) i received this error:

This doesn't work because the .assign you're performing needs to have enough values to fit the original dataframe:
In [81]: df = pd.DataFrame({'x': np.random.rand(61800), 'y':np.random.rand(61800), 'z':np.random.rand(61800)})
In [82]: df.assign(count=np.repeat(range(10),10))
ValueError: Length of values does not match length of index
In this case, everything works fine if we do 10 groups repeated 6,180 times:
In [83]: df.assign(count=np.repeat(range(10),6180))
Out[83]:
x y z count
0 0.781364 0.996545 0.756592 0
1 0.609127 0.981688 0.626721 0
2 0.547029 0.167678 0.198857 0
3 0.184405 0.484623 0.219722 0
4 0.451698 0.535085 0.045942 0
... ... ... ... ...
61795 0.783192 0.969306 0.974836 9
61796 0.890720 0.286384 0.744779 9
61797 0.512688 0.945516 0.907192 9
61798 0.526564 0.165620 0.766733 9
61799 0.683092 0.976219 0.524048 9
[61800 rows x 4 columns]
In [84]: extract = df.assign(count=np.repeat(range(10),6180)).groupby('count',as_index=False).agg(['mean','min', 'max'])
In [85]: extract
Out[85]:
x y z
mean min max mean min max mean min max
count
0 0.502338 0.000230 0.999546 0.501603 0.000263 0.999842 0.503807 0.000113 0.999826
1 0.500392 0.000059 0.999979 0.499935 0.000012 0.999767 0.500114 0.000230 0.999811
2 0.498377 0.000023 0.999832 0.496921 0.000003 0.999475 0.502887 0.000028 0.999828
3 0.504970 0.000637 0.999680 0.500943 0.000256 0.999902 0.497370 0.000257 0.999969
4 0.501195 0.000290 0.999992 0.498617 0.000149 0.999779 0.497895 0.000022 0.999877
5 0.499476 0.000186 0.999956 0.503227 0.000308 0.999907 0.504688 0.000100 0.999756
6 0.495488 0.000378 0.999606 0.499893 0.000119 0.999740 0.495924 0.000031 0.999556
7 0.498443 0.000005 0.999417 0.495728 0.000262 0.999972 0.501255 0.000087 0.999978
8 0.494110 0.000014 0.999888 0.495197 0.000074 0.999970 0.493215 0.000166 0.999718
9 0.496333 0.000365 0.999307 0.502074 0.000110 0.999856 0.499164 0.000035 0.999927