Question

我有一个看起来像这样的数组：

a = ['UCI_99648;102568',  'UCI_99648;102568',  'UCI_99648;102568;99651', 'UCI_99651', 'UCI_99652', 'SIB_99658;102568;506010;706080', NaN]

我想找出多少个s具有单个数字，例如UCI_99651，UCI_99652

因此，预期结果为2。

我如何在python中做到这一点。

注意：：我的实际数据非常大，数字可以是任何数字，并且如示例中所述，可能涉及缺失值。

Answer 1

假设所有字符串的结构均遵循上述示例的结构，则列表理解如下：

l = ['UCI_99648;102568',  'UCI_99648;102568',  'UCI_99648;102568;99651', 
     'UCI_99651', 'UCI_99652', 'SIB_99658;102568;506010;706080', 'NaN']

[i for i in l if ';' not in i and i != 'NaN']

输出

['UCI_99651', 'UCI_99652']

Answer 2

由于您已标记了熊猫，因此可以采用另一种方式：

s=pd.Series(a).dropna()
s[s.str.split(';').str.len().eq(1)]

3    UCI_99651
4    UCI_99652

Answer 3

您可以尝试以下操作。希望这可以解决您的问题。

p = [word.split(";")[0] for word in uci if word != 'NaN']
print(Counter(p))
#Counter({'UCI_99648': 3, 'UCI_99651': 1, 'UCI_99652': 1, 'SIB_99658': 1})
#To filter only one occurance you can try below.
b = [word for word in p if p.count(word)==1]
print(b)

有关更多信息，请参见此处的列表理解文档。

http://dataunbox.com/course/24/119/Python%20for%20dummies

Answer 4

根据需要实施NaN检查-使用numpy或pandas。

a = ['UCI_99648;102568',  'UCI_99648;102568',  'UCI_99648;102568;99651', 'UCI_99651', 'UCI_99652', 'SIB_99658;102568;506010;706080', 'NaN']


b = [i.split(';')[0] for i in a if i != 'NaN' and i.startswith('UCI_')]

b = [x for x in b if b.count(x)==1]

print(b)

#[UCI_99651, UCI_99652]

Answer 5

您可以使用正则表达式提取数字。例如，如下所示：

import re
import numpy as np
from collections import Counter


def count_strings_with_unq_nums(list_of_strings):

    # Initialize two lists - one to keep track of where the numbers occur and another to isolate unique occurences of numbers
    all_nums_nested = []
    all_nums_flat = []

    # Loop through all strings and extract integers within
    for s in list_of_strings:
        try:
            nums = re.findall(r'\d+', s)
            all_nums_nested.append(nums)
            all_nums_flat.extend(nums)
        except:
            continue

    # Count occurences of all extracted numbers
    num_counter = Counter(all_nums_flat)

    # Loop through nested list to find strings where unique numbers occur
    unq_items = []
    for key, val in num_counter.items():
        if val == 1:
            for n, num_list in enumerate(all_nums_nested):
                if key in num_list:
                    unq_items.append(list_of_strings[n])

    # Return the number of strings containing unique numbers.        
    return len(set(unq_items))

if __name__ == '__main__':
    a = ['UCI_99648;102568',  'UCI_99648;102568',  'UCI_99648;102568;99651', 'UCI_99651', 'UCI_99652', 'SIB_99658;102568;506010;706080', np.NaN]

    print(count_strings_with_unq_nums(a))

>>> 2

计算字符串中的字符串

5 个答案: