I have checked similar questions on SO with the SettingWithCopyWarning error raised using .loc but I still don't understand why I have the error in the following example.
It appears line 3, I succeed to make it disappear with .copy() but I would like to understand why .loc didn't work specifically here.
Does making a conditional slice creates a view even if it's .loc ?
df = pd.DataFrame( data=[0,1,2,3,4,5], columns=['A'])
df.loc[:,'B'] = df.loc[:,'A'].values
dfa = df.loc[df.loc[:,'A'] < 4,:] # here .copy() removes the error
dfa.loc[:,'C'] = [3,2,1,0]
Edit : pandas version is 1.2.4
dfa = df.loc[df.loc[:,'A'] < 4,:]<br>
dfa is a slice of the df dataframe, still referencing the dataframe, a view..copy creates a separate copy, not just a view of the first dataframe.
dfa.loc[:,'C'] = [3,2,1,0]
When it's a view not a copy, you are getting the warning : A value is trying to be set on a copy of a slice from a DataFrame.
.loc is locating the conditions you give it, but it's still a view that you're setting values to if you don't make it a copy of the dataframe.
Related
I am doing a simple parquet file reading and running a query to find the un-matched rows from left table. Please see the code snippet below.
argTestData = '<path to parquet file>'
tst_DF = spark.read.option('header', True).parquet(argTestData)
argrefData = '<path to parquet file>'
refDF = spark.read.option('header', True).parquet(argrefData)
cond = ["col1", "col2", "col3"]
fi = tst_DF.join(refDF, cond , "left_anti")
So far things are working. However, as a requirement, I need to get the elements list if the above gives count > 0, i.e. if the value of fi.count() > 0, then I need the elements name. So, I tried below code, but it is throwing error.
if fi.filter(col("col1").count() > 0).collect():
fi.show()
error
TypeError: 'Column' object is not callable
Note:
I have 3 columns as a joining condition which is in a list and assigned to a variable cond, and I need to get the un-matched records for those 3 columns, so the if condition has to accommodate them. OfCourse there are many other columns due to join.
Please suggest where am I making mistakes.
Thank you
If I understand correctly, that's simply :
fi.select(cond).collect()
The left_anti already get the records which do not match (exists in tst_DF but not in refDF).
You can add a distinct before the collect to remove duplicates.
Did you import the column function?
from pyspark.sql import functions as F
...
if fi.filter(F.col("col1").count() > 0).collect():
fi.show()
When I attempt to query my input table as a view, I get the error com.palantir.foundry.spark.api.errors.DatasetPathNotFoundException. My code is as follows:
def Median_Product_Revenue_Temp2(Merchant_Segments):
Merchant_Segments.createOrReplaceTempView('Merchant_Segments_View')
df = spark.sql('select * from Merchant_Segments_View limit 5')
return df
I need to dynamically query this table, since I am trying to calculate the median using percentile_approx across numerous fields, and I'm not sure how to do this without using spark.sql.
If I try to avoid using spark.sql to calculate median across numerous fields using something like the below code, it results in the error Missing Transform Attribute: A module object does not have an attribute percentile_approx. Please check the spelling and/or the datatype of the object.
import pyspark.sql.functions as F
exprs = {x: percentile_approx("x", 0.5) for x in df.columns if x is not exclustion_list}
df = df.groupBy(['BANK_NAME','BUS_SEGMENT']).agg(exprs)
try createGlobalTempView. It worked for me.
eg:
df.createGlobalTempView("people")
(Don't know the root cause why localTempView dose not work )
I managed to avoid using dynamic sql for calculating median across columns using the following code:
df_result = df.groupBy(group_list).agg(
*[ F.expr('percentile_approx(nullif('+col+',0), 0.5)').alias(col) for col in df.columns if col not in exclusion_list]
)
Embedding percentile_approx in an F.expr bypassed the issue I was encountering in the second half of my post.
I want to map through the rows of df1 and compare those with the values of df2 , by month and day, across every year in df2,leaving only the values in df1 which are larger than those in df2, to add into a new column, 'New'. df1 and df2 are of the same size, and are indexed by 'Month' and 'Day'. what would be the best way to do this?
df1=pd.DataFrame({'Date':['2015-01-01','2015-01-02','2015-01-03','2015-01-``04','2005-01-05'],'Values':[-5.6,-5.6,0,3.9,9.4]})
df1.Date=pd.to_datetime(df1.Date)
df1['Day']=pd.DatetimeIndex(df1['Date']).day
df1['Month']=pd.DatetimeIndex(df1['Date']).month
df1.set_index(['Month','Day'],inplace=True)
df1
df2 = pd.DataFrame({'Date':['2005-01-01','2005-01-02','2005-01-03','2005-01-``04','2005-01-05'],'Values':[-13.3,-12.2,6.7,8.8,15.5]})
df2.Date=pd.to_datetime(df1.Date)
df2['Day']=pd.DatetimeIndex(df2['Date']).day
df2['Month']=pd.DatetimeIndex(df2['Date']).month
df2.set_index(['Month','Day'],inplace=True)
df2
df1 and df2
df2['New']=df2[df2['Values']<df1['Values']]
gives
ValueError: Can only compare identically-labeled Series objects
I have also tried
df2['New']=df2[df2['Values'].apply(lambda x: x < df1['Values'].values)]
The best way to handle your problem is by using numpy as a tool. Numpy has an attribute called "where"that helps a lot in cases like this.
