EDIT: Upon request I provide an example that is closer to the real data I am working with.
So I have a table data that looks something like
value0 value1 value2
run step
0 0 0.12573 -0.132105 0.640423
1 0.1049 -0.535669 0.361595
2 1.304 0.947081 -0.703735
3 -1.265421 -0.623274 0.041326
4 -2.325031 -0.218792 -1.245911
5 -0.732267 -0.544259 -0.3163
1 0 0.411631 1.042513 -0.128535
1 1.366463 -0.665195 0.35151
2 0.90347 0.094012 -0.743499
3 -0.921725 -0.457726 0.220195
4 -1.009618 -0.209176 -0.159225
5 0.540846 0.214659 0.355373
(think: collection of time series) and a second table valid_range
start stop
run
0 1 3
1 2 5
For each run I want to drop all rows that do not satisfy start≤step≤stop.
I tried the following (table generating code at the end)
for idx in valid_range.index:
slc = data.loc[idx]
start, stop = valid_range.loc[idx]
cond = (start <= slc.index) & (slc.index <= stop)
data.loc[idx] = data.loc[idx][cond]
However, this results in:
value0 value1 value2
run step
0 0 NaN NaN NaN
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 NaN NaN NaN
5 NaN NaN NaN
1 0 NaN NaN NaN
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 NaN NaN NaN
5 NaN NaN NaN
I also tried data.loc[idx].drop(slc[cond].index, inplace=True) but it didn't have any effect...
Generating code for table
import numpy as np
from pandas import DataFrame, MultiIndex, Index
rng = np.random.default_rng(0)
valid_range = DataFrame({"start": [1, 2], "stop":[3, 5]}, index=Index(range(2), name="run"))
midx = MultiIndex(levels=[[],[]], codes=[[],[]], names=["run", "step"])
data = DataFrame(columns=[f"value{k}" for k in range(3)], index=midx)
for run in range(2):
for step in range(6):
data.loc[(run, step), :] = rng.normal(size=(3))
)
First, merge data and valid range based on 'run', using the merge method
>>> data
value0 value1 value2
run step
0 0 0.12573 -0.132105 0.640423
1 0.1049 -0.535669 0.361595
2 1.304 0.947081 -0.703735
3 -1.26542 -0.623274 0.041326
4 -2.32503 -0.218792 -1.24591
5 -0.732267 -0.544259 -0.3163
1 0 0.411631 1.04251 -0.128535
1 1.36646 -0.665195 0.35151
2 0.90347 0.0940123 -0.743499
3 -0.921725 -0.457726 0.220195
4 -1.00962 -0.209176 -0.159225
5 0.540846 0.214659 0.355373
>>> valid_range
start stop
run
0 1 3
1 2 5
>>> merged = data.reset_index().merge(valid_range, how='left', on='run')
>>> merged
run step value0 value1 value2 start stop
0 0 0 0.12573 -0.132105 0.640423 1 3
1 0 1 0.1049 -0.535669 0.361595 1 3
2 0 2 1.304 0.947081 -0.703735 1 3
3 0 3 -1.26542 -0.623274 0.041326 1 3
4 0 4 -2.32503 -0.218792 -1.24591 1 3
5 0 5 -0.732267 -0.544259 -0.3163 1 3
6 1 0 0.411631 1.04251 -0.128535 2 5
7 1 1 1.36646 -0.665195 0.35151 2 5
8 1 2 0.90347 0.0940123 -0.743499 2 5
9 1 3 -0.921725 -0.457726 0.220195 2 5
10 1 4 -1.00962 -0.209176 -0.159225 2 5
11 1 5 0.540846 0.214659 0.355373 2 5
Then select the rows which satisfy the condition using eval. Use the boolean array to mask data
>>> cond = merged.eval('start < step < stop').to_numpy()
>>> data[cond]
value0 value1 value2
run step
0 2 1.304 0.947081 -0.703735
1 3 -0.921725 -0.457726 0.220195
4 -1.00962 -0.209176 -0.159225
Or if you want, here is a similar approach using query
res = (
data.reset_index()
.merge(valid_range, on='run', how='left')
.query('start < step < stop')
.drop(columns=['start','stop'])
.set_index(['run', 'step'])
)
I would go on groupby like this:
(df.groupby(level=0)
.apply(lambda x: x[x['small']>1])
.reset_index(level=0, drop=True) # remove duplicate index
)
which gives:
big small
animal animal attribute
cow cow speed 30.0 20.0
weight 250.0 150.0
falcon falcon speed 320.0 250.0
lama lama speed 45.0 30.0
weight 200.0 100.0
Related
I have this dataset, which contains some NaN values:
df = pd.DataFrame({'Id':[1,2,3,4,5,6], 'Name':['Eve','Diana',np.NaN,'Mia','Mae',np.NaN], "Count":[10,3,np.NaN,8,5,2]})
df
Id Name Count
0 1 Eve 10.0
1 2 Diana 3.0
2 3 NaN NaN
3 4 Mia 8.0
4 5 Mae 5.0
5 6 NaN 2.0
I want to test if the column has a NaN value (0) or not (1) and creating two new columns. I have tried this:
df_clean = df
df_clean[['Name_flag','Count_flag']] = df_clean[['Name','Count']].apply(lambda x: 0 if x == np.NaN else 1, axis = 1)
But it mentions that The truth value of a Series is ambiguous. I want to make it avoiding redundancy, but I see there is a mistake in my logic. Please, could you help me with this question?
