[英]Append a String to a String in a List
I am reading an excel table: 我正在阅读一张excel表:
import pandas as pd
df = pd.read_excel('file.xlsx', usecols = 'A,B,C')
print(df)
Now I want to create a list with every row in the table as string. 现在我想创建一个列表,其中表中的每一行都是字符串。 In addition I want to add a 'X' at the end of every string in the list:
另外我想在列表中的每个字符串的末尾添加一个'X':
keylist = []
list1, list2, list3 = df['A'].tolist(), df['B'].tolist(), df['C'].tolist()
for i in zip(list1, list2, list3):
val = map(str, i)
keylist.append('/'.join(val))
keylist += 'X'
print(keylist)
Everything works except the 'adding a X' part. 一切都有效,除了“添加X”部分。 This results in:
这导致:
['blue/a/a1', 'X', 'blue/a/a2', 'X', ....
But what I want is: 但我想要的是:
['blue/a/a1/X', 'blue/a/a2/X',
Thanks beforehand. 先谢谢。
I think better is: 我认为更好的是:
d = {'A': ['blue', 'blue', 'blue', 'red', 'red', 'red', 'yellow',
'yellow', 'green', 'green', 'green'],
'B': ['a', 'a', 'b', 'c', 'c', 'c', 'd', 'e', 'f', 'f', 'g'],
'C': ['a1', 'a2', 'b1', 'c1', 'c2', 'c3', 'd1', 'e1', 'f1', 'f2', 'g1']}
df = pd.DataFrame(d)
print (df)
A B C
0 blue a a1
1 blue a a2
2 blue b b1
3 red c c1
4 red c c2
5 red c c3
6 yellow d d1
7 yellow e e1
8 green f f1
9 green f f2
10 green g g1
keylist = df.apply(lambda x: '/'.join(x), axis=1).add('/X').values.tolist()
print (keylist)
['blue/a/a1/X', 'blue/a/a2/X', 'blue/b/b1/X', 'red/c/c1/X', 'red/c/c2/X',
'red/c/c3/X', 'yellow/d/d1/X', 'yellow/e/e1/X',
'green/f/f1/X', 'green/f/f2/X', 'green/g/g1/X']
Or if only few columns: 或者如果只有几列:
keylist = (df['A'] + '/' + df['B'] + '/' + df['C'] + '/X').values.tolist()
Some timing s: 一些时间 :
#[110000 rows x 3 columns]
df = pd.concat([df] * 10000, ignore_index=True)
In [364]: %%timeit
...: (df['A'] + '/' + df['B'] + '/' + df['C'] + '/X').values.tolist()
...:
60.2 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [365]: %%timeit
...: df.apply(lambda x: '/'.join(x), axis=1).add('/X').tolist()
...:
2.48 s ± 39.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [366]: %%timeit
...: list1, list2, list3 = df['A'].tolist(), df['B'].tolist(), df['C'].tolist()
...: for i in zip(list1, list2, list3):
...: val = map(str, i)
...: keylist.append('/'.join(val))
...: keylist[-1] += '/X'
...:
192 ms ± 78.5 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [367]: %%timeit
...: df.iloc[:,0].str.cat([df[c] for c in df.columns[1:]],sep='/').tolist()
...:
61.1 ms ± 540 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [368]: %%timeit
...: df.assign(New='X').apply('/'.join,1).tolist()
...:
2.51 s ± 76.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [369]: %%timeit
...: ['{0}/{1}/{2}/X'.format(i, j, k) for i, j, k in df.values.tolist()]
74.6 ms ± 2.27 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
You are doing += do the keylist which adds to that list, you need to do it to the val
array. 你正在做+ =做添加到该列表的密钥列表,你需要对
val
数组做。
for i in zip(list1, list2, list3):
val = map(str,i)
val += 'X' # you can combine this and the above if you want to look like:
#val = map(str, i) + 'X'
keylist.append("/".join(val))
print(keylist)
Here is one way using a list comprehension with str.format
: 以下是使用
str.format
的列表str.format
一种方法:
res = ['{0}/{1}/{2}/X'.format(i, j, k) for i, j, k in df.values.tolist()]
# ['blue/a/a1/X', 'blue/a/a2/X', 'blue/b/b1/X', 'red/c/c1/X', ...]
There is no need, as in this solution, to split into 3 lists and zip
them. 如此解决方案中一样,不需要拆分为3个列表并将其
zip
。
Base on pandas
基于
pandas
df.assign(New='X').apply('/'.join,1).tolist()
Out[812]: ['blue/a/a1/X', 'blue/a/a2/X', 'blue/b/b1/X']
You could add /X
to last item in list
everytime in the loop: 您可以在循环
list
每次将/X
添加到list
中的最后一项:
for i in zip(list1, list2, list3):
val = map(str, i)
keylist.append('/'.join(val))
keylist[-1] += '/X'
# ['blue/a/a1/X', 'blue/a/a2/X',....]
You can use the cat
string operation to join the columns into a single series with a specified sep
argument. 您可以使用
cat
string操作将列连接到具有指定sep
参数的单个系列。 Then simply convert the new series into a list 然后只需将新系列转换为列表即可
df
A B C
0 blue a a1
1 blue a a2
2 blue b b1
3 red c c1
4 red c c2
5 red c c3
6 yellow d d1
7 yellow e e1
8 green f f1
9 green f f2
10 green g g1
df.iloc[:,0].str.cat([df[c] for c in df.columns[1:]],sep='/').tolist()
['blue/a/a1', 'blue/a/a2', 'blue/b/b1', 'red/c/c1', 'red/c/c2', 'red/c/c3', 'yellow/d/d1', 'yellow/e/e1', 'green/f/f1', 'green/f/f2', 'green/g/g1']
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