Is it possible to get the row number (i.e. "the ordinal position of the index value") of a DataFrame row without adding an extra row that contains the row number (the index can be arbitrary, i.e. even a MultiIndex)?
>>> import pandas as pd
>>> df = pd.DataFrame({'a': [2, 3, 4, 2, 4, 6]})
>>> result = df[df.a > 3]
>>> result.iloc[0]
a 4
Name: 2, dtype: int64
# but how can I get the original row index of iloc[0] in df?
I could have done df['row_index'] = range(len(df)) which would maintain the original row number, but I am wondering if Pandas has a built-in way of doing this.
Access the .name attribute and use get_loc:
In [10]:
df.index.get_loc(result.iloc[0].name)
Out[10]:
2
Looking this from a different side:
for r in df.itertuples():
getattr(r, 'Index')
Where df is the data frame. May be you want to use a conditional to get the index when a condition are met.
Related
I have a csv file and I need to print the duplicate values in a column "hash" . so I made this using pandas but I'm not sure what its printing , it seems like its printing an entire row of duplicates which I don't need , I need only duplicate vaults in that column " hash"
the script I made:
import pandas as pd
df = pd.read_csv("combined_values.csv")
dups = df[df.duplicated("hash")]
print(dups)
I also tried this one but it seems to print all the "hash" column
import pandas as pd
df = pd.read_csv("combined_values.csv")
dups = df["hash"].duplicated
print(dups)
We can try with Series.duplicated to create a boolean index then use loc to select from the DataFrame where hash is duplicated, and just the hash column:
s = df.loc[df['hash'].duplicated(), 'hash']
We can set keep=False if all duplicates are wanted:
s = df.loc[df['hash'].duplicated(keep=False), 'hash']
With some sample data:
import pandas as pd
df = pd.DataFrame({
'a': [1, 2, 3, 4],
'b': [5, 6, 7, 8],
'hash': [4, 5, 4, 6]
})
s = df.loc[df['hash'].duplicated(), 'hash']
s:
2 4
Name: hash, dtype: int64
Or keeping all duplicates:
s = df.loc[df['hash'].duplicated(keep=False), 'hash']
s:
0 4
2 4
Name: hash, dtype: int64
I have a numpy array and an empty dataframe:
element = numpy.array([1,2,3])
df = pandas.DataFrame(columns = ["Col"])
I want to insert element in the first row of df. The following code:
df["Col"] = element
Gives me a dataframe 3x1 whose elements are 1, 2 and 3. I want a dataframe 1x1 whose element is the array. How can I get this result?
Thanks in advance!
Use DataFrame.loc or DataFrame.at for specify label for set array to DataFrame:
df.loc[0, "Col"] = element
print (df)
Col
0 [1, 2, 3]
df.at[0, "Col"] = element
Wrap element in a list.
>>> df['Col'] = [element]
>>> df
Col
0 [1, 2, 3]
Imagine I have a Pandas DataFrame:
# create df
df = pd.DataFrame({'id': [1,1,1,2,2,2],
'val': [5,4,6,3,2,3]})
Lets assume it is ordered by 'id' and an imaginary, not shown, date column (ascending).
I want to create another column where each row is a list of 'val' at that date.
The ending DataFrame will look like this:
df = pd.DataFrame({'id': [1,1,1,2,2,2],
'val': [5,4,6,3,2,3],
'val_list': [[5],[5,4],[5,4,6],[3],[3,2],[3,2,3]]})
I don't want to use a loop because the actual df I am working with has about 4 million records. I am imagining I would use a lambda function in conjunction with groupby (something like this):
df['val_list'] = df.groupby('id')['val'].apply(lambda x: x.runlist())
This raises an AttributError because the runlist() method does not exist, but I am thinking the solution would be something like this.
Does anyone know what to do to solve this problem?
Let us try
df['new'] = df.val.map(lambda x : [x]).groupby(df.id).apply(lambda x : x.cumsum())
Out[138]:
0 [5]
1 [5, 4]
2 [5, 4, 6]
3 [3]
4 [3, 2]
5 [3, 2, 3]
Name: val, dtype: object
I have a dataframe and a list, which includes a part of columns' name from my dataframe as follows:
my_frame:
col1, col2, col3, ..., coln
2, 3, 4, ..., 2
5, 8, 5, ..., 1
6, 1, 8, ..., 9
my_list:
['col1','col3','coln']
Now, I want to create an array with the size of my original dataframe (total number of columns) which consists only zero and one. Basically I want the array includes 1 if the there is a similar columns name in "my_list", otherwise 0. My desired output should be like this:
my_array={[1,0,1,0,0,...,1]}
This should help u:
import pandas as pd
dictt = {'a':[1,2,3],
'b':[4,5,6],
'c':[7,8,9]}
df = pd.DataFrame(dictt)
my_list = ['a','h','g','c']
my_array = []
for column in df.columns:
if column in my_list:
my_array.append(1)
else:
my_array.append(0)
print(my_array)
Output:
[1, 0, 1]
If u wanna use my_array as a numpy array instead of a list, then use this:
import pandas as pd
import numpy as np
dictt = {'a':[1,2,3],
'b':[4,5,6],
'c':[7,8,9]}
df = pd.DataFrame(dictt)
my_list = ['a','h','g','c']
my_array = np.empty(0,dtype = int)
for column in df.columns:
if column in my_list:
my_array = np.append(my_array,1)
else:
my_array = np.append(my_array,0)
print(my_array)
Output:
[1 0 1]
I have used test data in my code for easier understanding. U can replace the test data with ur actual data (i.e replace my test dataframe with ur actual dataframe). Hope that this helps!
I want to Fill in these missing numbers in column b with the consecutive values 1 and 2.
This is what I have done:
import pandas as pd
import numpy as np
df = pd.DataFrame({'a': [1, 4, 7,8,4],
'b': [1, np.nan, 3, np.nan, 5]})
df['b'].fillna({'b':[1,2]}, inplace=True)
but nothing is done.
One way is to use loc with an array:
df.loc[df['b'].isnull(), 'b'] = [1, 2]
What you're attempting is possible but cumbersome with fillna:
nulls = df['b'].isnull()
df['b'] = df['b'].fillna(pd.Series([1, 2], index=nulls[nulls].index))
You may be looking for interpolate but the above solutions are generic given an input list or array.
If, on the other hand, you want to fill nulls with a sequence 1, 2, 3, etc, you can use cumsum:
# fillna solution
df['b'] = df['b'].fillna(df['b'].isnull().cumsum())
# loc solution
nulls = df['b'].isnull()
df.loc[nulls, 'b'] = nulls.cumsum()
You can't feed fillna a list of values, as stated here and in the documentation. Also, if you're selecting the column, no need to tell fillna which column to use. You could do:
df.fillna({'b':1}, inplace=True)
Or
df['b'].fillna(1, inplace=True)
By the way, inplace is on the way to deprecation in Pandas, the preferred way to do this is, for example
df = df.fillna({'b':1})
You can interpolate. Example:
s = pd.Series([0, 1, np.nan, 3])
s.interpolate()
0 0
1 1
2 2
3 3
If I understand wording " consecutive values 1 and 2" correctly, the solution may be:
from itertools import isclice, cycle
filler = [1, 2]
nans = df.b.isna()
df.loc[nans, 'b'] = list(islice(cycle(filler), sum(nans)))