I am stuck at a failry simple looping exercise through lists and getting error "TypeError: 'list' object is not callable".
I have three lists with n number of records. I want to write first record from all lists in the same line and want to repeat this procedure for n number of records, it will result in n number of lines. Following are lists that I want to use:
lst1 = ['1','2','4','5','3']
lst2 = ['3','4','3','4','3']
lst3 = ['0.52','0.91','0.18','0.42','0.21']
istring=""
lst=0
for i in range(0,10): # range is simply upper limit of number of records in lists
entry = lst1(lst)
istring = istring + entry.rjust(11) # first entry from each list will be cat here
lst=lst+1
Any startup would be really helpful.
This works for any size of lists:
for i in zip(lst1, lst2, lst3):
for j in i:
print j.rjust(11),
print
1 3 0.52
2 4 0.91
4 3 0.18
5 4 0.42
3 3 0.21
>>> lst1 = ['1','2','4','5','3']
>>> lst2 = ['3','4','3','4','3']
>>> lst3 = ['0.52','0.91','0.18','0.42','0.21']
>>> a = zip(lst1, lst2, lst3)
>>> istring = ""
>>> for entry in a:
... istring += entry[0].rjust(11)
... istring += entry[1].rjust(11)
... istring += entry[2].rjust(11) + "\n"
...
>>> print istring
1 3 0.52
2 4 0.91
4 3 0.18
5 4 0.42
3 3 0.21
Try entry = lst1[lst] instead of entry = lst1(lst)
() usually denotes calling a function, whereas
[] usually denotes accessing an element of something.
A list is not a function.
Also, while you can keep your own index, a for loop makes this unnecessary
x = [1,2,3,4,5,7,9,11,13,15]
y = [2,4,6,8,10,12,14,16,18,20]
z = [3,4,5,6,7,8,9,10,11,12]
for i in range(0,10):
print x[i], y[i], z[i]
1 2 3
2 4 4
3 6 5
4 8 6
5 10 7
7 12 8
9 14 9
11 16 10
13 18 11
15 20 12
Related
I've a dataframe which looks like this:
wave mean median mad
0 4050.32 -0.016182 -0.011940 0.008885
1 4208.98 0.023707 0.007189 0.032585
2 4508.28 3.662293 0.001414 7.193139
3 4531.62 -15.459313 -0.001523 30.408377
4 4551.65 0.009028 0.007581 0.005247
5 4554.46 0.001861 0.010692 0.027969
6 6828.60 -10.604568 -0.000590 21.084799
7 6839.84 -0.003466 -0.001870 0.010169
8 6842.04 -32.751551 -0.002514 65.118329
9 6842.69 18.293519 -0.002158 36.385884
10 6843.66 0.006386 -0.002468 0.034995
11 6855.72 0.020803 0.000886 0.040529
As it's clearly evident in the above table that some of the values in the column mad and median are very big(outliers). So i want to remove the rows which have these very big values.
For example in row3 the value of mad is 30.408377 which very big so i want to drop this row. I know that i can use one line
to remove these values from the columns but it doesn't removes the complete row
df[np.abs(df.mad-df.mad.mean()) <= (3*df.mad.std())]
But i want to remove the complete row.
How can i do that?
Predicates like what you've given will remove entire rows. But none of your data is outside of 3 standard deviations. If you tone it down to just one standard deviation, rows are removed with your example data.
