在列表中用逗号分隔元素

Question

I have a list of strings that contains elements of the type

List=['name1,vol', 'name1,price','name2, vol', 'name2,price'.... ] 

I would like to extract a list only of "names" which are the parts that actually change as the second components in each element have a fix pattern (here:vol, price). Notice that the "names" can obviously have different length. To sum up, I'd like to extract something like:

List_names=['name1', 'name2' ] 

How can I do that?

What if I have something of the type:

List_tricky=[('name1', 'vol'), ('name1', 'price'),('name2', 'vol'), ('name2', 'price').... ] 

Answer 1

Something like this?

List=['name1,vol', 'name1,price','name2, vol', 'name2,price']

names = []

for string in List:

    name = string.split(',')[0]

    names.append(name)

print(names)

For your 'tricky' case, you can try:

# initialize variables:
names = []

# iterate over each point (tuple):
for point in List:

    # get name:
    name = point[0]

    # append to list:
    names.append(name)

print(names)

Answer 2

与@Daniel Sokol 的回答类似的逻辑，您可以使用单行：

list2 = [x.split(',')[0] for x in List]

Answer 3

You could turn it into a dict then back into a list using str.split . (No loop required as it does it efficiently for you) Use functools.partial to apply the split to each string instead of a lambda:

from functools import partial
list(dict(map(partial(str.split, sep=','), List)))

---

This works for either input but way more simple for the list of tuples:

>>> l = ['name1,vol', 'name1,price','name2, vol', 'name2,price'.... ]
>>> list(dict(map(partial(str.split, sep=','), List)))
['name1', 'name2']

---

>>> l = [('name1', 'vol'), ('name1', 'price'),('name2', 'vol'), ('name2', 'price').... ] 
>>> list(dict(l))
['name1', 'name2']

Answer 4

To add on top of @Alireza Tajadod's already wonderful answer, you might want to apply conversion to a set, then back to a list to remove any possible duplication items, as suggested by @Cryptoharf84 in the comments.

names_list = list(set([entry.split(',')[0] for entry in List]))

The same logic with list comprehension can be applied to the trickier case.

names_list_2 = list(set([entry[0] for entry in List_tricky]))

To make list comprehension more explicit, you can also do the following:

names_list_3 = list(set([name for name, _ in List_tricky]))

The _ indicates that we are discarding the second value of the unpacked tuple.

Sets are useful because converting a list with duplicate elements into a set effectively removes any duplications.

Answer 5

As a tip, look for naming conventions in python. But never name variables starting with upper case, nor with existing class names.

I will try something like:

list_names = [s.split(',')[0].strip() for s in List]
list_unique_names(set(list_names))

split returns a list of "chunks" of the original string, and strip to remove whitespaces on beginning/end of the resulting string.

Answer 6

I will change your data structure to dict instead of list

d={'name1': ('vol', 'price'),'name2': ('vol', 'price'), .... }

In order to get just the names: d.keys()

Answer 7

You can also use a .map() function:

# Case 1: List
all_names = map(lambda x :a.split(',')[0], List)

# Case 2: List_tricky
all_names = [i[0] for i in List_tricky]

# After the code is the same
unique_names = set(all_names)
List_names = list(unique_names)
print(List_names)