I have a dataframe with the values as below:
df = pd.DataFrame({'Column4': ['NaN;NaN;1;4','4;8','nan']} )
print (df)
Column4
0 NaN;NaN;1;4
1 4;8
2 nan
I tried with the code below to get the sum.
df['Sum'] = df['Column4'].apply(lambda x: sum(map(int, x.split(';'))))
I am getting the error message as
ValueError: invalid literal for int() with base 10: 'NaN'
Use Series.str.split with expand=True for DataFrame, convert to floats and sum per rows - pandas by default exclude missing values:
df['Sum'] = df['Column4'].str.split(';', expand=True).astype(float).sum(axis=1)
print (df)
Column4 Sum
0 NaN;NaN;1;4 5.0
1 4;8 12.0
2 nan 0.0
Your solution should be changed:
f = lambda x: sum(int(y) for y in x.split(';') if not y in ('nan','NaN'))
df['Sum'] = df['Column4'].apply(f)
because if convert to float get mssing values for NaNs with another numeric:
df['Sum'] = df['Column4'].apply(lambda x: sum(map(float, x.split(';'))))
print (df)
Column4 Sum
0 NaN;NaN;1;4 NaN
1 4;8 12.0
2 nan NaN
Related
I need to change the format for values in a column in a dataframe. If I have a dataframe in that format:
df =
sector funding_total_usd
1 NaN
2 10,00,000
3 3,90,000
4 34,06,159
5 2,17,50,000
6 20,00,000
How to change it to that format:
df =
sector funding_total_usd
1 NaN
2 10000.00
3 3900.00
4 34061.59
5 217500.00
6 20000.00
This is my code:
for row in df['funding_total_usd']:
dt1 = row.replace (',','')
print (dt1)
This is the error that I got "AttributeError: 'float' object has no attribute 'replace'"
I need really to your help in how to do that?
Here's the way to get the decimal places:
import pandas as pd
import numpy as np
df= pd.DataFrame({'funding_total_usd': [np.nan, 1000000, 390000, 3406159,21750000,2000000]})
print(df)
df['funding_total_usd'] /= 100
print(df)
funding_total_usd
0 NaN
1 1000000.0
2 390000.0
3 3406159.0
4 21750000.0
funding_total_usd
0 NaN
1 10000.00
2 3900.00
3 34061.59
4 217500.00
To solve your comma problem, please run this as your first command before you print. It will remove all your commas for the float values.
pd.options.display.float_format = '{:.2f}'.format
I have a pandas dataframe as below:
import pandas as pd
df = pd.DataFrame({'ORDER':["A", "A"], 'col1':[np.nan, np.nan], 'col2':[np.nan, 5]})
df
ORDER col1 col2
0 A NaN NaN
1 A NaN 5.0
I want to create a column 'new' as sum(col1, col2) ignoring Nan only if one of the column as Nan,
If both of the columns have NaN value, it should return NaN as below
I tried the below code and it works fine. Is there any way to achieve the same with just one line of code.
df['new'] = df[['col1', 'col2']].sum(axis = 1)
df['new'] = np.where(pd.isnull(df['col1']) & pd.isnull(df['col2']), np.nan, df['new'])
df
ORDER col1 col2 new
0 A NaN NaN NaN
1 A NaN 5.0 5.0
Do sum with min_count
df['new'] = df[['col1','col2']].sum(axis=1,min_count=1)
Out[78]:
0 NaN
1 5.0
dtype: float64
Use the add function on the two columns, which takes a fill_value argument that lets you replace NaN:
df['col1'].add(df['col2'], fill_value=0)
0 NaN
1 5.0
dtype: float64
Is this ok?
df['new'] = df[['col1', 'col2']].sum(axis = 1).replace(0,np.nan)
I need to delete the row completely in a dataframe having "None" value in all the columns. I am using the following code -
df.dropna(axis=0,how='all',thresh=None,subset=None,inplace=True)
This does not bring any difference to the dataframe. The rows with "None" value are still there.
How to achieve this?