The expected table is:
Id Name Count Name_flag Count_flag
0 1 Eve 10.0 1 1
1 2 Diana 3.0 1 1
2 3 NaN NaN 0 0
3 4 Mia 8.0 1 1
4 5 Mae 5.0 1 1
5 6 NaN 2.0 0 1
Multiply boolean mask by 1:
df[['Name_flag','Count_flag']] = df[['Name', 'Count']].isna() * 1
>>> df
Id Name Count Name_flag Count_flag
0 1 Eve 10.0 0 0
1 2 Diana 3.0 0 0
2 3 NaN NaN 1 1
3 4 Mia 8.0 0 0
4 5 Mae 5.0 0 0
5 6 NaN 2.0 1 0
For your problem of The truth value of a Series is ambiguous
For apply, you cannot return a scalar 0 or 1 because you have a series as input . You have to use applymap instead to apply a function elementwise. But comparing to NaN is not an easy thing:
Try:
df[['Name','Count']].applymap(lambda x: str(x) == 'nan') * 1
We can use isna and convert the boolean to int:
df[["Name_flag", "Count_flag"]] = df[["Name", "Count"]].isna().astype(int)
Id Name Count Name_flag Count_flag
0 1 Eve 10.00 0 0
1 2 Diana 3.00 0 0
2 3 NaN NaN 1 1
3 4 Mia 8.00 0 0
4 5 Mae 5.00 0 0
5 6 NaN 2.00 1 0
I am trying to impute/fill values using rows with similar columns' values.
For example, I have this dataframe:
one | two | three
1 1 10
1 1 nan
1 1 nan
1 2 nan
1 2 20
1 2 nan
1 3 nan
1 3 nan
I wanted to using the keys of column one and two which is similar and if column three is not entirely nan then impute the existing value from a row of similar keys with value in column '3'.
Here is my desired result:
one | two | three
1 1 10
1 1 10
1 1 10
1 2 20
1 2 20
1 2 20
1 3 nan
1 3 nan
You can see that keys 1 and 3 do not contain any value because the existing value does not exists.
I have tried using groupby+fillna():
df['three'] = df.groupby(['one','two'])['three'].fillna()
which gave me an error.
I have tried forward fill which give me rather strange result where it forward fill the column 2 instead. I am using this code for forward fill.
df['three'] = df.groupby(['one','two'], sort=False)['three'].ffill()
If only one non NaN value per group use ffill (forward filling) and bfill (backward filling) per group, so need apply with lambda:
df['three'] = df.groupby(['one','two'], sort=False)['three']
.apply(lambda x: x.ffill().bfill())
print (df)
one two three
0 1 1 10.0
1 1 1 10.0
2 1 1 10.0
3 1 2 20.0
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
But if multiple value per group and need replace NaN by some constant - e.g. mean by group:
print (df)
one two three
0 1 1 10.0
1 1 1 40.0
2 1 1 NaN
3 1 2 NaN
4 1 2 20.0
5 1 2 NaN
6 1 3 NaN
7 1 3 NaN
df['three'] = df.groupby(['one','two'], sort=False)['three']
.apply(lambda x: x.fillna(x.mean()))
print (df)
one two three
0 1 1 10.0
1 1 1 40.0
2 1 1 25.0
3 1 2 20.0
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
You can sort data by the column with missing values then groupby and forwardfill:
df.sort_values('three', inplace=True)
df['three'] = df.groupby(['one','two'])['three'].ffill()
I am trying to impute/fill values using rows with similar columns' values.
For example, I have this dataframe:
one | two | three
1 1 10
1 1 nan
1 1 nan
1 2 nan
1 2 20
1 2 nan
1 3 nan
1 3 nan
I wanted to using the keys of column one and two which is similar and if column three is not entirely nan then impute the existing value from a row of similar keys with value in column '3'.
Here is my desired result:
one | two | three
1 1 10
1 1 10
1 1 10
1 2 20
1 2 20
1 2 20
1 3 nan
1 3 nan
You can see that keys 1 and 3 do not contain any value because the existing value does not exists.
I have tried using groupby+fillna():
df['three'] = df.groupby(['one','two'])['three'].fillna()
which gave me an error.
