Using Python's Pandas to find average values by bins

Question

I just started using pandas to analyze groundwater well data over time.

My data in a text file looks like (site_no, date, well_level):

485438103132901 19800417    -7.1

485438103132901 19800506    -6.8

483622101085001 19790910    -6.7

485438103132901 19790731    -6.2

483845101112801 19801111    -5.37

484123101124601 19801111    -5.3

485438103132901 19770706    -4.98

I would like an output with average well levels binned by 5 year increments and with a count:

site_no   avg 1960-end1964  count    avg 1965-end1969  count    avg 1970-end1974 count

I am reading in the data with:

names = ['site_no','date','wtr_lvl']
df = pd.read_csv('D:\info.txt', sep='\t',names=names)

I can find the overall average by site with:

avg = df.groupby(['site_no'])['wtr_lvl'].mean().reset_index()

My crude bin attempts use:

a1 = df[df.date > 19600000]
a2 = a1[a1.date < 19650000]
avga2 = a2.groupby(['site_no'])['wtr_lvl'].mean()

My question: how can I join the results to display as desired? I tried merge, join, and append, but they do not allow for empty data frames (which happens). Also, I am sure there is a simple way to bin the data by the dates. Thanks.

Indomitability

Answer 1

The most concise way is probably to convert this to a timeseris data and them downsample to get the means:

In [75]:

print df
                         ID  Level
1                                 
1980-04-17  485438103132901  -7.10
1980-05-06  485438103132901  -6.80
1979-09-10  483622101085001  -6.70
1979-07-31  485438103132901  -6.20
1980-11-11  483845101112801  -5.37
1980-11-11  484123101124601  -5.30
1977-07-06  485438103132901  -4.98
In [76]:

df.Level.resample('60M', how='mean') 
#also may consider different time alias: '5A', '5BA', '5AS', etc:
#see: http://pandas.pydata.org/pandas-docs/stable/timeseries.html#offset-aliases
Out[76]:
1
1977-07-31   -4.980
1982-07-31   -6.245
Freq: 60M, Name: Level, dtype: float64

Alternatively, you may use groupby together with cut:

In [99]:

print df.groupby(pd.cut(df.index.year, pd.date_range('1960', periods=5, freq='5A').year, include_lowest=True)).mean()
                        ID     Level
[1960, 1965]           NaN       NaN
(1965, 1970]           NaN       NaN
(1970, 1975]           NaN       NaN
(1975, 1980]  4.847632e+14 -6.064286

And by ID also:

In [100]:

print df.groupby(['ID', 
                  pd.cut(df.index.year, pd.date_range('1960', periods=5, freq='5A').year, include_lowest=True)]).mean()
                              Level
ID                                 
483622101085001 (1975, 1980]  -6.70
483845101112801 (1975, 1980]  -5.37
484123101124601 (1975, 1980]  -5.30
485438103132901 (1975, 1980]  -6.27

Nobody

Answer 2

so what i like to do is create a separate column with the rounded bin number:

    bin_width = 50000
    mult = 1. / bin_width
    df['bin'] = np.floor(ser * mult + .5) / mult

then, just group by the bins themselves

    df.groupby('bin').mean()

another note, you can do multiple truth evaluations in one go:

    df[(df.date > a) & (df.date < b)]

Sherri

ser is not defined here. I changed the last line to this which worked perfectly: df['bin'] = [ np.trunc( x * mult + mult ) for x in range( len( df ) ) ] – Settles