找到差异。 groupby 在 pandas 中的最大值和最小值？

Question

If I have date frame as below of 3 year of rainfall from 2015-2017 for three stations, could you help me how find diff. between maximum and minimum for every station?

Answer 1

Code below uses groupby() with axis=1 to get min() and max() for each row. The restuls are then combined using .merge() :

Option-1 :

Using the non-repeating names in the column 'Name'

# Import libraries
import pandas as pd

# Create DataFrame
df = pd.DataFrame({
    'Name':['Baghdad', 'Basra', 'Mousl'],
    'R2015':[300,190,350],
    'R2016':[240,180,540],
    'R2017':[290,160,490]
})

# Convert column to index
df = df.set_index('Name')

# Get min and max
df_min = df.groupby(['min']*df.shape[1],axis=1).min()
df_max = df.groupby(['max']*df.shape[1],axis=1).max()

# Combine
df_min_max = df_min.merge(df_max, on='Name')

# Get difference
df_min_max['diff'] = abs(df_min_max['min'] - df_min_max['max'])

# Output
df_min_max

Option-2 :

If the DataFrame had names in column Name repeating, then below should work. Here, added Baghdad as an additional repeating row. Here, groupby() of groupby() is used.

# Import libraries
import pandas as pd

# Create DataFrame
df = pd.DataFrame({
    'Name':['Baghdad', 'Basra', 'Mousl','Baghdad'],
    'R2015':[300,190,350,780],
    'R2016':[240,180,540,455],
    'R2017':[290,160,490,23]
})

# Convert column to index
df = df.set_index('Name')

# Get min and max
df_min = df.groupby(['min']*df.shape[1],axis=1).min().groupby(['Name']).min()
df_max = df.groupby(['max']*df.shape[1],axis=1).max().groupby(['Name']).max()

# Combine
df_min_max = df_min.merge(df_max, on='Name')

# Get difference
df_min_max['diff'] = abs(df_min_max['min'] - df_min_max['max'])

# Output
df_min_max