Here's an example using your data:
import pandas as pd
import numpy as np
columns = ["wave", "mean", "median", "mad"]
data = [
[4050.32, -0.016182, -0.011940, 0.008885],
[4208.98, 0.023707, 0.007189, 0.032585],
[4508.28, 3.662293, 0.001414, 7.193139],
[4531.62, -15.459313, -0.001523, 30.408377],
[4551.65, 0.009028, 0.007581, 0.005247],
[4554.46, 0.001861, 0.010692, 0.027969],
[6828.60, -10.604568, -0.000590, 21.084799],
[6839.84, -0.003466, -0.001870, 0.010169],
[6842.04, -32.751551, -0.002514, 65.118329],
[6842.69, 18.293519, -0.002158, 36.385884],
[6843.66, 0.006386, -0.002468, 0.034995],
[6855.72, 0.020803, 0.000886, 0.040529],
]
df = pd.DataFrame(np.array(data), columns=columns)
print("ORIGINAL: ")
print(df)
print()
res = df[np.abs(df['mad']-df['mad'].mean()) <= (df['mad'].std())]
print("REMOVED: ")
print(res)
this outputs:
ORIGINAL:
wave mean median mad
0 4050.32 -0.016182 -0.011940 0.008885
1 4208.98 0.023707 0.007189 0.032585
2 4508.28 3.662293 0.001414 7.193139
3 4531.62 -15.459313 -0.001523 30.408377
4 4551.65 0.009028 0.007581 0.005247
5 4554.46 0.001861 0.010692 0.027969
6 6828.60 -10.604568 -0.000590 21.084799
7 6839.84 -0.003466 -0.001870 0.010169
8 6842.04 -32.751551 -0.002514 65.118329
9 6842.69 18.293519 -0.002158 36.385884
10 6843.66 0.006386 -0.002468 0.034995
11 6855.72 0.020803 0.000886 0.040529
REMOVED:
wave mean median mad
0 4050.32 -0.016182 -0.011940 0.008885
1 4208.98 0.023707 0.007189 0.032585
2 4508.28 3.662293 0.001414 7.193139
3 4531.62 -15.459313 -0.001523 30.408377
4 4551.65 0.009028 0.007581 0.005247
5 4554.46 0.001861 0.010692 0.027969
6 6828.60 -10.604568 -0.000590 21.084799
7 6839.84 -0.003466 -0.001870 0.010169
10 6843.66 0.006386 -0.002468 0.034995
11 6855.72 0.020803 0.000886 0.040529
Observe that rows indexed 8 and 9 are now gone.
Be sure you're reassigning the output of df[np.abs(df['mad']-df['mad'].mean()) <= (df['mad'].std())] as shown above. The operation is not done in place.
Doing df[np.abs(df.mad-df.mad.mean()) <= (3*df.mad.std())] will not change the dataframe.
But assign it back to df, so that:
df = df[np.abs(df.mad-df.mad.mean()) <= (3*df.mad.std())]
I want to count areas of interest in my dataframe column 'which_AOI' (ranging from 0 -9). I would like to have a new column with the results added to a dataframe depending on a variable 'marker' (ranging from 0 - x) which tells me when one 'picture' is done and the next begins (one marker can go on for a variable length of rows). This is my code so far but it seems to be stuck and runs on without giving output. I tried reconstructing it from the beginning once but as soon as i get to 'if df.marker == num' it doesn't stop. What am I missing?
(example dataframe below)
## AOI count of spec. type function (in progress):
import numpy as np
import pandas as pd
path_i = "/Users/Desktop/Pilot/results/gazedata_filename.csv"
df = pd.read_csv(path_i, sep =",")
#create a new dataframe for AOIs:
d = {'marker': []}
df_aoi = pd.DataFrame(data=d)
### Creating an Aoi list
item = df.which_AOI
aoi = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] #list for search
aoi_array = [0, 0 , 0, 0, 0, 0, 0, 0, 0, 0] #list for filling
num = 0
for i in range (0, len (df.marker)): #loop through the dataframe
if df.marker == num: ## if marker = num its one picture
for index, item in enumerate(aoi): #look for item (being a number in which_AOI) in aoi list
if (item == aoi[index]):
aoi_array[index] += 1
print (aoi)
print (aoi_array)
se = pd.Series(aoi_array) # make list into a series to attach to dataframe
df_aoi['new_col'] = se.values #add list to dataframe
aoi_array.clear() #clears list before next picture
else:
num +=1
index pos_time pos_x pos_y pup_time pup_diameter marker which_AOI fixation Picname shock
1 16300 168.608779907227 -136.360855102539 16300 2.935715675354 0 7 18 5 save
2 16318 144.97673034668 -157.495513916016 16318 3.08838820457459 0 8 33 5 save
3 16351 152.92560577392598 -156.64172363281298 16351 3.0895299911499 0 7 17 5 save
4 16368 152.132453918457 -157.989685058594 16368 3.111008644104 0 7 18 5 save
5 16386 151.59835815429702 -157.55587768554702 16386 3.09514689445496 0 7 18 5 save
6 16404 150.88092803955098 -152.69479370117202 16404 3.10009074211121 1 7 37 5 save
7 16441 152.76554107666 -142.06188964843798 16441 3.0821495056152304 1 7 33 5 save
Not 100% clear based on your question but it sounds like you want to count the number of rows for each which_AOI value in each marker.
