I have a geopandas dataframe where there are NaN in a column. I want to impute the NaN with the average value of its neighbors. I made up the following example and would appreciate it if somebody can help me out with the final steps. Thanks.
# Load libraries
import numpy as np
import pandas as pd
import geopandas as gpd
from libpysal.weights.contiguity import Queen
# Make data
world = gpd.read_file(gpd.datasets.get_path('naturalearth_lowres'))
africa = world[world['continent'] == 'Africa']
africa.reset_index(inplace=True, drop=True)
africa.loc[[2,8,15,22,30,35,39,43],'pop_est'] = np.nan # Make NaN value for pop_est
africa
# Generate weight
w = Queen.from_dataframe(africa)
w.neighbors[2] # Check neighbors of index 2
For example, index 2 has a missing value on population estimate and its neighbors are [0, 35, 36, 48, 49, 50, 27, 28, 31]. I want to use the mean of population estimate from [0, 35, 36, 48, 49, 50, 27, 28, 31] to replace NaN. Thanks.
I finally figured out how to do it.
Related
I have written this code that pulls and excel file into python reorganizes the columns and tries to remove any rows where the data in that row is older than 2012.
import pandas as pd
import numpy as np
import datetime
#Read the excel with the data from IHS
Coal_Exports = pd.read_excel("U:\drive\Coal Exports.xlsx")
#Rearrange the data by columns
Coal_Exports = pd.melt(Coal_Exports, id_vars= ["Export Country", "Import Country", "Coal type", "Date Last Updated", "Unit"],var_name="Date", value_name="value")
#set the Date as the Index
Coal_Exports = Coal_Exports.set_index(keys="Date")
#Delete pre 2012 Rows
Coal_Exports = Coal_Exports[(Coal_Exports.ix[0] > "01/01/2012")]
#Send the reformatted data to excel
Coal_Exports.to_excel ("U:\drive\Formatted Coal Exports.xlsx")
print "Done"
My issue is deleting the rows pre 2012,
I get this error:
---------------------------------------------------------------------------
IndexingError Traceback (most recent call last)
<ipython-input-25-8eb1513ad7c2> in <module>()
14 #Delete pre 2012 Rows
15 #Coal_Exports = Coal_Exports.drop[(Coal_Exports.set_index(keys="Date")<('01/01/2012') )]
---> 16 Coal_Exports = Coal_Exports[(Coal_Exports.ix[0] > "01/01/2011")]
17
18
C:\Users\xxxxx\AppData\Local\Continuum\Anaconda2\lib\site-packages\pandas\core\frame.pyc in __getitem__(self, key)
1989 if isinstance(key, (Series, np.ndarray, Index, list)):
1990 # either boolean or fancy integer index
-> 1991 return self._getitem_array(key)
1992 elif isinstance(key, DataFrame):
1993 return self._getitem_frame(key)
C:\Users\xxxxx\AppData\Local\Continuum\Anaconda2\lib\site-packages\pandas\core\frame.pyc in _getitem_array(self, key)
2029 # check_bool_indexer will throw exception if Series key cannot
2030 # be reindexed to match DataFrame rows
-> 2031 key = check_bool_indexer(self.index, key)
2032 indexer = key.nonzero()[0]
2033 return self.take(indexer, axis=0, convert=False)
C:\Users\xxxxx\AppData\Local\Continuum\Anaconda2\lib\site-packages\pandas\core\indexing.pyc in check_bool_indexer(ax, key)
1799 mask = com.isnull(result._values)
1800 if mask.any():
-> 1801 raise IndexingError('Unalignable boolean Series key provided')
1802
1803 result = result.astype(bool)._values
IndexingError: Unalignable boolean Series key provided
The dates in the excel file are of the format 31/02/2011 etc.
I think it's something to do with the format of dates in excel being different from pandas but i'm not sure, can anyone help with this?
I have a dataframe that gets read in from csv and has extraneous data.
Judgment on what is extraneous is made by evaluating one column, SystemStart.
Any data per row that is in a column with a heading of date value lower than SystemStart for that row, is set to nan. For example, index = 'one' has a SystemStart date of '2016-1-5', and when the pd.date_range is set up, it has no nan values to populate. index= 'three' is '2016-1-7' and hence has two nan values replacing the original data.
I can go row-by-row and throw np.nan values at all columns, but that is slow. Is there a faster way?
I've created a representative dataframe below, and am looking to get the same result without iterative operations, or a way to speed up those operations. Any help would be greatly appreciated.
