pandas select percentile
In [48]: cols = list('abc')
In [49]: df = DataFrame(randn(10, len(cols)), columns=cols)
In [50]: df.a.quantile(0.95)
Out[50]: 1.5776961953820687pandas select percentile
In [48]: cols = list('abc')
In [49]: df = DataFrame(randn(10, len(cols)), columns=cols)
In [50]: df.a.quantile(0.95)
Out[50]: 1.5776961953820687calculate percentile pandas dataframe
import pandas as pd
import random
A = [ random.randint(0,100) for i in range(10) ]
B = [ random.randint(0,100) for i in range(10) ]
df = pd.DataFrame({ 'field_A': A, 'field_B': B })
df
# field_A field_B
# 0 90 72
# 1 63 84
# 2 11 74
# 3 61 66
# 4 78 80
# 5 67 75
# 6 89 47
# 7 12 22
# 8 43 5
# 9 30 64
df.field_A.mean() # Same as df['field_A'].mean()
# 54.399999999999999
df.field_A.median()
# 62.0
# You can call `quantile(i)` to get the i'th quantile,
# where `i` should be a fractional number.
df.field_A.quantile(0.1) # 10th percentile
# 11.9
df.field_A.quantile(0.5) # same as median
# 62.0
df.field_A.quantile(0.9) # 90th percentile
# 89.10000000000001how to sort values of pandas dataframe for iqr
def mod_outlier(df):
df1 = df.copy()
df = df._get_numeric_data()
q1 = df.quantile(0.25)
q3 = df.quantile(0.75)
iqr = q3 - q1
lower_bound = q1 -(1.5 * iqr)
upper_bound = q3 +(1.5 * iqr)
for col in col_vals:
for i in range(0,len(df[col])):
if df[col][i] < lower_bound[col]:
df[col][i] = lower_bound[col]
if df[col][i] > upper_bound[col]:
df[col][i] = upper_bound[col]
for col in col_vals:
df1[col] = df[col]
return(df1)Copyright © 2021 Codeinu
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