You can accomplish this using groupby
df_aoi = df.groupby(['marker','which_AOI']).size().unstack('which_AOI',fill_value=0)
In:
pos_time pos_x pos_y pup_time pup_diameter marker \
0 16300 168.608780 -136.360855 16300 2.935716 0
1 16318 144.976730 -157.495514 16318 3.088388 0
2 16351 152.925606 -156.641724 16351 3.089530 0
3 16368 152.132454 -157.989685 16368 3.111009 0
4 16386 151.598358 -157.555878 16386 3.095147 0
5 16404 150.880928 -152.694794 16404 3.100091 1
6 16441 152.765541 -142.061890 16441 3.082150 1
which_AOI fixation Picname shock
0 7 18 5 save
1 8 33 5 save
2 7 17 5 save
3 7 18 5 save
4 7 18 5 save
5 7 37 5 save
6 7 33 5 save
Out:
which_AOI 7 8
marker
0 4 1
1 2 0
A dataframe stores some values in columns, passing those values to a function I get another dataframe. I'd like to concatenate the returned dataframe's columns to the original dataframe.
I tried to do something like
i = pd.concat([i, i[['cid', 'id']].apply(lambda x: xy(*x), axis=1)], axis=1)
but it did not work with error:
ValueError: cannot copy sequence with size 2 to array axis with dimension 1
So I did like this:
def xy(x, y):
return pd.DataFrame({'x': [x*2], 'y': [y*2]})
df1 = pd.DataFrame({'cid': [4, 4], 'id': [6, 10]})
print('df1:\n{}'.format(df1))
df2 = pd.DataFrame()
for _, row in df1.iterrows():
nr = xy(row['cid'], row['id'])
nr['cid'] = row['cid']
nr['id'] = row['id']
df2 = df2.append(nr, ignore_index=True)
print('df2:\n{}'.format(df2))
Output:
df1:
cid id
0 4 6
1 4 10
df2:
x y cid id
0 8 12 4 6
1 8 20 4 10
The code does not look nice and should work slowly.
Is there pandas/pythonic way to do it properly and fast working?
python 2.7
Option 0
Most directly with pd.DataFrame.assign. Not very generalizable.
df1.assign(x=df1.cid * 2, y=df1.id * 2)
cid id x y
0 4 6 8 12
1 4 10 8 20
Option 1
Use pd.DataFrame.join to add new columns
This shows how to adjoin new columns after having used apply with a lambda
df1.join(df1.apply(lambda x: pd.Series(x.values * 2, ['x', 'y']), 1))
cid id x y
0 4 6 8 12
1 4 10 8 20
Option 2
Use pd.DataFrame.assign to add new columns
This shows how to adjoin new columns after having used apply with a lambda
df1.assign(**df1.apply(lambda x: pd.Series(x.values * 2, ['x', 'y']), 1))
cid id x y
0 4 6 8 12
1 4 10 8 20
Option 3
However, if your function really is just multiplying by 2
df1.join(df1.mul(2).rename(columns=dict(cid='x', id='y')))
Or
df1.assign(**df1.mul(2).rename(columns=dict(cid='x', id='y')))
Im trying to write values to a csv file such that for every two iterations, the result is in the same row and then the next the values print to a new row. Any help would be greatly appreciated. Thank you!
This is what I have so far:
import csv
import math
savePath = '/home/dehaoliu/opencv_test/Engineering_drawings_outputs/'
with open(str(savePath) +'outputsTest.csv','w') as f1:
writer=csv.writer(f1, delimiter='\t',lineterminator='\n',)
temp = []
for k in range(0,2):
temp = []
for i in range(0,4):
a = 2 +i
b = 3+ i
list = [a,b]
temp.append(list)
writer.writerow(temp)
The result I am getting now is
[2 3][3 4][4 5][5 6]
[2 3][3 4][4 5][5 6]
But I would like to get this (without the brackets) where each number in a row is in a separate column:
2 3 3 4
4 5 5 6
Try the following:
import csv
import math
savePath = '/home/dehaoliu/opencv_test/Engineering_drawings_outputs/'
with open(str(savePath) +'outputsTest.csv','w') as f1:
writer=csv.writer(f1, delimiter='\t',lineterminator='\n',)
temp = [2, 3]
for i in range(2):
temp = [x + i for x in temp]
additional = [y+1 for y in temp]
writer.writerow(temp + additional)
temp = additional[:]
This should return:
# 2 3 3 4
# 4 5 5 6
You start with a temporary containing the numbers 2 and 3. Then, you loop from 0 to 2 (excluding). At every iteration, you increment the values of the temporary by the current index and subsequently create an additional list with these new values of your temporary list. Once that's done, you join the two lists together and write the result out to your file. At this point, you can set your temporary list to be equal to the values of the additional list, before moving on to the next iteration.