There Nones should be strings, so use replace first:
df = df.replace('None', np.nan).dropna(how='all')
df = pd.DataFrame({
'a':['None','a', 'None'],
'b':['None','g', 'None'],
'c':['None','v', 'b'],
})
print (df)
a b c
0 None None None
1 a g v
2 None None b
df1 = df.replace('None', np.nan).dropna(how='all')
print (df1)
a b c
1 a g v
2 NaN NaN b
Or test values None with not equal and DataFrame.any:
df1 = df[df.ne('None').any(axis=1)]
print (df1)
a b c
1 a g v
2 None None b
You should be dropping in the axis 1. Use the how keyword to drop columns with any or all NaN values. Check the docs
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,3], 'b':[-1, 0, np.nan], 'c':[np.nan, np.nan, np.nan]})
df
a b c
0 1 -1.0 NaN
1 2 0.0 NaN
2 3 NaN 5.0
df.dropna(axis=1, how='any')
a
0 1
1 2
2 3
df.dropna(axis=1, how='all')
a b
0 1 -1.0
1 2 0.0
2 3 NaN
I was wondering how I can remove rows which have a negative value but keep the NaNs. At the moment I am using:
DF = DF.ix[DF['RAF01Time'] >= 0]
But this removes the NaNs.
Thanks in advance.
You need boolean indexing with another condition with isnull:
DF = DF[(DF['RAF01Time'] >= 0) | (DF['RAF01Time'].isnull())]
Sample:
DF = pd.DataFrame({'RAF01Time':[-1,2,3,np.nan]})
print (DF)
RAF01Time
0 -1.0
1 2.0
2 3.0
3 NaN
DF = DF[(DF['RAF01Time'] >= 0) | (DF['RAF01Time'].isnull())]
print (DF)
RAF01Time
1 2.0
2 3.0
3 NaN
Another solution with query:
DF = DF.query("~(RAF01Time < 0)")
print (DF)
RAF01Time
1 2.0
2 3.0
3 NaN
You can just use < 0 and then take the inverse of the condition.
DF = DF[~(DF['RAF01Time'] < 0)]
I have two dataframes (df1 and df2) that each have the same rows and columns. I would like to take the maximum of these two dataframes, element-by-element. In addition, the result of any element-wise maximum with a number and NaN should be the number. The approach I have implemented so far seems inefficient:
def element_max(df1,df2):
import pandas as pd
cond = df1 >= df2
res = pd.DataFrame(index=df1.index, columns=df1.columns)
res[(df1==df1)&(df2==df2)&(cond)] = df1[(df1==df1)&(df2==df2)&(cond)]
res[(df1==df1)&(df2==df2)&(~cond)] = df2[(df1==df1)&(df2==df2)&(~cond)]
res[(df1==df1)&(df2!=df2)&(~cond)] = df1[(df1==df1)&(df2!=df2)]
res[(df1!=df1)&(df2==df2)&(~cond)] = df2[(df1!=df1)&(df2==df2)]
return res
Any other ideas? Thank you for your time.
A more readable way to do this in recent versions of pandas is concat-and-max:
import scipy as sp
import pandas as pd
A = pd.DataFrame([[1., 2., 3.]])
B = pd.DataFrame([[3., sp.nan, 1.]])
pd.concat([A, B]).max(level=0)
#
# 0 1 2
# 0 3.0 2.0 3.0
#
You can use where to test your df against another df, where the condition is True, the values from df are returned, when false the values from df1 are returned. Additionally in the case where NaN values are in df1 then an additional call to fillna(df) will use the values from df to fill those NaN and return the desired df:
In [178]:
df = pd.DataFrame(np.random.randn(5,3))
df.iloc[1,2] = np.NaN
print(df)
df1 = pd.DataFrame(np.random.randn(5,3))
df1.iloc[0,0] = np.NaN
print(df1)
0 1 2
0 2.671118 1.412880 1.666041
1 -0.281660 1.187589 NaN
2 -0.067425 0.850808 1.461418
3 -0.447670 0.307405 1.038676
4 -0.130232 -0.171420 1.192321
0 1 2
0 NaN -0.244273 -1.963712
1 -0.043011 -1.588891 0.784695
2 1.094911 0.894044 -0.320710
3 -1.537153 0.558547 -0.317115
4 -1.713988 -0.736463 -1.030797
In [179]:
df.where(df > df1, df1).fillna(df)
Out[179]:
0 1 2
0 2.671118 1.412880 1.666041
1 -0.043011 1.187589 0.784695
2 1.094911 0.894044 1.461418
3 -0.447670 0.558547 1.038676
4 -0.130232 -0.171420 1.192321