This is how the sentence works:
df1['new column that will contain the comparison results'] = np.where(condition,'value if true','value if false').
First import the library:
import numpy as np
Using the condition provided by you:
df2['New'] = np.where(df2['Values'] > df1['Values'], df2['Values'],'')
So, I think that solves your problem... You can change the value passed to the False condition to every thin you want, this is only an example.
Tell us if it worked!
Let´s try two possible solutions:
The first solution is to sort the index first.
df1.sort_index(inplace=True)
df2.sort_index(inplace=True)
Perform a simple test to see if it works!
df1 == df2
it is possible to raise some kind of error, so if that happens, try this correction instead:
df1.sort_index(inplace=True, axis=1)
df2.sort_index(inplace=True, axis=1)
The second solution is to drop the indexes and reset it:
df1.sort_index(inplace=True)
df2.sort_index(inplace=True)
Perform a simple test to see if it works!
df1 == df2
See if it works and tell us the result.
New to Python. I'm importing a CSV, then if any data is missing I need to return a CSV with an additional column to indicate which rows are missing data. Colleague suggested that I import CSV into a dataframe, then create a new dataframe with a "Comments" column, fill it with a comment on the intended rows, and append it to the original dataframe. I'm stuck at the step of filling my new dataframe, "dferr", with the correct number of rows that would match up to "dfinput".
Have Googled, "pandas csv return error column where data is missing", but haven't found anything related to creating a new CSV that marks bad rows. I don't even know if the proposed way is the best way to go about this.
import pandas as pd
dfinput = None
try:
dfinput = pd.read_csv(r"C:\file.csv")
except:
print("Uh oh!")
if dfinput is None:
print("Ack!")
quit(10)
dfinput.reset_index(level=None, drop=False, inplace=True, col_level=0,
col_fill='')
dferr = pd.DataFrame(columns=['comment'])
print("Empty DataFrame", dferr, sep='\n')
Expected results: "dferr" would have an index column with number of rows equal to "dfinput", and comments on the correct rows where "dfinput" has missing values.
Actual results: "dferr" is empty.
My understanding of 'missing data' here would be null values. It seems that for every row, you want the names of null fields.
df = pd.DataFrame([[1,2,3],
[4,None,6],
[None,8,None]],
columns=['foo','bar','baz'])
# Create a dataframe of True/False, True where a criterion is met
# (in this case, a null value)
nulls = df.isnull()
# Iterate through every row of *nulls*,
# and extract the column names where the value is True by boolean indexing
colnames = nulls.columns
null_labels = nulls.apply(lambda s:colnames[s], axis=1)
# Now you have a pd.Series where every entry is an array
# (technically, a pd.Index object)
# Pandas arrays have a vectorized .str.join method:
df['nullcols'] = null_labels.str.join(', ')
The .apply() method in pandas can sometimes be a bottleneck in your code; there are ways to avoid using this, but here it seemed to be the simplest solution I could think of.
EDIT: Here's an alternate one-liner (instead of using .apply) that might cut down computation time slightly:
import numpy as np
df['nullcols'] = [colnames[x] for x in nulls.values]
This might be even faster (a bit more work is required):
np.where(df.isnull(),df.columns,'')
I've been trying to use reindex instead of loc in pandas as from 0.24 there is a warning about reindexing with lists.
The issue I have is that I use loc to change the values of my dataframes.
Now if use reindex i lose this and if I try to be smart I even get a bug.
Contemplate the following case:
df = pd.DataFrame(data=pd.np.zeros(4, 2), columns=['a', 'b'])
ds = pd.Series(data=[1]*3)
I want to change a subset of values (while retaining the others), so df keeps the same shape.
So this is the original behaviour which works (and changes the values in a subset of df['a'] to 1)
df.loc[range(3), 'a'] = ds
But when I'm using the reindex I fail to change anything:
df.reindex(range(3)).loc['a'] = ds
Now when I try something like this:
df.loc[:, 'a'].reindex(range(3)) = ds
I get a SyntaxError: can't assign to function call error message.
For reference I am using pandas 0.24 and python 3.6.8
The quick answer from #coldspeed was the easiest, though the behaviour of the warning is misleading.
So reindex returns a copy when loc doesn't.
From the pandas docs:
A new object is produced unless the new index is equivalent to the current one and copy=False.
So saying reindex is an alternative to loc as per the warning is actually misleading.
Hope this helps people who face the same situation.