I have tried forward fill which give me rather strange result where it forward fill the column 2 instead. I am using this code for forward fill.
df['three'] = df.groupby(['one','two'], sort=False)['three'].ffill()
If only one non NaN value per group use ffill (forward filling) and bfill (backward filling) per group, so need apply with lambda:
df['three'] = df.groupby(['one','two'], sort=False)['three']
.apply(lambda x: x.ffill().bfill())
print (df)
one two three
0 1 1 10.0
1 1 1 10.0
2 1 1 10.0
3 1 2 20.0
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
But if multiple value per group and need replace NaN by some constant - e.g. mean by group:
print (df)
one two three
0 1 1 10.0
1 1 1 40.0
2 1 1 NaN
3 1 2 NaN
4 1 2 20.0
5 1 2 NaN
6 1 3 NaN
7 1 3 NaN
df['three'] = df.groupby(['one','two'], sort=False)['three']
.apply(lambda x: x.fillna(x.mean()))
print (df)
one two three
0 1 1 10.0
1 1 1 40.0
2 1 1 25.0
3 1 2 20.0
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
You can sort data by the column with missing values then groupby and forwardfill:
df.sort_values('three', inplace=True)
df['three'] = df.groupby(['one','two'])['three'].ffill()
daychange SS
0.017065 0
-0.009259 100
0.031542 0
-0.004530 0
0.000709 0
0.004970 100
-0.021900 0
0.003611 0
I have two columns and I want to calculate the sum of next 5 'daychange' if SS = 100.
I am using the following right now but it does not work quite the way I want it to:
df['total'] = df.loc[df['SS'] == 100,['daychange']].sum(axis=1)
Since pandas 1.1 you can create a forward rolling window and select the rows you want to include in your dataframe. With different arguments my notebook kernel got terminated: use with caution.
indexer = pd.api.indexers.FixedForwardWindowIndexer(window_size=5)
df['total'] = df.daychange.rolling(indexer, min_periods=1).sum()[df.SS == 100]
df
Out:
daychange SS total
0 0.017065 0 NaN
1 -0.009259 100 0.023432
2 0.031542 0 NaN
3 -0.004530 0 NaN
4 0.000709 0 NaN
5 0.004970 100 -0.013319
6 -0.021900 0 NaN
7 0.003611 0 NaN
Exclude the starting row with SS == 100 from the sum
That would be the next row after rows with SS == 100. As all rows are computed you can use
df['total'] = df.daychange.rolling(indexer, min_periods=1).sum().shift(-1)[df.SS == 100]
df
Out:
daychange SS total
0 0.017065 0 NaN
1 -0.009259 100 0.010791
2 0.031542 0 NaN
3 -0.004530 0 NaN
4 0.000709 0 NaN
5 0.004970 100 -0.018289
6 -0.021900 0 NaN
7 0.003611 0 NaN
Slow hacky solution using indices of selected rows
This feels like a hack, but works and avoids computing unnecessary rolling values
df['next5sum'] = df[df.SS == 100].index.to_series().apply(lambda x: df.daychange.iloc[x: x + 5].sum())
df
Out:
daychange SS next5sum
0 0.017065 0 NaN
1 -0.009259 100 0.023432
2 0.031542 0 NaN
3 -0.004530 0 NaN
4 0.000709 0 NaN
5 0.004970 100 -0.013319
6 -0.021900 0 NaN
7 0.003611 0 NaN
For the sum of the next five rows excluding the rows with SS == 100 you can adjust the slices or shift the series
df['next5sum'] = df[df.SS == 100].index.to_series().apply(lambda x: df.daychange.iloc[x + 1: x + 6].sum())
# df['next5sum'] = df[df.SS == 100].index.to_series().apply(lambda x: df.daychange.shift(-1).iloc[x: x + 5].sum())
df
Out:
daychange SS next5sum
0 0.017065 0 NaN
1 -0.009259 100 0.010791
2 0.031542 0 NaN
3 -0.004530 0 NaN
4 0.000709 0 NaN
5 0.004970 100 -0.018289
6 -0.021900 0 NaN
7 0.003611 0 NaN
7 0.003611 0 NaN
How can I replace all the non-NaN values in a pandas dataframe with 1 but leave the NaN values alone? This almost does what I'm looking for. The problem is it also makes NaN values 0. Then I have to reset them to NaN after.
I would like this
a b
0 NaN QQQ
1 AAA NaN
2 NaN BBB
to become this
a b
0 NaN 1
1 1 NaN
2 NaN 1
This code is almost what I want
newdf = df.notnull().astype('int')
The above code does this
a b
0 0 1
1 1 0
2 0 1
One way would be to select all non-null values from the original data frame and set them to one:
df[df.notnull()] = 1
This solution on your data:
df = pd.DataFrame({'a': [np.nan, 'AAA', np.nan], 'b': ['QQQ', np.nan, 'BBB']})
df[df.notnull()] = 1
df
a b
0 NaN 1
1 1 NaN
2 NaN 1
You can use np.where() with DataFrame.isna() to accomplish this
df=pd.DataFrame(data=[[1,np.NaN,5],
['q',np.NaN,np.NaN],
['7',{'a':1},np.NaN]],
columns=['a','b','c'])
a b c
0 1 NaN 5.0
1 q NaN NaN
2 7 {'a': 1} NaN
df1=pd.DataFrame(np.where(df.isna(),df,1), columns=df.columns)
a b c
0 1 NaN 1
1 1 NaN NaN
2 1 1 NaN