I hope this helps.
The way you present it you can do it with a simple seed and increment:
import csv
import os
save_path = "/home/dehaoliu/opencv_test/Engineering_drawings_outputs/"
with open(os.path.join(save_path, "outputsTest.csv"), "w") as f:
writer = csv.writer(f, delimiter="\t", lineterminator="\n")
temp = [2, 3, 3, 4] # init seed
increment = len(temp) // 2 # how many pairs we have, used to increase our seed each row
for _ in range(2): # how many rows do you need, any positive integer will do
writer.writerow(temp) # write the current value
temp = [x + increment for x in temp] # add 'increment' to the elements
Resulting in:
2 3 3 4
4 5 5 6
But if your seed is: temp = [2, 3, 3, 4, 4, 5] and you decide to generate 4 rows, it will still adapt:
2 3 3 4 4 5
5 6 6 7 7 8
8 9 9 10 10 11
11 12 12 13 13 14
Here is my minimal working example:
list1 = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] #len = 21
list2 = [1,1,1,0,1,0,0,1,0,1,1,0,1,0,1,0,0,0,1,1,0] #len = 21
list3 = [0,0,1,0,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1] #len = 21
list4 = [1,0,0,1,1,0,0,0,0,1,0,1,1,1,1,0,1,0,1,0,1] #len = 21
I have four lists and I want to "clean" my list 1 using the following rule: "if any of list2[i] or list3[i] or list4[i] are equal to zero, then I want to eliminate the item I from list1. SO basically I only keep those elements of list1 such that the other lists all have ones there.
here is the function I wrote to solve this
def clean(list1, list2,list3,list4):
for i in range(len(list2)):
if (list2[i]==0 or list3[i]==0 or list4[i]==0):
list1.pop(i)
return list1
however it doesn't work. If you apply it, it give the error
Traceback (most recent call last):line 68, in clean list1.pop(I)
IndexError: pop index out of range
What am I doing wrong? Also, I was told Pandas is really good in dealing with data. Is there a way I can do it with Pandas? Each of these lists are actually columns (after removing the heading) of a csv file.
EDIT
For example at the end I would like to get: list1 = [4,9,11,15]
I think the main problem is that at each iteration, when I pop out the elements, the index of all the successor of that element change! And also, the overall length of the list changes, and so the index in pop() is too large. So hopefully there is another strategy or function that I can use
This is definitely a job for pandas:
import pandas as pd
df = pd.DataFrame({
'l1':list1,
'l2':list2,
'l3':list3,
'l4':list4
})
no_zeroes = df.loc[(df['l2'] != 0) & (df['l3'] != 0) & (df['l4'] != 0)]
Where df.loc[...] takes the full dataframe, then filters it by the criteria provided. In this example, your criteria are that you only keep the items where l2, l3, and l3 are not zero (!= 0).
Gives you a pandas dataframe:
l1 l2 l3 l4
4 4 1 1 1
9 9 1 1 1
12 12 1 1 1
18 18 1 1 1
or if you need just list1:
list1 = df['l1'].tolist()
if you want the criteria to be where all other columns are 1, then use:
all_ones = df.loc[(df['l2'] == 1) & (df['l3'] == 1) & (df['l4'] == 1)]
Note that I'm creating new dataframes for no_zeroes and all_ones and that the original dataframe stays intact if you want to further manipulate the data.
Update:
Per Divakar's answer (far more elegant than my original answer), much the same can be done in pandas:
df = pd.DataFrame([list1, list2, list3, list4])
list1 = df.loc[0, (df[1:] != 0).all()].astype(int).tolist()
Here's one approach with NumPy -
import numpy as np
mask = (np.asarray(list2)==1) & (np.asarray(list3)==1) & (np.asarray(list4)==1)
out = np.asarray(list1)[mask].tolist()
Here's another way with NumPy that stacks those lists into rows to form a 2D array and thus simplifies things quite a bit -
arr = np.vstack((list1, list2, list3, list4))
out = arr[0,(arr[1:] == 1).all(0)].tolist()
Sample run -
In [165]: arr = np.vstack((list1, list2, list3, list4))
In [166]: print arr
[[ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20]
[ 1 1 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 0 1 1 0]
[ 0 0 1 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 1]
[ 1 0 0 1 1 0 0 0 0 1 0 1 1 1 1 0 1 0 1 0 1]]
In [167]: arr[0,(arr[1:] == 1).all(0)].tolist()
Out[167]: [4, 9, 12, 18]