import pandas as pd
import numpy as np
start_date = '2016-1-05'
end_date = '2016-1-7'
dates = pd.date_range(start_date, end_date, freq='D')
dt_dates = pd.to_datetime(dates, unit='D')
ind = ['one', 'two', 'three']
df = pd.DataFrame(np.random.randint(0,100,size=(3, 3)), columns = dt_dates, index = ind)
df['SystemStart'] = pd.to_datetime(['2016-1-5', '2016-1-6', '2016-1-7'])
print 'Initial Dataframe: \n', df
for msn in df.index:
zero_date_range = pd.date_range(start_date, df.loc[msn,'SystemStart'] - pd.Timedelta(days=1), freq='D')
# we set zeroes for all columns in the index element in question - this is a horribly slow way to do this
df.loc[msn, zero_date_range] = np.NaN
print '\nAltered Dataframe: \n', df
Below are the df outputs, Initial and Altered:
Initial Dataframe:
2016-01-05 00:00:00 2016-01-06 00:00:00 2016-01-07 00:00:00 \
one 24 23 65
two 21 91 59
three 62 77 2
SystemStart
one 2016-01-05
two 2016-01-06
three 2016-01-07
Altered Dataframe:
2016-01-05 00:00:00 2016-01-06 00:00:00 2016-01-07 00:00:00 \
one 24.0 23.0 65
two NaN 91.0 59
three NaN NaN 2
SystemStart
one 2016-01-05
two 2016-01-06
three 2016-01-07
First thing I do is make sure SystemStart is datetime
df.SystemStart = pd.to_datetime(df.SystemStart)
Then I strip out SystemStart to a separate series
st = df.SystemStart
Then I drop SytstemStart from my df
d1 = df.drop('SystemStart', 1)
Then I convert the columns I have left to datetime
d1.columns = pd.to_datetime(d1.columns)
Finally I use numpy broadcasting to mask the appropriate cells and join SystemStart back in.
d1.where(d1.columns.values >= st.values[:, None]).join(st)
My dataframe looks like this :
fruits = pd.DataFrame({'orange': [10, 20], 'apple': [30, 40], 'banana': [50, 60]})
apple banana orange
0 30 50 10
1 40 60 20
And I have this vector (its also a dataframe)
sold = pd.DataFrame({'orange': [1], 'apple': [2], 'banana': [3]})
apple banana orange
0 2 3 1
I want to subtract this vector to each row of the initial dataframe to obtain a dataframe which looks like this
apple banana orange
0 28.0 47.0 9.0
1 38.0 57.0 19.0
I tried :
print fruits.subtract(sold, axis = 0)
And the output is
apple banana orange
0 28.0 47.0 9.0
1 NaN NaN NaN
It worked only for the first line. I could create a dataframe filled with the vector for each row. Is there a more efficient way to subtract this vector ? I don't want to use a loop.
convert the df to a series using squeeze and pass axis=1:
In [6]:
fruits.sub(sold.squeeze(), axis=1)
Out[6]:
apple banana orange
0 28 47 9
1 38 57 19
The conversion is necessary as by design arithmetic operations between dfs will align on indices and columns, by passing a Series this allows you to subtract from each row in the df the row from the other df.
Try:
fruits.sub(sold.iloc[0, :])
What you tried before didn't work because sold is a dataframe and the subtraction will try to align both columns and index. sold.iloc[0, :] gets at the first row and is a series thus will work as you intended.
I have many columns in a data frame and I have to find the difference of time in two column named as in_time and out_time and put it in the new column in the same data frame.
The format of time is like this 2015-09-25T01:45:34.372Z.
I am using Pandas DataFrame.
I want to do like this:
df.days = df.out_time - df.in_time
I have many columns and I have to increase 1 more column in it named days and put the differences there.
You need to convert the strings to datetime dtype, you can then subtract whatever arbitrary date you want and on the resulting series call dt.days:
In [15]:
df = pd.DataFrame({'date':['2015-09-25T01:45:34.372Z']})
df
Out[15]:
date
0 2015-09-25T01:45:34.372Z
In [19]:
df['date'] = pd.to_datetime(df['date'])
df['day'] = (df['date'] - dt.datetime.now()).dt.days
df
Out[19]:
date day
0 2015-09-25 01:45:34.372 -252
Well, it all kinda depends on the time format you use. I'd recommend using datetime.
If in_time and out_time are currently strings, convert them with datetime.strptime():
from datetime import datetime
f = lambda x: datetime.strptime(x, '%Y-%m-%dT%H:%M:%S.%fZ')
df.in_time = df.in_time.apply(f)
df.out_time = df.out_time.apply(f)
and then you can simply subtract them, and assign the result to a new column named 'days':
df['days'] = df.out_time - df.in_time
Example: (3 seconds and 1 day differences)
In[5]: df = pd.DataFrame({'in_time':['2015-09-25T01:45:34.372Z','2015-09-25T01:45:34.372Z'],
'out_time':['2015-09-25T01:45:37.372Z','2015-09-26T01:45:34.372Z']})
In[6]: df
Out[6]:
in_time out_time
0 2015-09-25T01:45:34.372Z 2015-09-25T01:45:37.372Z
1 2015-09-25T01:45:34.372Z 2015-09-26T01:45:34.372Z
In[7]: type(df.loc[0,'in_time'])
Out[7]: str
In[8]: df.in_time = df.in_time.apply(lambda x: datetime.strptime(x, '%Y-%m-%dT%H:%M:%S.%fZ'))
In[9]: df.out_time = df.out_time.apply(lambda x: datetime.strptime(x, '%Y-%m-%dT%H:%M:%S.%fZ'))
In[10]: df # notice that it looks exactly the same, but the type is different
Out[10]:
in_time out_time
0 2015-09-25 01:45:34.372 2015-09-25T01:45:37.372Z
1 2015-09-25 01:45:34.372 2015-09-26T01:45:34.372Z
In[11]: type(df.loc[0,'in_time'])
Out[11]: pandas.tslib.Timestamp
And the creation of the new column:
In[12]: df['days'] = df.out_time - df.in_time
In[13]: df
Out[13]:
in_time out_time days
0 2015-09-25 01:45:34.372 2015-09-25 01:45:37.372 0 days 00:00:03
1 2015-09-25 01:45:34.372 2015-09-26 01:45:34.372 1 days 00:00:00
Now you can play with the output format. For example, the portion of seconds difference:
In[14]: df.days = df.days.apply(lambda x: x.total_seconds()/60)
In[15]: df
Out[15]:
in_time out_time days
0 2015-09-25 01:45:34.372 2015-09-25 01:45:37.372 0.05
1 2015-09-25 01:45:34.372 2015-09-26 01:45:34.372 1440.00
Note: Regarding the in_time and out_time format, notice that I made some assumptions (for example, that you're using a 24H clock (thus using %H and not %I)). To play with the format have a look at: strptime() documentation.
