RMS is calculated by moving average.
Inpandas, you can write like this.

import pandas as pd
import numpy as np
window_size = 3
rms = lambda d: np.sqrt ((d ** 2) .sum ()/d.size)
data = pd.read_csv ('in.csv')
data ['rms'] = data.iloc [:, 0] .rolling (window = window_size, min_periods = 1, center = True) .apply (rms)
data.to_csv ('out.csv', index = None)

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[Supplement]

A brief explanation.

First, if there is a one-dimensional array (pandas.Series) such as [1,2,3,4,5,6,7,8,9], RMS is

import pandas as pd
import numpy as np
row_data = pd.Series ([1,2,3,4,5,6,7,8,9])
res = np.sqrt ((row_data ** 2) .sum ()/row_data.size)
print (res)
# 5.627314338711377

can be obtained, so the expression to calculate this RMS is functionalized withlambda

import pandas as pd
import numpy as np
row_data = pd.Series ([1,2,3,4,5,6,7,8,9])
rms = lambda d: np.sqrt ((d ** 2) .sum ()/d.size)
res = rms (row_data)
print (res)
# 5.627314338711377

It becomes.

Next, in the case of the code of this question, it is required to calculate the RMS of a certain interval instead of calculating the RMS of the entire column, so this part isSeries.rolling ( ).
To explainSeries.rolling ()with a simple sample,for a one-dimensional array such as [1,2,3,4,5,6,7,8,9] Apply rolling (). sum ()

row_data = pd.Series ([1,2,3,4,5,6,7,8,9])
res = row_data.rolling (window = 3) .sum ()
print (res)
# 0 NaN
# 1 NaN
# 2 6.0
# 3 9.0
# 4 12.0
# 5 15.0
# 6 18.0
# 7 21.0# 8 24.0
#dtype: float64

As in

, you can find the total for each fixed interval (this time, the interval specified by window = 3).
In this case, I want to apply my own function (rms), not the total for each fixed interval, so useSeries.rolling (). Apply ()

row_data = pd.Series ([1,2,3,4,5,6,7,8,9])
rms = lambda d: np.sqrt ((d ** 2) .sum ()/d.size)
res = row_data.rolling (window = 3) .apply (rms)
print (res)
# 0 NaN
# 1 NaN
# 2 2.160247
# 3 3.109126
# 4 4.082483
# 5 5.066228
# 6 6.055301
# 7 7.047458
# 8 8.041559
#dtype: float64

It becomes.

Lastly, regarding the rolling parameters, in the above example, the result ofindex = 0,1isNaN. This is because the number of input data for calculating this part does not satisfy the number of sections (Window = 3) and cannot be calculated. Therefore,min_periods = 1is passed as a parameter, and it is specified to calculate if there is at least one input data. In the above example, the result of the input dataindex = 0,1,2is inindex = 2. Therefore, by passingcenter = True, the result of index = 0,1,2 is output to index = 1.

df_iwhr1of the caller isDataFrame, soa2in thewindow_rmsfunctionshapebecomes(12345,1)and a presentation error has occurred.
In this presentation code, all rows are targeted, so it is not necessary to use.iloc, anddata ['semit']is used to set the column value toPass it as a Series.
Reference: ValueError: object too deep for desired array while using convolution

import numpy as np
import pandas as pd
def window_rms (a, window_size):
    a2 = np.power (a, 2)
    window = np.ones (window_size)/float (window_size)
    return np.sqrt (np.convolve (a2, window, "same"))
WINDOW_SIZE = 3
data = pd.DataFrame ({'semit': [i + 1 for i in range (10)], 'iwhr1': [(i + 1) * 10 for i in range (10)]})
print (data)
RMS1 = window_rms (data ['semit'], WINDOW_SIZE)
data ['RMS'] = RMS1
print (data)

Because it has been converted into a DataFrame, it is a code that makes full use of it.

data ['sq'] = np.sqrt (df_iwhr1) # square all elements
data ['mean'] = data.rolling (window = WINDOW_SISE) .mean () ['sq'] # Moving average
data ['rms'] = np.sqrt (data ['mean']) #Square root of the result

Since the progress of the calculation is also in the data, please take out what you need and output it to a csv file