Note2: It would obviously be better if you can design your program to use datetime from the beginning (instead of using strings and converting them).
First of all, you need to convert in_time and out_time columns to datetime type.
for col in ('in_time', 'out_time') : # Looping a tuple is faster than a list
df[col] = pd.to_datetime(df[col])
You can check the type using dtypes:
df['in_time'].dtypes
Should give: datetime64[ns, UTC]
Now you can substract them and get the difference time using dt.days or from numpy using np.timedelta64.
Example:
import numpy as np
df['days'] = (df['out_time'] - df['in_time']).dt.days
# Or
df['days'] = (df['out_time'] - df['in_time']) / np.timedelta64(1, 'D')
I have several dataframes which have the same look but different data.
DataFrame 1
bid
close
time
2016-05-24 00:00:00 NaN
2016-05-24 00:05:00 0.000611
2016-05-24 00:10:00 -0.000244
2016-05-24 00:15:00 -0.000122
DataFrame 2
bid
close
time
2016-05-24 00:00:00 NaN
2016-05-24 00:05:00 0.000811
2016-05-24 00:10:00 -0.000744
2016-05-24 00:15:00 -0.000322
I need to build a list of the dataframes, then pass that list of dataframes to a function that can take a list of dataframes and converts it to a numpy array. So below, each entry in the matrix is the elements of the dataframe ('bid
close') column. Notice I don't need the index 'time' column
data = np.array([dataFrames])
returns this (example not actual data)
[[-0.00114415 0.02502565 0.00507831 ..., 0.00653057 0.02183072
-0.00194293] `DataFrame` 1 is here ignore that the data doesn't match above
[-0.01527224 0.02899528 -0.00327654 ..., 0.0322364 0.01821731
-0.00766773] `DataFrame` 2 is here ignore that the data doesn't match above
....]]
Try
master_matrix = pd.concat(list_of_dfs, axis=1)
master_matrix = master_matrix.values.reshape(master_matrix.shape, order='F')
if each row in the final matrix corresponds to the same date
master_matrix = pd.concat(list_of_dfs, axis=1).values
otherwise.
Edit to address the newly added example.
In this case, you can use np.vstack on columns returned from each dataframe.
import pandas as pd
import numpy as np
from io import StringIO
df1 = pd.read_csv(StringIO(
'''
time bid_close
2016-05-24 00:00:00 NaN
2016-05-24 00:05:00 0.000611
2016-05-24 00:10:00 -0.000244
2016-05-24 00:15:00 -0.000122
'''), sep=r' +')
df2 = pd.read_csv(StringIO(
'''
time bid_close
2016-05-24 00:00:00 NaN
2016-05-24 00:05:00 0.000811
2016-05-24 00:10:00 -0.000744
2016-05-24 00:15:00 -0.000322
'''), sep=r' +')
dfs = [df1, df2]
out = np.vstack(df.iloc[:,-1].values for df in dfs)
Result:
In [10]: q.out
Out[10]:
array([[ nan, 0.000611, -0.000244, -0.000122],
[ nan, 0.000811, -0.000744, -0.000322]])
Setup
import pandas as pd
import numpy as np
df1 = pd.DataFrame([1, 2, 3, 4],
index=pd.date_range('2016-04-01', periods=4),
columns=pd.MultiIndex.from_tuples([('bid', 'close')]))
df2 = pd.DataFrame([5, 6, 7, 8],
index=pd.date_range('2016-03-01', periods=4),
columns=pd.MultiIndex.from_tuples([('bid', 'close')]))
print df1
bid
close
2016-04-01 1
2016-04-02 2
2016-04-03 3
2016-04-04 4
print df2
bid
close
2016-03-01 5
2016-03-02 6
2016-03-03 7
2016-03-04 8
Solution
df = np.concatenate([d.T.values for d in [df1, df2]])
print df
[[1 2 3 4]
[5 6 7 8]]
Note
The indices were not required to line up. This just takes the raw np.array from each dataframe and uses np.concatenate to do the rest.